Download Math 112 Quadratics Notes

Math 112 Quadratics- Notes
Resource: Mathematical Modeling Book 3
Instructional time frame for test #1
1 day of start up with introduction and handout of text books.
8 days instruction
1 day to test
10 days to end of test
Assignments referred to as handouts are attached at the end of this
document.
Note(1) Sequence- a series of numbers that are related in some way by a rule
or pattern.
e.g. 1,3,5,7,9,… the rule here is to add 2 to the previous term to get the
next.
Note(2) Term- is a member or item in a sequence. We can represent a term
using a letter and a subscript.
Eg.t1 = term 1 and from the sequence above:
t1 = 1
t2 = 3
t3 = 5
t4 = 7
t5 = 9
Note(3) General term- is represented by the letter t and the variable n so that:
tn = term n or the nth term.
If n = 7 then tn = t7
If n = 8 then tn = t8
Note(4) Infinite sequence- a sequence that continues indefinitely.
Notation: { t1 , t2 , t3 , t4 , t5 , …}
ellipsis (3 dots) signifies continuance.
Note(5) Finite sequence- a sequence that terminates.
Notation: { t1 , t2 , t3 , t4 , t5 , …, tn}
tn signifies a termination
Asn(A1) Investigation #1 p. 2
Asn(A2) Answer questions 1,2,6 p. 3
Math 112 Quadratics- Notes
Note(6) Arithmetic sequence- a pattern of numbers in which each term is
found by adding or subtracting a number to (or from) the previous term.
e.g. 5, 7, 9, 11, … the next number is obtained by adding 2 to the
previous term. It is, therefore, an arithmetic sequence.
Note(7) Common difference- if the differences between any two numbers in
a sequence is always the same we say that we have a common difference.
e.g. 5, 7, 9, 11,… We have a common difference of +2
  
2 2 2
e.g. 1, 4, 9, 16… We do not have a common difference.
  
3 5 7
Note: If a sequence does not have a common difference (at the first level) it
is not an arithmetic sequence.
Note(8) Arithmetic sequence formula:
If t1 = the first term of a sequence and
If d = common difference then we can construct a formula that can
be used for predicting any term in the sequence as follows:
tn = t1 + (n-1)d
Where: t1 = the first term
d = the common difference
n = the term number
(n-1) one less than the term number.
Asn(A3) Answer questions 7 to 12 p. 4,5 (Question 9 to 12 to be submitted
for marking).
Math 112 Quadratics- Notes
Note(9) Graphing properties of an arithmetic sequence- An arithmetic
sequence has the following graphing properties:
a) it is linear- forms a straight line
b) the graph is discrete (no meaning between points)
c) the slope of the line is equivalent to the common difference so that
D1 = slope of the graph = d the common difference
d) there is a constant rate of growth because the slope between any two
points is the same.
e) the domain of an arithmetic sequence is D= {x x  N }
Asn(A4) A,B,C,D,E p. 7
Asn(A5) Q 17, 20 (Graph for Q 17 only) p. 8,9
Asn(A6) Q 22,23 p. 9 and Q 27,29 p. 10
Math 112 Quadratics- Notes
Note(10) Quadratic sequence- is a sequence of numbers formed from an
function containing a power of at least the second order.
The general formula that generates a quadratic sequence is:
tn = an2 + bn + c
Where:
tn = term value
n = term number
D2 = 2a or a = D2
2
Note(11) Properties of a quadratic sequence include:
1) The common difference = D2
2) The multiplication of two arithmetic sequences generates a quadratic
sequence. E.g. n(n-1) = n2 – n
3) In the equation tn = an2 + bn + c
a = ½ D2
4) One of the variables must be of at least the order two.
Note(12) Graphing properties of a quadratic sequence:
1) The graph of a quadratic sequence forms a parabola
2) The slope of a quadratic is not a constant
3) The graph is discrete. (not continuous)
Note(13) Power sequences- a sequence where one of the variables is of the
second order or higher. Examples of power sequences and properties are:
Quadratic:
tn = ax2 +bx +c
1) Common difference at D2
2) D2 = 2a or a = D2
2
Cubic:
tn = ax3 +bx2 +cx +d
1) Common difference at D3
2) D3 = 6a or a = D3
6
Quartic:
tn = ax4 +bx3 +cx2 +dx +e
1) Common difference at D4
2) D4 =24a or a = D4
24
Asn(A7) Handout on sequences.
Advanced. Handout on algebraic method. Q 38, 39, 40 p. 12,13
Math 112 Quadratics- Notes
Note(14) Analyzing data- Given a table of values it is possible to determine
the sequence of data as being either arithmetic or quadratic. First enter the
data of the independent variable under L1 and the data for the dependent
variable under L2. You may then determine the nature of the sequence by
one of the steps given below:
1) Plot the data using: 2nd y= 1: Plot1 (choose the desired options)
Set the Window parameters then press Graph. If the data forms a straight
line then the relationship is arithmetic. If the data is curved or parabolic it is
likely quadratic or one of the other power sequences (cubic, quartic, etc.)
OR
2) Determine the first and second differences and place results under L3 and
L4 (if necessary). Place cursor at top of L3 and press: 2nd Stat Ops
7  List followed by name of the list for which you want differences taken.
(Example L2) and repeat the procedure at the top of L4. If D1 is a constant
the graph is arithmetic and if D2 is a constant the graph is quadratic
Note (15) Determining the line of best fit- Once you have determined
whether the sequence of numbers is arithmetic or quadratic you may find the
line of best fit by using linear regression or quadratic regression.
Linear regression:
Quadratic regression:
Stat Calc 4: LinReg Enter
Stat Calc 5: QuadReg Enter
(For the golf ball question from Asn(A9) at the end of this document
Quadregression gives a function y = -4.9x2 + 28.7x + 9)
Note(16) Graphing your line of best fit- The function generated by the
calculator may be graphed by entering the function in your calculator.
1) Press Y=
2) After \Y1 = (enter the function)
3) Press Window and set the values for Xmin, Xmax, Ymin, Ymax
4) Press Graph
Asn(A8) Q 8a) 5 a) b) c) 9 a) b) c) p. 16,17 and Q 17 p. 19
Math 112 Quadratics- Notes
For notes 17 to 21 refer to Q 1 of Asn(A9) – the flight of a golf ball.
Note(17) Determining the maximum or minimum value of a graph- The line
of best fit for a quadratic always generates a curve that will have a maximum
or minimum value. There are three ways of obtaining this maximum or
minimum value:
1) Press Trace then move the “x” displayed on your graph left or right until
it is centered over the max (or min) value. Read the coordinates for this
point on the bottom of your screen.
2) Press 2nd Graph and read through the second column of Y values to find
the highest (or lowest) value from the tables provided.
3) Press 2nd Trace then choose option 4: maximum . You will be required
to set the Left and Right boundaries (pressing Enter each time) then press
Enter a third time upon which the calculator will “look” between the two
boundaries that you have set to locate the maximum (or minimum) value.
(2.9 , 51)
Left boundary
Right boundary
The boundaries must be pointing towards each other. This requires the
left boundary be set before the right boundary. Pressing enter a third time
engages the calculator to search for the maximum value of the graph.
Math 112 Quadratics- Notes
Note(18) Axis of symmetry- A vertical line drawn through the vertex of the
graph that defines the maximum (or minimum) value of a graph is called the
axis of symmetry. The equation of the axis of symmetry equals the x value
of the vertex. (2.9, 51).
x = 2.9 is the equation of the
axis of symmetry.
The graph is split in two by this axis of symmetry so that the left side
is a mirror image of the right side of the graph or we may say the graph is
symmetrical about the line x = 2.9
Note(19) Maximum value of the graph- The maximum value of the graph
is the y-coordinate in the vertex:
(2.9, 51 )
y = 51 is the maximum value
for the graph.
Math 112 Quadratics- Notes
Note(20) The Domain and Range of a graph- the domain and range of a
graph can be determined from the table of values if one is given.
e.g. From the golf ball question Asn(A9) at the end of these notes:
Time
Height
1
32.8
2
46.8
3
51
4
45.4
5
30
The domain of the graph is the values that are determined by the
independent variable, which in this case, is the time of the golf ball in flight.
The lowest value here could be 0 seconds and the highest 6.2 seconds. The
domain may be written two ways:
1) Interval notation
2) Set notation
x  [0, 6.2]
D = [ x 0  x  6.2, x R]
The range of a graph are the values determined by the independent
variable and given to the dependent value by some function which in this
case is the height of the golf ball. The lowest value here is 0 m as this is the
height of the ball when it lands on the green and the highest value is 51 m as
this is the maximum value of the ball in flight given by the vertex.
The range may be written two ways:
1) Interval notation
2) Set notation
y  [0,51]
R= [y 0< y < 51, y  R]
Math 112 Quadratics- Notes
Note(21) Interpreting the special points of a quadratic- There are 3 key
points of a quadratic that may have special meaning:
1) vertex
2) y-intercept
3) x-intercept
In the flight of a golf ball there are three questions that are often
asked:
1) What is the highest point in its flight? This can be read from the graph
and is the y-coordinate of the vertex which is the maximum value of the
graph.
2) How far off the ground was the football when it was thrown? This
corresponds to the y value of the y-intercept.
3) After how many seconds will the ball hit the ground? This corresponds to
the x value of the x-intercept.
The graph below illustrates these key points and what they represent.
Vertex (2.9, 51)
Maximum value = 51
y intercept (0,9)
Height above
ground is 9 m
x intercept (6.2 , 0)
time to hit ground
is 6.2 seconds
From the graph above:
1) The maximum height of the football is 51 meters (y value in vertex)
2) The height above ground when hit is 9 meters (y value of y-intercept)
3) The time it takes to hit the ground is 6.2 seconds (x value of x-intercept)
Math 112 Quadratics- Notes
Using the graphing calculator these values may be obtained using the
following keystrokes:
1) 2nd trace 4: maximum
(maximum height of football)
nd
2) 2 trace 1: value
(height above ground when thrown)
nd
3) 2 trace 2: zero
(time it takes to hit the ground)
Asn(A9) Handout of 2 questions with table of values that require Quadratic
Regression Analysis, maximum value, y intercept and x intercept
determination. (see end of this document)
Test #1
Asn(A10) handout on campfire sheet.
Asn(A11) Q 24,25,26 p. 22 Advanced Q 23 a) b) p. 22
Asn(A12) Handout of problems
Math 112 Quadratics- Notes
Asn(B1) Transformations handout
Asn(B2) Practice of transformations (to be submitted for marking)
Note(22) Forms of quadratic functions- Quadratic functions are found
written in 3 main forms:
1. transformational form:
1 (y - k) = ( x - h)2
a0
a
e.g. 1 (y + 2) =(x - 3)2
3
2. standard form:
y = a(x - h)2 + k
a0
e.g. y = 3(x - 3)2 - 2
3. general form:
y = ax2 + bx + c
a0
e.g. y = 3x2 - 18x + 25
Each form provides useful immediate information about the graph
such as:
1) the location of the vertex
2) the location of the y-intercept
3) vertical stretch of the graph
4) vertical and horizontal translations of the graph
5) maximum/minimum
It is useful to learn each form of graph as each reveals its own
characteristics of the quadratic function. Each may be rewritten in the other
form by algebraic method.
Note(23) The transformational form- This form of the function has the
general formula:
1 (y - k) = ( x - h)2
a0
a
The meaning of the letters a, k., and h are:
a= vertical stretch of the graph. (take the reciprocal of 1/a)
k= the vertical translation of the graph (multiply by -1)
h= the horizontal translation of the graph (multiply by -1)
Math 112 Quadratics- Notes
Note(24) The standard form- this form of the quadratic function has the
general formula:
y= a(x-h)2 + k where a  0
The meaning of the letters a, h, and k are the same:
a = vertical stretch factor.
h = horizontal translation
k = vertical translation
(read as is)
(multiply by -1)
(read as is)
The advantage of the standard form of the function is:
1) It gives the vertex of the function….
(h, k)
2) It gives the vertical stretch factor as……. a
3) It can be easily be entered into the graphing calculator
Asn(B3) Writing a quadratic in standard form. (see end of this document)
Math 112 Quadratics- Notes
Note(25) The general form- This form of the quadratic function has the
formula:
y = ax2 + bx + c
a0
The meaning of the letters a and c are:
a = vertical stretch factor
b = (unknown)
c = y- intercept value
The advantage of the general form of the function is:
1) It gives the vertical stretch factor as…….. a
2) It gives the y-intercept of the graph as…...c
3) It is the form given by the calculator and may easily be entered.
½ (y+1) = (x +2)2 transformation format
y = 2(x+2)2 -1
standard format
To write in general format the square (x+2)2 must be expanded:
y = 2(x+2)2 -1
y = 2[(x+2)(x+2)] - 1
y = 2[x2 +2x +2x + 4] - 1
y = 2[x2 +4x + 4] - 1
y = 2x2 + 8x + 8 - 1
expand (x+2)2
use FOIL to expand
add middle two “x” terms
distribute the 2 in front
simplify the constants
y = 2x2 + 8x + 7
a = 2 the vertical stretch
c = +7 the y-intercept (0,7)
Asn(B4) Writing a quadratic in general form. (see end of this document)
Math 112 Quadratics- Notes
Note(26) In order to rewrite the general form of an function into
transformation form it is necessary to recognize a “perfect square trinomial”.
To make a perfect square trinomial a binomial is squared.
e.g.(x +3)2 = x2 + 3x + 3x +9
= x2 + 6x +9
Perfect square trinomial
A perfect square trinomial can be factored into two identical factors:
e.g. x + 6x +9
= (x +3) (x +3)
= (x+3)2
2
We recognize (x+3)2 as the factored form of the “perfect square
trinomial” and as the “x- part” of the transformational form of a some
quadratic function.
e.g ½ (y +3) = (x+3)2
From the above information we conclude that the general form of an
function must be written in such a way as to contain a perfect square
trinomial which is then factored into a form similar to (x+3)2 above. It is
necessary, therefore, to create a perfect square trinomial as the first step in
rewriting.
From this example “ x2 + 6x +9 “ we compare the 6 with 9 and notice
that if we take ½ of 6 then square the answer we obtain 9. I.e. (6/2)2 = 9.
From this we have a way of predicting the third term of any perfect square
trinomial.
In general “c” in ax2 + bx + c can be found as c = (b/2)2 .
Eg. Find “c” in x2 + 10x + c.
c = (b/2)2
c = (10/2)2
c = (5)2
c = 25
so that x2 + 10x + 25 is a perfect square trinomial with factors:
(x+5)(x+5)
(x+5)2
Asn(B5) Q 4 p. 29
jumbled equations
Asn(B6) Q 3 p. 29
reflection on x- axis
Asn(B7) Extra practice. (summary handout)
Math 112 Quadratics- Notes
Note(27) Writing the general form into transformational form requires that
we find the value of “c” for a quadratic. This is called “completing the
square”.
e.g Rewrite this function in transformational form by completing the square.
y = x2 + 6x + 12
Move the constant to left hand side by
subtracting 12 from both sides.
Find the value of c to complete the
square. c = (b/2)2 so (6/2)2 so c = 9
Factor the trinomial, and simplify
constants on left side.
Transformational form.
y - 12 = x2 + 6x
y - 12 + 9 = x2 + 6x + 9
(y-3) = (x+3)2
Some quadratics written in general form require GCF factoring first.
e.g. Rewrite this function in transformational form by completing the square.
y = 2x2 -8x + 5
Move constant to left side of the by
subtracting 5 from both sides.
Factor out 2 on right hand side.
y -5 = 2x2 - 8x
y - 5 = 2[x2 - 4x]
y - 5 +2 4 = 2[x2 - 4x + 4
y - 5 + 8 = 2[(x -2)2 ]
1 (y +3) = 1 2(x - 2)2
2
2
½ (y +3) = (x -2)2
]
Find the value of c to complete the
square. c = (b/2)2 so (-4/2)2 so c = 4
Add 2 times 4 (equivalent to 8) to
both sides of the equation.
Complete the square and simplify
constants on the left side.
Multiply both sides by the reciprocal
of 2 or ½
Transformational form
Asn(B8) Q 15, 17, 20 p. 33 Advanced: Q 32 p. 35
Asn(B9) Q 19, 22, 26 a) and c) p. 33, 34
Asn(B10) Handout on completing the square. Advanced Q31 p. 35
Math 112 Quadratics- Notes
Note(27) (repeated). Writing the general form into transformational form
requires that we find the value of “c” for a quadratic. This is called
“completing the square”.
An alternative method of completing the square is provided here for the
same example given on the previous page.
Divide all terms by “a” the first term
coefficient. “a” =2 in this case
y = 2x2 -8x + 5
y = 2x2 -8x + 5
2 2
2 2
2
y = 2x -8x + 5
2 2
2 2
2
1 y = x -4x + 5
2
2
1y -5
= x2 -4x + 5 - 5
2 2
2
2
1y -5
2 2
Reduce all terms if possible
Simplify
Subtract 5/2 from both sides
= x2 -4x + 5 - 5
2
2
1y -5+ 4
2
2
= x2 -4x + 4
Simplify right side.
.
1y -5+ 4
= (x -2)(x -2) .
2
2
1y -5+ 4
= (x -2)2 .
2
2
1y -5+ 8
= (x -2)2 .
2 2 2
1y +3
= (x -2)2
2
2
1 (y + 2  3 ) = (x -2)2
2
1 2
1 (y +3)
2
= (x -2)2
Make a perfect square trinomial using
(b/2)2 = (-4/2)2 = (-2)2 = (-2 -2) = +4
Factor the right side.
Write as a single binomial squared.
Find a common denominator
for -5/2 and 4. (Rewrite 4 as 8/2).
Write left side in brackets.
Use the reciprocal of ½ (outside the
bracket) to determine the value
needed inside
Transformational form
Math 112 Quadratics- Notes
Note(28) Determining the equation of a quadratic from a graph.
Method #1:
1) Find the coordinates of the vertex (h , k)
2) Place k and h into the transformation formula
3) Solve for the unknown stretch factor in front (?) of the transformation
formula by substituting in a second point from the graph.
e.g. Find the equation of the quadratic illustrated below:
(3,4)
(4,2)
transformation formula:
1/a(y - k) = (x - h)2
Substitute in vertex (3,4)
2
(?)(y - 4) = (x - 3)
Replace 1/a with (?) for simpler solving
2
(?)(y - 4) = (x - 3)
Substitute in another point (4,2)
2
(?)(2 - 4) = (4 - 3)
(?)(-2) = (+1)2
Solve for (?)
(-2)(?) = 1
(?) = - 1
2
Replace (?) with -1 to obtain formula and use vertex (3,4)
2
-1 (y - 4) = (x - 3)2
Function in transformational format
2
Asn(B11) Q 39, 40 p. 38 39
Math 112 Quadratics- Notes
Note(29) Determining the equation of a quadratic from a graph.
Method #2:
1) Find the coordinates of the vertex (k , h)
2) Locate another point through which the graph passes.
3) Determine the stretch factor “a” by comparing to an equivalent point of
the basic quadratic. (Divide equivalent y coordinates.)
4) Determine the sign of the quadratic by the direction the graph opens.
5) Enter k, h, a, and the sign into the transformational formula.
e.g. Find the equation of the quadratic illustrated below:
(3,4)
(4,2)
transformation formula:
1/a(y - k) = (x - h)2
1/a(y - 4) = (x - 3)2
1 (y -4) = (x - 3)2
2
Substitute in vertex (3,4)
Determine the stretch factor using point
(4,2):
For y = x2
(over 1, up 1)
For this graph
(over 1, down 2)
Stretch factor:
2 divided by 1 = 2
Determine the sign (+) or (-) in front
of 1/a. Since the graph turns down
the sign is negative.
Transformational format
-1 (y - 4) = (x - 3)2
2
Asn(B12) Q 1 p. 61 in curriculum document.
Test #2
Math 112 Quadratics- Notes
Note(30) Quadratic functions come in the form:
y = ax2 + bx + c
where a  0
Very often a quadratic function will touch or cross the x- axis. The
point(s) where the function crosses are significant as they often have special
meaning. These points contain the zeroes or roots of the quadratic.
y = x2 +1
no x intercepts
no zeroes of the quadratic function
no Real roots, the roots are imaginary
y = x2
one x intercept
one zero of the quadratic function
one root of the quadratic equation
Math 112 Quadratics- Notes
y = x2 -3
two x intercepts
two zeroes of the quadratic function
two roots of the quadratic equation
Roots or zeroes
For the above graph the points on the x-axis are called intercepts.
The x values of these points are called:
1) the roots of the quadratic equation or
2) the zeroes of the quadratic function
Math 112 Quadratics- Notes
Note(31) Quadratic Equation- a quadratic equation comes in the form:
0 = ax2 + bx + c where a 0 and y= 0
When the quadratic function has y set to 0 it is called a quadratic
equation. Solving for the variable “x” gives us the values of the “roots” of
the equation or “zeroes” of the function.
For the quadratic function y = x2 +2x –8, the roots can be found by
solving the quadratic equation when y is set to “0”. We solve for variable in
the equation: 0 = x2 +2x –8 and obtain the roots: x = 2 and x = -4. These
values may be obtained a number of ways….
Method #1 Graphing calculator method
Enter the function after y1= x2 + 2x –8
Press 2nd trace option 2:zeroes. Set left boundary, enter, and right
boundary, enter, then press enter to guess the value of the root.
Method #2 Factoring method
Set y = 0 then factor the quadratic equation
0 = x2 +2x –8
0 = (x-2) (x+4)
set x-2 = 0
set x+4 = 0
x=2
x = -4
The roots of the equation are 2 and –4
The submarine will surface in 2 hours. (-4 is an inadmissible time)
Method #3 completing the square method
0
= x2 +2x –8
Add 8 to both sides to move constant
2
8
= x + 2x
Find the missing 3rd term of the trinomial
8 + 1 = x2 +2x + 1
(b/2)2 = (2/2)2 = (1)2 = 1
9
= (x+1)2
Take the square root of both sides
3
= (x+1)
Subtract 1 from both sides
-1  3
=x
The roots are:
x = -1 +3
and
x = -1 – 3
x = +2
x = -4
The submarine will surface in 2 hours. (-4 is an inadmissible time)
Asn(C1) Q 8 a c d ,9 a b ,10 a) g) p. 45
Math 112 Quadratics- Notes
Method #4 using the quadratic formula
The quadratic formula obtained from completing the square using the
general form of the quadratic y = ax2 +bx +c is:
x = -b  (b)2 – 4ac
2a
For the equation y = x2 +2x –8 the values of a, b, and c are a =1 b = 2 c=
-8
Substituting into the formula we have:
x = - (2 )  (2 )2 – 4( 1 )(-8 )
x = -2 + 6 = 4 = +2
2( 1 )
2
2
x = -2 4+32
2
x = -2 – 6 = -8 = -4
x = -2  36
x = -2 6
2
2
2
2
The roots are +2 and –4
The submarine will surface in 2 hours. (-4 is an inadmissible time)
Asn(C2) Q 24 p. 48 Advanced: Derive the quadratic formula by hand
Math 112 Quadratics- Notes
Note(32) Using the quadratic formula to find the two roots of the equation
gives us the x values of the intercepts. Averaging these values will give us
the center point of the parabola along the x- axis through which the axis of
symmetry runs. The vertex shares this same x-value. The average for the two
x values expressed in general form is:
-b + (b)2 – 4ac + -b - (b)2 – 4ac
2a
2a
=
-b
2a
2
This expression (–b/2a) is the midpoint of the parabola along the x
axis and therefore the x value in the vertex. We may find the vertex by
solving for x then substituting this value into the quadratic function to find
the y value. e.g. Find the vertex of the parabola y = x2 +2x – 8.
Midpoint = -b = -(2) = -2 = -1
Coordinate of midpoint (-1, 0)
2a 2(1) 2
Since the vertex is directly below/above the midpoint, the coordinate of the
vertex is: (-1 , ?)
Now substitute –1 into the quadratic function to find the y value of the
vertex.
y = x2 +2x -8
= (-1 )2 + 2(-1 ) -8
= 1–2–8
y = -9
The vertex has coordinates (-1,9)
Asn(C2) cont’d Q 25 p. 48
Math 112 Quadratics- Notes
Note(33) The quadratic formula will sometimes produce an answer that does
not belong to the Real number system. In such a case the quadratic has no
Real roots so that the graph has no x-intercepts. The roots are complex
numbers and belong to the imaginary number system.
E.g. Find the roots of the equation 0 = x2 –6x +13
Use the quadratic formula:
x = -b  (b)2 – 4ac
2a
x = - (-6 )  (-6 )2 – 4(1 )(+13
2(1 )
x = +6   36 – 52
2
x = +6  -16
2
x = +6   16  -1
2
x = +6  16 -1
2
x = +6  4 -1
2
x = +6  4i
2
x = +3  2i
)
a= 1 b = -6 c = +13
Rewrite -16 as 16  -1
Separate 16 and –1 into two radicals
Take the square root of 16
Write -1 as i, an imaginary number
Reduce by dividing each term by 2
The roots of the quadratic are:
X = +3 + 2i and x = +3 – 2i
Note: These are imaginary roots and therefore there are no x-intercepts.
Asn(C3) Q 27,28,29 p. 48,49
Asn(C4) Q 30 a,c,e,g 31 a,b p. 49 Advanced Q 32 a , c, e p. 49
Math 112 Quadratics- Notes
Note(34) Problem solving often requires that we solve for the vertex and/or
the roots of the equation. The vertex gives the maximum or minimum value
of a question whereas the roots give the “time to hit the ground” or the value
when there is “no profit” or “width” when the area is zero etc.
Eg. a golf ball is hit from a 1 meter platform inside a golf dome and follows
a path given by y = -6x2 +12x +1 where x = time (s) and y = height (m)
1) What maximum height will it reach? (Hint: find the vertex)
Vertex formula = -b = - (12) = -12 = +1 = vertex = (+1 , ?)
2a
2(-6)
-12
To find the y value in the vertex we substitute +1 into the formula as
follows:
y = -6x2 +12x + 1
y = -6(+1)2 +12(+1) +1
y = -6(+1) + 12(+1) +1
y = -6 +12 +1
y = +7
The vertex has coordinates (1,7) which we interpret as after 1 second
the golf ball reaches a maximum of 7 meters.
2) When will it hit the ground? (Hint: It hits the ground at the x- intercept)
Find the zeroes or roots of the quadratic equation:
x = -b  (b)2 – 4ac
2a
-12  13
-12 +13 = 1 = -0.8
2
x = - (12 )  (12 ) – 4(-6 )(1)
-12
-12
-12
2(-6)
x = -(12)  144 +24)
-12 –13 = -25 = +2.08
-12
-12
-12
x = -12   168
-12
The golf ball will hit the ground after 2.08 seconds. The first answer is
inadmissible as it is not sensible to have a negative answer.
Asn(C5) Q 22 p. 33, Q 26 p. 34, Q 5 p. 44, Q 39,40,41 p. 53
Asn(C6) Q 34,35,36,38 p. 52,53
Math 112 Quadratics- Notes
Note(35) The discriminant of the quadratic formula is given by the
expression b2 – 4ac. The value of the expression can assist us in determining
the nature of the roots for a quadratic equation.
e.g Find the discriminant for the following quadratic by using the expression
b2 –4ac . Also find the roots of the equation.
Example #1 y = x2 –6x +13
a = 1 b = -6 c = 13
The value of the discriminant is:
b2 –4ac
(-6 )2 – 4(1 )(13)
36 – 52
-16
Note: The discriminant is negative
Find the roots:
x = -b  (b)2 – 4ac
2a
x = -(-6 )  -16
(from above)
2(1)
x = +6  16  -1
-16 may be written as 16  -1
2
x = +6  16 -1
16  -1 may be separated into 16 -1
2
x = +6  4i
Take the square root of 16 and replace -1 with i
2
x = +6  4i
Separate into two fractions and reduce
2 2
x = 3  2i
which gives the 2 roots…… 3 + 2i and 3 – 2i
Conclusion: When the discriminant is negative we obtain 2 complex roots.
The graph obtained indicates that there are no x-intercepts:
Example #2
Math 112 Quadratics- Notes
y = x +4x + 4
a = 1 b = +4 c = 4
2
The value of the discriminant is:
b2 –4ac
(+4 )2 – 4(1 )(4)
16  16
0
Note: The discriminant is zero
Find the roots:
x = -b  (b)2 – 4ac
2a
x = - (4 )   0
(from above)
2(1)
x = -4
2
x = -2
One answer
By factoring we discover that there are two roots (not one) but they are
identical. We can factor the trinomial to see the two roots:
0 = x2 +4x +4
0 = (x +2)(x + 2)
First factor:
We set x +2 = 0
Solve: x = -2
Second factor:
We set x +2 = 0
Solve: x = -2
Conclusion: When the discriminant is zero we obtain 2 real roots that are
identical. The graph obtained indicates that there is one x-intercept:
(-2 , 0)
Math 112 Quadratics- Notes
Example #3
y = x2 -x -6
a = 1 b = -1
c = -6
The value of the discriminant is:
b2 –4ac
(-1)2 – 4(1 )(-6)
1 +24
25
Note: The discriminant is positive
Find the roots:
x = -b  (b)2 – 4ac
2a
x = -(-1 )   25
(from above)
2(1)
x = +1  5
2
x = 6 = +3
or x = -4 = -2 The roots are +3 and -2
2
2
Conclusion: When the discriminant is positive we obtain 2 real roots. The
graph obtained indicates that there are two x-intercepts.
(0,-2)
(0,+3)
Summary of the discriminant:
If b2 – 4ac < 0 The graph has two imaginary roots and no x-intercepts.
If b2 – 4ac = 0 The graph has two identical real roots and one x-intercept.
If b2 – 4ac > 0 The graph has two real roots and two x-intercepts.
Note: All quadratics have two roots. It is the x-intercepts that vary.
Asn(C7) Q 53,56,57, 58 a b, 59 p.56,57
Asn(C8) Handout on various problems involving quadratics. Selected.
Test #3
Asn(C9) Advanced Q 62,64,63 p. 57
Advanced Q Investigation #7 65,66,67 p. 58
Advanced Q 68 to 72 p. 59 selected questions.
Assignment section for handouts
Asn(A7) Extra practice sheet (to be submitted for marking)
1. Identify the type of sequence as arithmetic, quadratic, cubic or none of
these:
A)
B)
C)
D)
E)
F)
G)
[-2, -5, -8, -14, …]
[-3, -10 , -21, -36, -55, -78, -105, …]
[–1, -14, -51, -124, -245, -426, -679,…]
[-1/3, 2/3, 3, 20/3, 35/3, 18, 77/3, 104/3, 45, …]
[11, 101, 1001, 10001, 100001, …]
[1/5, 2/5, 4/5, 7/5, 11/5, 16/5, …]
[0.0, 3.2, 18.4, 75.6, 252.8, 700.0, 1663.2, 3508.4, 6745.6, …]
2. Find the first four terms of each sequence:
A) tn = -2n – 5
B) tn = -3n + ¾
C) tn = -n2 –5n +3
D) tn = 1/3 n2
3. What is D1 for the sequence defined by tn = 5/7n +4/5
4. Dan had 80 m of fencing to enclose a rectangular yard on three sides. He
used his house as the fourth side. By experimenting with different widths as
shown in the table, Dan found he could enclose different areas. What type
of function could he use to model the data/
Width
Area
1
78
2
152
3
222
4
288
5
350
6
408
7
462
8
512
Asn(A9) Determining the special points of a quadratic
1. The table shows the height of a golf ball (in metres) after various lengths
of time measured in seconds. The tee-off is on a cliff.
Time (s)
1
2
3
4
5
Height (m) 32.8 46.8 51.0 45.4 30.0
a) Draw a sketch of your graph
b) Does the data suggest a linear or quadratic function? Explain
c) What is the maximum height of the golf ball in its flight?
d) What is the height of the cliff?
e) How long will it take to hit the fairway?
f) State the Domain and Range of the graph
2. The table below gives the height of a baseball (in metres) for various
times measured in seconds.
Time (s)
1
2
3
4
5
6
Height (m) 32.3 53.3 64.5 65.9 57.5 39.3
a) Draw a sketch of your graph.
b) What is the value of D2?
c) What is the maximum height of the baseball in its flight?
d) How long does it take to reach this height?
e) How far off the ground is the baseball when it is hit?
f) After how many second will it hit the ground?
g) State the Domain and Range of the graph
Asn(B3) Write the following in standard form and interpret the standard
form by identifying the vertex and vertical stretch factor.
1. ½ (y-1)
= (x – 1)2
2. 2(y – 1) = (x + 2)2
3. ¼(y
= (x+2)2
4. 3(y-2)
= (x)2
5. 2y
= x2
Asn(B4) Write in general form and identify the y- intercept and vertical
stretch factor.
1. ½(y-1)
= (x –1)2
2. 2(y+1)
= (x+3)2
3. ¼(y)
= (x+2)2
4. 3(y+2)
= x2
5. 2y
= x2
Math 112 Rate of Change- Notes
Note(36) Rate of change is a concept that is used in many different contexts.
A few examples are given below:
1. Birth rate
2. Heart rate
3. Population growth rate
4. Employment rate
5. Global warming
6. Rate of travel (speed)
7. Production rates
births/1000 per year
beats/ minute
people/year
people/year
degrees Celsius/ year
km/hour
units/hour
Each of these rates can be calculated by dividing the first quantity by
the second which is given as a time unit.
Eg. #1 A car travels 220km in 2 hours. What is the rate of travel in
km/hour?
Rate = D/t = 220/2 = 110km/hr
E.g. #2 Your heart beats 20 times in 15 seconds. What is your heart rate in
beats/min?
Heart rate = 20 beats/15 sec = 1.3 beats/sec = 80 beats/min.
Asn(D1) Handout
Math 112 Rate of Change- Notes
Note(37) Function notationThe general notation of y = ax2 + bx + c is often seen written in
function notation as: f(x) = ax2 + bx + c
Since y = f(x) we may write the general form of a point (x,y) as …..
(x, f(x) ) so that (x,y) = (x , f(x) )
For the 2 general points (x1,y1) and (x2, y2) the slope is given by:
Slope = Δ y = y2 - y1
Δx
x2 – x 1
The same 2 points written in function notation are:
(x1 , f(x1) ) and (x2 , f(x2) ) and have the slope given by:
Slope = Δ f(x) = f(x2)- f(x1)
Δx
x 2 – x1
For points (a, f(a)) and (b, f(b)) the average rate of change is often seen
written as:
Δ f(x) = f(b) – f(a)
Δ x
b-a
Math 112 Rate of Change- Notes
Note(38) The line drawn between two points on a curve is known as the
secant for the two points. The slope of the secant is equal to the average rate
of change.
B
A
Secant through points A and B
The average rate of change for a function can be calculated by using
the formula:
Δ f(x) = f(x2)- f(x1)
Δx
x2 – x1
E.g. John leaves for work to go to Toronto at 6 am and arrives at 8 am . The
odometer is reset to read 0 km at the start. At the end of his trip it reads 240
km. What is his average rate of travel?
Points are (6, 0) and (8,240) where x = time and y = distance
Average rate = Δ f(x) = f(x2)- f(x1) = 240 – 0 = 240 = 120 km/hr
of change
Δx
x2 – x1
8 –6
2
Asn(D2) Handout
Asn(D3) Q 13 a,b p. 79 Q 15 a,b,c p. 80 Q 16 a to e p. 81 Q 17 a,b,c p.81
Q 24 a to c p. 83
Asn(D4) Handout. Curriculum Document
Math 112 Rate of Change- Notes
Note(39) The average rate of change for a straight line is a constant
regardless of points used to calculated.
e.g The function f(x) = 2x + 1 has ordered pairs such as:
A (0,1)
B(1,3)
C(2,5)
The graph would appear as follows:
C
Ave rateAB = 3 – 1 = 2 = +2
1–0 1
B
Ave rateBC = 5 –3 = 2 = +2
2–1 1
A
The average rates are both +2 therefore
constant. (A straight line has only one slope)
Note(40) The average rate of change for a quadratic is not a constant but is
continually changing.
e.g. The path of a football is given by the function f(x) = -1/2 x2 + 4x -2
and has ordered pairs such as:
A (1, 1.5) B(2,4)
C(3, 5.5)
D (4,6)
E(5, 5.5)
F(6,4)
G(7,1.5)
Ave rateAB = 4.0 -1.5
2-1
Ave rateBC = 5.5 –4.0
3-2
Ave rateCD = 6.0 –5.5
4–3
Ave rateDE =5.5 –6.0
5–4
Ave rateEF = 4.0 - 5.5
6–5
Ave rateFG = 1.5 - 4.0
7-6
= +2.5
= +1.5
Ave rate at vertex = 0
= +0.5
D
= -0.5
C
E
B
F
G
A
F
= -1.5
= -2.5
Math 112 Rate of Change- Notes
Note(41) The average rate of change for a football’s flight is:
1) Positive as it gains height
2) Zero at the maximum height
3) Negative as it loses height
Note(42) Instantaneous rate of change is a concept that asks for the rate of
change for a single point in time. It is not an average.
An example of where this can be illustrated in real life is asking for the
speed at any given time when a car is accelerating. By looking at your
speedometer we can read the instantaneous rate of speed at any time. The
speedometer measures instantaneous rate of change.
Asn(D5) Handout for calculating the average rate of change as P approaches
Q where the function is y = x2 and the point of interest is (5,25)
Note(43) We can approximate the instantaneous rate of change for a
function by using 2 points either side of a given point of interest that are
very, very close to the given point
E.g. Find the instantaneous rate of change for the function y =x2 when x = 5
Step #1 Find two points on either side of (5,25) the point of interest that fit
the curve y= x2
If we choose the increment (.01) we can obtain points close to (5,25)
by:
Adding
.01 to 5
to obtain the point (5.01, ? )
Subtracting .01 from 5 to obtain the point (4.99, ? )
We may now find the y value of each point above by substitution:
y = x2
y = x2
so that (5.01)2 = 25.1001 which gives us (5.01, 25.1001)
so that (4.99)2 = 24.9001 which gives us (4.99, 24.9001)
Math 112 Rate of Change- Notes
The two points are equidistant from the point of interest (5,25) and if
connected with a straight line form a secant line that is very close to the
tangent at the point (5,25). This line is parallel to the tangent at this point.
Since parallel lines have equal slopes the slope of the secant would equal
the slope of the tangent at point (5,25).
Step #2 Using the formula for the slope of a function we obtain an
approximation for the instantaneous rate of change at point (5,25)
Point (5.01, 25.1001)
Point (4.99, 24.9001)
Instantaneous rate = f(b) – f(a) = 25.1001 – 24.9001 = .2 = 10
of change
ba
5.01 – 4.99
.02
The instantaneous rate of change for the function y =x2 at the point
(5,25) is 10
Asn(D6) Approximating instantaneous rate of change. (overhead).
Asn(D7) Questions on instantaneous rate of change. (handout NFLD)
Asn(D8) Garden Question (handout).
Asn(D9) Curriculum Document (handout).
Math 112 Rate of Change- Notes
Note(44) Problem solving may require that you find one of the following
points of interest:
Point of interest: When you find these points:
a) vertex
b) x-intercepts
c) y-intercepts
If the question asks for a maximum (height, area)
If the question asks for time in air
If the question asks for initial starting point
Point of interest: Formulas to use:
a) vertex
(-b , )
2a
b) x-intercepts
x = -b   b2 –4ac
2a
c) y-intercepts
Once you find the x coordinate substitute
into the quadratic to obtain the y- coordinate
Read this from the quadratic. It is the value
c in the function if it is written in general
format y= ax2 + bx + c
Example:
Babe Ruth was renowned for hitting high fly balls in baseball. One day he
made contact with a ball 1 metre above ground and hit the ball at a speed of
27.5 m/sec. Using the general formula h = H + vit – 4.9t2 substitute for H
and vi to make a quadratic function to fit this situation.
a) What maximum height did the ball reach?
b) When did it reach this height?
c) How high off the ground was the ball hit?
d) How long was the ball in the air?
e) What was the ball’s average velocity between 1 and 2 seconds?
f) What is the height of the ball at 4 seconds?
g) What was the speed of the ball at 4 seconds?
Math 112 Rate of Change- Notes
To solve this problem we need to substitute for the values H and vi
H = initial height = 1 metre (we are careful not to confuse H for h)
Vi = initial velocity = 27.5 m/sec
By substitution we have:
h = 1 + 27.5t –4.9t2
Written in descending order of power we have:
h = -4.9t2 + 27.5t + 1
For the purposes of the quadratic formula and vertex formula we
identify the letters a, b, and c as:
a = -4.9
b = 27.5
c=1
We are now ready to answer the question a) through f)
a) What maximum height did the ball reach?
This question requires that you solve for the vertex. The y value of
the vertex gives the maximum height. Use the formula:
–b = -27.5 = -27.9 = + 2.846938776
2a 2(-4.9) -9.8
We are careful to avoid rounding off too early so choose 2.85 as the x
coordinate. Now we substitute this value into the quadratic
h = -4.9t2 + 27.5t + 1
h = -4.9(2.85)2 + 27.5(2.85) + 1
h = 39.57 m
The vertex is therefore (2.85, 39.57). The y-coordinate is 39.57m, the
maximum height of the ball.
Statement: The maximum height is 39.75 metres
Math 112 Rate of Change- Notes
b) When did it reach this height?
This question requires that you have the vertex of your quadratic
which you have already found in a) above (2.85, 39.57). The x coordinate
is the answer to this question 2.85
Statement: The time required to reach this maximum height is 2.85
seconds.
c) How high off the ground was the ball hit?
This question does not require a formula. The c value in the quadratic
gives this answer, as it is the y-intercept value.
h = -4.9t2 + 27.5t + 1
The value of c is +1.
Statement: The ball was hit 1 metre off the ground.
d) How long was the ball in the air?
The question is asking for the time, which is given along the x-axis.
Therefore, we need the distance along the x-axis which is given by the x
coordinate of the x intercept. We use the quadratic formula.
x = -b   b2 –4ac
2a
x = -27.5  (27.5)2 –4(-4.9)(1)
2(-4.9)
x = -27.5   775.85
-9.8
x = -27.5  27.85
-9.8
x = -55.35 = 5.64
-9.8
x = +.35
-9.8
= -0.0357
The value 5.64 is the x value of the x-intercept (5.64,0) and
represents the time of the ball in the air.
Statement: The ball was in the air for 5.64 seconds.
Math 112 Rate of Change- Notes
e) What was the ball’s average velocity between 1 and 2 seconds ?
The word average suggests two items or points are involved. This
requires that we find or locate two points for which an average velocity is to
be calculated from. The numbers 1 and 2 are in fact the x-values of the two
points… (1, ) and (2, ).
Two find the y values of these points we substitute into the quadratic
as follows:
x = 1 or ( t = 1)
x = 2 or (t = 2)
h = -4.9t2 + 27.5t + 1
h = -4.9(1)2 + 27.5(1) + 1
h = 23.1
h = -4.9t2 + 27.5t + 1
h = -4.9(2)2 + 27.5(2) + 1
h = 35.4
Point (1, 23.1)
Point (2, 35.4)
The average rate of change is found using the equivalent of the slope
formula but for quadratic functions;
f(b) – f(a) = f(2) – f(1) = 35.4 – 23.1 = 12.3 = 12.3 m/sec
b- a
2–1
2–1
1
Statement: The average rate of change for t = 1 to 2 seconds is 12.3
m/sec.
Math 112 Rate of Change- Notes
f) What is the height of the ball at 4 seconds?
This question requires that we understand that 4 represents the time
and that we are asking for height so that we substitute 4 for t and solve for h.
Substitute t = 4 into the quadratic to obtain the y value
h = -4.9t2
+ 27.5t + 1
2
h = -4.9(4 ) +27.5(4 ) + 1
h = 32.6
Point (4, 32.6)
Statement: The height of the ball at 4 seconds is 32.6 metres.
g) What was the speed of the ball at 4 seconds?
This question is in fact asking for the instantaneous rate of change
when t = 4 seconds and must be distinguished from the question “What is
the height of the ball at 4 seconds?” The key word here is speed
Step #1 Find two points really close to (4,32.6) by adding and subtracting a
small increment (.01) to/from the x value of this point.
4 + .01 = 4.01
4 - .01 = 3.99
The two points are: (4.01, ) and (3.99, ). Now substitute to find the yvalues of each point. (Do not round off the y-values for accuracy’s sake).
Point (4.01, )
h = -4.9t2 + 27.5t + 1
h = -4.9(4.01)2 +27.5(4.01) + 1
h = 32.48251
Point (3.99,
)
2
h = -4.9t + 27.5t + 1
h = -4.9(3.99)2 +27.5(3.99) + 1
h = 32.71651
The points are: (4.01, 32.48251) and (3.99, 32.71651)
Step #2 Find instantaneous rate of change or speed use the formula
f(b) – f(a) = 32.71651 – 32.48251 = 0.231 = 11.55 m/s
b- a
3.99 – 4.01
0.02
Statement: The instantaneous rate of change (speed) at the time t = 4
seconds is 11.55 m/s.
Test #4
Math 112 Exponential Growth
Note(45) A geometric sequence is a sequence in which each term can be
generated by multiplying each successive term by a common factor.
e.g 2 , 4 , 8 , 16,
2 2 2
This geometric sequence is generated by multiplying
each term by the common factor 2
Note(46) A common ratio is a factor used to generate a geometric sequence.
The common ratio above is 2.
Note(47) We can find the common ratio of a sequence by dividing any two
successive terms as long as we use the current term as the dividend and the
previous term as the divisor.
e.g Find the common ratio for the following sequence:
3, 12, 48 ,192
12 = 4second term divided by first term
3
48 = 4third term divided by second term
12
Note(48) A recursive formula is a formula that is made by relating two
successive terms in some way such that other successive terms are related in
the same way.
To develop a recursive formula we first establish the pattern between
any two successive terms in a concrete way then generalize using variables.
E.g. Develop a recursive formula for this geometric sequence:
2 , 6, 18, 54
The common ratio is 6/2 = 3 so that b = 3 therefore:
2nd = three times the 1st
6 = (3)

2
Replace 2nd term by 6 and 1st by 2
18 = (3)

6
Replace 3rd term by 18 and 2nd by 6
tn+1 = (3)

tn
Replace next term by tn+1 and current
term by tn
The recursive formula for this sequence is tn+1 = (3)  tn
Math 112 Exponential Growth
Note(49) A general formula for a geometric sequence is much more
practical that the recursive formula as the general formula will give the value
for any term without the need for a previous term. The general formula for a
geometric sequence is given by y =abx
where:
a = the first term
b = the common ratio
x = real number
e.g Generate a geometric sequence where the first term is 3 and the common
ratio is 4.
a = 3 the first term
b = 4 the common ratio
Substituting the values of “a” and “b” into the formula we have:
y = a*bx
y = 3*4x
x value
substitution
y- value
Let x = 0
Let x = 1
Let x = 2
y = 3*4x = 3*40 = 3*1 =
y = 3*4x = 3*41 = 3*4 =
y = 3*4x = 3*42 = 3*16 =
3 the 1st number
12 the 2nd number
48 the 3rd number
The sequence generated is:
3 , 12, 48, etc.
Math 112 Exponential Growth
Note(50) We can determine the general formula for a geometric sequence…
y = a*bx by identifying the 1st term “a” and by calculating the common
ratio.
Eg. What is the general formula that will generate the sequence below:
2, 6, 18, 54, 162, etc.
a = 2 the first term
b = 6/2 = 3 the common ratio
Substituting the values for “a” and “b” into the general formula:
y = a*bx
y = 2*3x
The general formula for the sequence is y = 2*3x
Note(51) Graphing technology can generate the same sequence and ordered
pairs can be read from the resulting table.
Enter the equation after \y1= 2*3x
Press 2nd Table to see the ordered pairs
Press Graph to see the graph of the function
Asn(E1) Investigation #1 Q 3 and 8 p. 110 to 113
Math 112 Exponential Growth
Note(52) The graph of a geometric sequence is discrete with a domain
given by D = {x  x  N}
E.g Graph the following:
X
Y
1
2
2
4
3
8
4
16
Discrete = dotted line graphing
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a) What is the recursive formula for this sequence? By observing the pattern
in the second row we notice that the next term is 3 times the previous. By
example we have:
2 = (2)  1
the 2nd is double the 1st
4 = (2)  2
the 3rd is double the 2nd
tn+1 = (2)  tn
the next is double the previous
The recursive formula is : tn+1 = (2)  tn
b) What is the y intercept? The first term of the sequence has x = 0 and y =
1
The y intercept is (0,1) y = 1
c) What is the common ratio? The common ratio is found by dividing any
two successive terms. So in general the common ration is found by:
tn+1  tn = 2/1 or 4/2 = 2
d) What is the range of the graph?
R = {y y  2, y  N}
Math 112 Exponential Growth
Note(53) The graph of a geometric function is continuous with a domain
given by D = {x x  R}
E.G. Graph y = 3x
Continuous = solid line graphing
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a) What is the general formula for this graph? y = 3x
b) What is the y intercept? Given when x = 0 so y = 30 = 1
The y intercept is y = 1
c) What is the common ratio? The base is the common ratio so = 3
d) What is the Range for this graph? R = {y y>0, yR}
Math 112 Exponential Growth
Note(54) Growth and decay curves are determined by the “b” value in the
general formula y = a*bx
For the general formula y = a*bx
a) if b < 1 the curve is a decay curve. E.g. let b = (0.5)
e.g. y = 2* (0.5)x
b) If b>1 the curve is a growth curve. E.g. let b = (1.5)
e.g. y = 2(1.5)x
Asn(E2) Investigation #2 Q 17 to 21 p. 115
Math 112 Exponential Growth
Note(55) Function notation for geometric sequences can be written n the
form:
f(x) = a*bx
where a = first term
b = common ratio
x0
E.g. A geometric sequence is given by : 2, 6, 18, 54
a) Write this in function notation
b) Find f(3)
c) What is the value of f(5)?
a) a = 2 the first term.
b = 6/2 = 3 the common ratio
The function notation for this geometric sequence is f(x) = 2*3x
b) f(3) = 2*3x
f(3) = 2*33
f(3) = 2*27
f(3) = 54
Substitute in x = 3
c) f(5) = 2*3x
f(5) = 2*35
f(5) = 2*243
f(5) = 486
Substitute in x = 5
Exponential Growth
Note(56) Geometric sequences relate to many real life situations such as:
1) Population growth2) Radioactive decay3) Compound interest growth
Percentage increase or decrease (property appreciation or
depreciation) can be expressed in terms of a common ratio. The total
increase can be found by multiplying by a common ratio.
E.g.#1 The population of Fredericton is 40,000. If the city expects a 10%
increase by next year what will the new population be?
Method #1: Old way
10% of 40,000
= 4000
40,000 + 4000
= 44,000
Take 10% of 40,000
Add this answer to 40,000
The new population will be 44 000.
Method #2: New Way (Growth factor)
This may be done another way. If we add 10% to 100% we have:
100%
+10%
110% Change this to decimal format by dividing by 100 = (1.10)
(1.10) is a growth factor. We use this to multiply by 40,000 so using the
general formula we can substitute for a and b as follows:
y = a*bx
where a = 40,000 initial population
b= (1.10) growth factor which represents 110%
x = 1 year’s growth
1
y = 40 000*(1.10)
y = 44 000
The new population will be 44 000
Exponential Growth
E.g.#2 A $20,000 car depreciates by 18% per year after you buy it. What
will the car be worth in 5 years?
Since this is a depreciation question instead of adding the percent to
100% be instead subtract as follows:
100%
-18%
82%
Change this to decimal format by dividing by 100 = (.82)
(.82) is the decay factor. We use this to multiply by $20 000 so using the
general formula we can substitute for a and b as follows:
y = a*bx
where a = 20,000 initial car value
b= (.82) decay factor which represents 82%
x = 5 year’s growth
y = 20 000*(.82)5
y = 20 000*(0.370739843)
y = $7414.80
Asn(E3) Focus B Q 22 to 28 p.116 Q 30,31, 33 p. 118 Q. 10, 11 p. 113
Exponential Growth
Asn(E4) Investigation #3 Q 34 to 37 p. 119
Note(57) For exponential curves the growth factor or decay factor is affected
by the sign of the variable. Ie. when x > 0 vs when x < 0
(For each case below “a” = 1)
a) if x > 0 ie. x = [1,2,3,4, …]
y = 2x
y = 21
y=2
let x = 1
y = 2x
y = 22
y=4
let x = 2
y >1
+1
x>0
Observation: All y values will be greater that 1. (Never negative)
I.e. the range is R = [y  y >1, yR]
b) if x< 0 ie. x […,-2,-1]
y = 2x
y = 2-1
y=1
2
let x = -1
0 < y < +1
+1
x
y=2
let x = -2
-2
y=2
y=1
x<0
2
2
y=1
4
Observation: All y values will be less than 1 but greater than 0. (Never
negative).
Ie. the range is R= [y  0 < y < 1, y R]
Exponential Growth
Asn(E5) Investigation #4 Q 41 to 49 p. 122
Note(58) The value of “b” for the general formula y = a*bx determines the
rate of growth (or decay).
Compare: y1 = 5x and y1 = 6x when x < 0
1) The common ratio for y = 5x is 5 so that b = 5. This means the y values or
the graph will increase at a rate of 5 times for each increment of 1 on the x
variable.
2) The common ratio for y = 6x is 6 so that b = 6. This means the y values or
the graph will increase at a rate of 6 times for each increment of 1 on the x
variable.
x
y1
y2
-3
.008
.0046
-2
.04
.028
-1
.20
.17
0
Growing by a factor of 5
Growing by a factor of 6 (faster rate)
Conclusion: Even though the values for y2 are smaller than y1 the growth
rate is still faster. When x = 0 the values of y2 began to surpass those of y1
Compare: y1 = 5x and y1 = 6x when x ≥ 0
x
y1
y2
0
1
1
1
5
6
2
25
36
3
Growing by a factor of 5
Growing by a factor of 6 (faster rate)
Conclusion: The rate of growth is still faster for the graph y = 6x compared
to the graph y = 5x . (y2= y1 when x = 0) . When x >0 the y2 values are
greater than the y1 values.
Exponential Growth
Note(59) The rules for exponent operations:
The product rule
bm  bn = bm+n
The quotient rule
bm = bm-n
bn
b0 = 1
The zero exponent
b-m = 1
and
m
b
(ab)m = ambm
The negative exponent
The power of a product
1 = bn
b-n
a n = an
b
bn
(am)n = amn
The power of a quotient
The power of a power
Note(60) Success with evaluating powers is related to the understanding of
what the base is in each power.
E.G. Determine the base of each power then evaluate.
E.g.#1
Meaning
(2)4 =
(-2)4 =
-24 =
2  2  2  2 = +16
(-2)(-2)(-2)(-2) = +16
- 2 2  2  2 = -16
E.g.#2
20
=
0
(-2) =
-20 =
Answer
1
1
-1
The base:
The base is 2
The base is –2
The base is 2. The subtract sign is not
affected by the exponent 4.
The base:
The base is 2
The base is –2. The minus is affected by the 0 exp.
The base is 2. The minus is independent of 0 exp.
Asn(E6) Handout. (see next page of this printout)
Exponential Growth
Asn(E6)
1.Simplify to a single base:
a) 32  35
b) 36  33
c) (24)3
2. Simplify to a single base:
a) (-3)12
(-3)4  (-3)6
b) 25  2-4
2-2
23
3. Simplify and write as a single power:
a) (75  73 )2
b) (9-3  98)-2
4. Simplify these expressions:
a) w -4  w6
b) n-11  n -4
c) x -4 y -6
x -5 y4
d) 24 x3 y -2
12 x2 y3
e) (9a-1b0)(5a4b-3)
5. Evaluate: (No decimal answers)
a) -30
b) (-4)0
c) -2-2
d) -(-3)-2
e) 1
5-2
6. Simplify and substitute for a = -2 and b = 3
a) 18a-3 b12
a-5b14
Math 112 Exponential Growth
Note(61) Algebraic exponents are handled the same way as regular
exponents except you are required to group like terms (tiles) then simplify.
E.g. Simplify to a single base m:
(m 2a +b)(m a+3b) = m 2a+b+a+3b
= m 2a+a +b+3b
= m 3a +4b
Add the exponents
Group like terms
Add like terms
Asn(E7) Q22 a) to r) p. 169
Note(62) Fractional exponents may be written in root form as they are
equivalent.
Eg. 41/2 = 2
Evaluate on calculator 4^(1/2) = 2
4 = 2
Evaluate on calculator
The two powers above are equivalent so that we can say:
41/2 = 4
Eg. Write each in root form and check with calculator
91/2 = 9 = 3
161/2 = 16 = 4
251/2 = 25 = 5
In general: b1/ n = n b
Note(63) Cube roots appear in the form b 1/3 and the interpretation of this is
that we are looking for the 3rd root of the base.
E.g. Find the cube roots of each of the following:
Meaning:
1/3
3
8
=
8 = Find 3 identical factors that when multiplied, give 8.
(2)(2)(2) = 8. The answer is 2 so that 38 = 2
271/3
=
3
27 = Find 3 identical factors that when multiplied, give 27.
(3)(3)(3) = 27. The answer is 3 so that 327 = 3
64 = Find 3 identical factors that when multiplied give 64
(4)(4)(4) = 64. The answer is 4 so that 364 = 4
Math 112 Exponential Growth
Note(64) The fourth root is similar to the 3rd root and is exemplified below:
641/3
=
3
16 = Find 4 identical factors that when multiplied give 16
(2)(2)(2)(2) = 16. The answer is 2 so that 416 = 2.
Note(65) Rational exponents that have numerators and denominators may be
written in radical form as well.
161/4 =
4
E.g. Evaluate the following by calculator:
8 2/3 = 4
(38)2 = 4
8
2/3
Enter as 8 ^ (2/3)
Evaluate 38 by hand first then square the answer.
From the above we discover that the two are equivalent so that:
= (38)2
Write the following in radical form then evaluate:
16
= (416)3 Evaluate inside bracket to get 2 so (2)3 = 8
32 4/5 = (532)4 Evaluate inside bracket to get 2 so (2)4 = 16
64 2/3 = (364)2 Evaluate inside bracket to get 4 so (4)2 = 16
3/4
Note(66) Negative rational exponents require that the reciprocal be taken
first, then the base rewritten in radical form.
E.g. Evaluate the following without technology
16 -3/4 = 1
16 3/4
= 1
(416)3
= 1
(2)3
= 1
8
In general: b -m/n
Take the reciprocal of the base 16
Write the denominator in radical form
Take the fourth root of 16 inside the brackets.
Cube the base 2.
=
1
(  b )m
Asn(E8) Q 1,3,5 p. 164 Asn(E9) Q 7 to 10 p.164,165 Asn(E10) Q 54 p.123
Test # 5
Math 112 Exponential Growth
n
Note(67) The vertical stretch of an exponential function is determined by
the value of “a” in the general exponential function y = a(b)x + c
e.g. y = 3(2)x
a = 3 The vertical stretch is 3
This may be written in transformation form as 1/3y = (2)x
Note(68) The vertical translation of an exponential function is determined
by the value of “c” in the general exponential function y = a(b)x + c
e.g. y = 3(2)x + 1 c = 1 The vertical translation is 1 up
This may be written in transformation form as 1/3( y -1) = (2)x
Special note: The imaginary line from which the function starts is y = 1.
this line is called an asymptote.
Further note: The y- intercept is found 3 units up from this asymptote so
that the y-intercept is given by adding a to c. So 3+1= 4 is the y-intercept.
y-intercept = “a + c”
the asymptote has an equation defined by y = “c”
Math 112 Exponential Growth
Note(69) The horizontal stretch of an exponential function is determined
by the incremental change (factor) acting upon the x variable.
For the exponential function y =a(b)x/x + c x is the incremental change
with the variable x.
e.g. y = (2)x/3
x = 3 The horizontal stretch upon x is 3 units.
Comparing y = (2)x to y = (2)x/3 we can see the graph is stretched
horizontally by a factor of “3” units.
X
y1
y2
0
1
1
(x,y1)
(1,2)
(2,4)
1
2
1.26
(x,y2)
(3,2)
(6,4)
3
2
4
1.58
3
8
2
4
16
2.52
x is stretched by a factor of 3
x is stretched by a factor of 3
5
32
3.17
6
64
4
Math 112 Exponential Growth
Note(70) The horizontal translation of an exponential function is
determined by the constant associated with the variable x.
For the exponential function given by y = a(b)x + d + c the horizontal
translation is given by the variable “d” .
e.g. y = 1(2)x -3
d= -3 The horizontal translation is 3 units to the right.
Comparing the tables of values for y = 1(2)x and y = 1(2)x-3 we can see that
each point of the graph is translated horizontally 3 units right.
x
y1
y2
0
1
.1251
(x,y1)
(0,1)
(1,2)
(2,4)
+3
(x,y2)
(3,1)
(4,2)
(5,4)
1
2
.25
2
4
.5
3
8
1
4
16
2
x is translated 3 units right
x is translated 3 units right
x is translated 3 units right
5
32
4
6
64
8
Math 112 Exponential Growth
Note(71) The reflection about the x-axis is given by the sign in front of the
variable “a”.
e.g. y = -2(3)x + 4
The graph is a reflection on the x-axis.
y = -2(3)x + 4 is the mirror image of the graph defined by…..
y = 2(3)x + 4 and the image line is the asymptote defined by y = 4
Note(72) The reflection about the y-axis is given by the sign in front of the
variable x.
e.g. y = 2(3)-x + 4
The graph is a reflection on the y-axis
The function may be rewritten as y = 2(1/3)x + 4 and from this we can see
that it is a decay curve. We conclude that the mirror image of a growth
curve around the y-axis is a decay curve and vice versa.
Math 112 Exponential Growth
Note(73) Vertical Stretch Questions. When c= 0 in the expression y = a(b)x
+ c both the vertical stretch and the y-intercept are given by the value of “a”
Asn(F1) Investigation #5 A , Q 1 to 5 p. 128 Q 7 p. 129 Q 9 to 12 p.
129,130.
An exponential function can be generated from a table of values or
from the information given in a question.
Table of values: Look for the y intercept in the table or the “initial” or
starting value in the question to find “a”. Divide successive terms to find the
value of “b”.
Asn(F2) Q 19 to 23 p. 132,133 Advanced Q 24,25 p.133,134
Math 112 Exponential Growth
Note(74) Horizontal Stretch Questions. For exponential functions, it is often
common to ask “When will the value of a treasured object double or triple in
price?”
E.g. The exponential function y = 1.50(1.15)x generates the following table:
x1
y
0
1.5
1
1.73
2
3
4
6
7
8
9
5
10
1.98 2.28 2.62 3.00 3.47 3.99 4.59 5.27 6.00
From this table we observe that the initial cost of $1.50 for a treasured
toy (e.g. model car) doubles every 5 years. We may redo the chart in
increments of 5 years as follows: (We can obtain this same result on the
graphing calculator (set Δtbl = 5)
x1
y
0
1.5
5
3.00
10
6.00
From this new table we can obtain values for “a” and “b” quite
readily. “a” = 1.50 the starting value or y-intercept. “b” = 2 the common
ratio obtained by dividing successive terms. Putting this all together we
come up with an equivalent exponential function given by:
y = 1.50(2)x/5
where: a = 1.50 b = 2 and x = 5
Comparing we see that the two functions are equivalent so that:
1.50(2)x/5 = 1.50(1.15)x
Asn(F3) Q 26 to 30 p. 135 to 136
Asn(F4) Q 31 to 37 p. 136 to 138
Advanced Q 38 p. 138
Math 112 Exponential Growth
Note(75) Vertical Translation Questions. When c  0 in the expression
y=a(b)x/x + c the graph is vertically translated up or down by the value of
“c”. The asymptote of an exponential function is given by the letter c in the
general expression:
y = a*bx +c
c = location of the horizontal asymptote
E.g. For the function y = 3(2)x + 1
a=3
b=2
c=1
From the graphing calculator the graph appears as follows:
+4(0,4) the y intercept is 4 not 3
+1
y = +1 equation of horizontal asymptote
x-axis
We observe that the entire graph is vertically translated upward by 1
unit. The y-intercept is “4” not “3” so that in general the y-intercept is given
by: a + c
example: a = 3 and c = 1 for the function y = 3(2)x + 1 so that:
a+c=
3+1=4
Asn(F5) Q 43 to 46 p. 140,141 Advanced Q 1,2,3,4 p. 144; Q 6 to 16 p.
149,150
Transformation form of an exponential function is given by:
1/a(y – k) = b^ x/x -h
Where:
VS = a
HS = x
VT = +k
HT = +h
Math 112 Exponential Growth
Note(76) Exponential equations whose bases are equivalent may be solved
by equating the exponents.
Eg.#1 Solve for x
23 = 2x
Bases are equivalent so exponents must be equivalent
3=x
E.g.#2 Solve for x
34 = 32x
Bases are equivalent so exponents must be equivalent
4 = 2x
4 = 2x
Divide by 2 then cancel or reduce right side.
2 2
2 = x
E.g. #3 Solve for x
52 = 5 x/5
Bases are equivalent so exponents are equivalent
2 = x
5
52 = 5 x Multiplication by 5 then cancel or reduce right side
5
10 = x
Note(77) Exponential equations whose bases are not equivalent need to have
their bases reduced or rewritten so that they are equal.
E.g. #4 Solve for x
8 = 2x
Rewrite 8 as base 2
3
x
2 =2
Equate exponents
3=x
E.g. #5 Solve for x
16 = 1 x
Rewrite 16 as base 2
2
4
2 = 1 x
Rewrite ½ as base 2 with a negative exponent
2
4
2 = (2-1)x Carry –1 through brackets to outside by multiplication
24 = 2–x
Equate the exponents
4 = -x
Write the opposite (Flip tiles both sides)
-4 = x
Math 112 Exponential Growth
E.g. #6 Solve for x
1 = (9)x
Write 27 as base 3
27
1 = (9)x
Write fraction as base with negative exponent
3
3
3-3 = (9)x
Write 9 as base 3
-3
2 x
3 = (3 )
Multiply exponents 2 and x
-3
2x
3 =3
Equate exponents
-3 = 2x
Divide both sides by 2
-3 = 2x
Cancel or reduce right side
2 2
-3 = x
2
Note(78) Exponential equations may involve the variable with more than
one base. We follow the normal steps of equation solving.
E.g. Solve for x
22x +1= 2x+3
Equate exponents
2x+1 = x +3
Move variable to left. Subtract x
2x-x +1 = x –x +3 Simplify
x+1=3
Move constant to right. Subtract 1
x +1 –1 = 3 –1
Simplify
x=2
Note(79) Exponential equations in the form y = a*bx + c require several
steps prior to equating the exponents.
E.g. Solve for x
70 = 12(6)-x+2 – 2
Isolate base 6 by adding 2 to both sides
-x+2
70 +2 = 12(6) – 2 +2
Simplify
-x+2
72 = 12(6)
Divide by 12
-x+2
72 = 12(6)
Cancel or reduce
12
12
1
6 = (6)-x+2
Equate exponents
1 = -x +2
Subtract 2 from both sides
-1 = -x+2
Write the opposite or flip both sides
1 =x
Asn(F6) Q 12,13,14,16 p. 160
Asn(F7) Q 8,9,15,17,18,10,11 p. 158 to 161 Q. 19 p. 161
Math 112 Exponential Growth
Note(80) Unknown bases for exponential equations can be solved by
multiplying by the reciprocal of the exponent.
E.g. Solve for a
a3/2 = 64
a 3/22/3 = 64 2/3
a = (362)2
a = (4)2
a = 16
Multiply both bases by the reciprocal 2/3
Cancel or reduce exponents with base a
Take cube root of 64
Square base 4
E.g. Solve for b
b –2/3 = 16
Multiply both bases by the reciprocal –3/2
–2/3-3/2
–3/2
b
= 16
Cancel or reduce exponents with base b
–3/2
b = 16
Take reciprocal of base 16
b=
1
Take square root of base 16
2
3
( 16)
b= 1
Cube base 4
3
(4)
b= 1
64
Math 112 Exponential Growth
Note(81) Radioactive decay is a special case involving the exponential
function. The language associated with radioactive decay requires
interpretation.
E.g. The half-life of radium is 1620 years. The initial mass of the radium is
10 g. (The half-life of radium is the time it takes to lose ½ of its mass). The
common ratio is therefore ½ or the factor, which is used for the decay curve.
a) How much radium will be present in 2500 years?
a = 10 the initial mass
b=½
Δx = 1620 the change in years for each half-life
x = 2500 (time)
Substitution:
y = a*b x/Δx
y = 10*(1/2)x/1620
y = 10*(1/2)2500/1620
y = 3.4 g
Substitute x = 2500
Evaluate on calculator
3.4 grams of the original 10 grams
will be present in 2500 years.
b) When will the radium be 1/5 of its original mass?
1/5 of the original mass is 1/5 10 g = 2 grams. We substitute into the
equation in place of y and solve for x.
y = a*b xΔx
2 = 10*(1/2)x/1620
Step 1: Enter right side into the graphing calculator after \y1 =
Step 2: 2nd tableset. Set: tblstart to 0 and Δtbl to 1000 since Δtbl is 1620
Step 3: Scroll down until the y value is approximately 2
Step 4: Reset Δtbl to 10 and scroll to find x
x = 3760
Math 112 Exponential Growth
Note(82) The range of exponential functions are limited by the asymptote of
the function so that:
R = {y y > c, y R}
where c comes from y = a*bx +c
E.g. Find the range and domain of these functions:
y = 3x –2
y = 2x +1
+1
-2
D = {x x  R}
R = {y y >1, y R}
D = {x x R}
R = {y y > -2, y  R}
+3
D = {x x ≥ 0,x  R}
R = {y y ≥+3, y R}
Asn(F8) Advanced Q 22,23,24,26,27 p. 153 to 155
Remember to write in full transformation form before identifying
transformations. ½(y-3) = (2) –x/4 -2 =
½(y-3) = (2) -1/4(x +8)
Test #6
Math 112 Exponential Growth
Note(83) An inverse function is a function that when applied to the Range
values produced by the original function returns the original Domain values
from which they came.. The inverse function returns the original values of
the initial function.
e.g.
original function:
y = x2
let x = 5
2
y = (5)
y = 25so y = 25
inverse function:
x = y
let y = 25
x = 25
x=5
so x = 5
We have the original value back of x = 5
Because the final value 5 is the same as the initial value 5 we say that
the function x = y is the inverse function of the original function y = x2
Note(84) The inverse function of an exponential function is the logarithmic
function.
e.g.
original function:
y = 10x
let x = 3
3
y = 10
y = 1000
so y = 1000
inverse function:
x = log10(y) let y = 1000
x = log10(1000)
x=3
so x = 3
Because the final value 3 is the same as the initial value 3 we say that
the function x = log10(y) is the inverse of the original function y = 10x.
Asn(G1) Q 1 to 10 p. 172 to 174
Math 112 Exponential Growth
Note(85) Inverse functions are symmetrical about the line y = x. A table of
values for the function y = 10x and its inverse y = Log2(x) are given below:
y = 10
x
x
1
2
3
y
10
100
1000
Inverse (switch variables)
x = 10 y or y = log 10 (x)
x
10
100
1000
y
1
2
3
(Show that the y values in second table can also be found by applying the
function y = log 10 (x) to x values in second table. This will prove that x =
10 y can be rewritten as y = log 10 (x ) )
+8
+7
+6
+5
+4
+3
+2
+1
y = 10x (original function)
y = x (axis of symmetry)
x = 10 y can be written as…
y = log10 (x) (inverse function)
1 2 3 4 5 6 7 8
Note: The above graph can be illustrated on the graphing calculator.
Note(86) From the above diagram and tables we conclude that functions in
the form 10y = x can be written in the form log10 (x) = y. A “proof” is given
below by evaluation on the calculator.
E.g. if 103 = 1000 then log10(1000) = 3 (evaluate on calculator)
if 104 = 10000 then log10(10000) = 4
if 10y = x
then log10(x) = y
e
In general: if b = a then logb(a) = e Example: if 23 = 8 then log28 = 3
Asn(G2) Q 11 to 15 p. 174,175 Advanced Q 16,17 p. 175
Math 112 Exponential Growth
Note(87) Logarithm of a product
If we enter the following on your calculator we obtain the result to the right.
log(23) = .778
log(2) + log(3) = .788
We generalize as follows: (Other examples may be done)
log(23) = log(2) + log(3)
In general:
log(mn) = log(m) + log(n)
Note(88) Logarithm of quotient
If we enter the following on your calculator we obtain the result to the right.
log(42) = .301
log(4) - log(2) = .301
We generalize as follows: (Other examples may be done)
log(42) = log(4) - log(2)
In general:
log(mn) = log(m) - log(n)
Note(89) Logarithm of a power
If we enter the following on your calculator we obtain the result to the right.
log(42) = 1.204
2log(4) = 1.204
We generalize as follows: (Other examples may be done)
log(42) = 2log(4)
In general:
log(mn) = n*log(m)
Asn(G3) Q 1 to 3 p. 177,178 Advanced Q 7,8 p. 178
Math 112 Exponential Growth
Note(90) Solving exponential equations using logarithms.
E.g. Solve for x
9x = 27
log10(9x) = log10(27)
x log10(9) = log10(27)
x log10(9) = log10(27)
log10(9) log10(9)
x = (1.4314)
(.9542)
x = 1.5
Apply log10 to both sides
Apply power log rule to left side only
Divide both sides by log10(9)
Evaluate
Therefore we can shorten the original equation in a few steps as follows:
9x = 27
x = log10(27)
log10(9)
x = 1.5
In general:
If be = a then e = log10(a)
log10(b)
Asn(G3) cont’d Q 4,5,6 p. 177,178
Math 112 Exponential Growth
Note(91) Log identities can be derived by equating two equal expressions
for an exponential equation.
E.g. 9x = 27 Solving for x we have:
9x = 27 In logarithmic form we have:
x = log10(27)
log10 (9)
x = log9(27)
Since we have two expressions for x, we may equate the two as:
log9(27) = log10(27) = 1.5
log10(9)
In general:
Logb(a) = log10(a)
log10(b)
Special Note: The significance of the above rule cannot be overemphasized
as it indicates that if a log is not written in base 10 it can be rewritten in
base 10 and evaluated on any calculator.
SUMMARY OF LOG RULES AND IDENTITIES
1. If be = a then logb(a) = e
Exponent form to Log form
2. logb(a) = log10(a)
log10(b)
e
3. If b = a then e = log10 (a)
log10(b)
4. log(mn) = log(m) + log(n)
Calculator log form
5. log(mn) = log(m) - log(n)
Log of a quotient
6. log(mn) = n*log(m)
Log of a power
Exponent form to calculator form
Log of a product
Asn(G4) Q 9 to 11 p. 180,181 Q 12 in reverse order.
Math 112 Exponential Growth
Note(92) Problem solving requires interpretation and proper application of
logarithm identities.
E.g#1. A bacteria colony’s original population is 1000. It doubles every
hour. When will it reach a population of 5000?
y = a*bx/Δx
a = 1000
original population
b=2
doubles
Δx = 1
every hour
y = 5000
future population
Substituting we have:
5000 = 1000(2)x
Solve for x
x
5000 = 1000(2)
Isolate base 2 by dividing both sides by 1000
1000 1000
5 = (2)x
x = log10(5)
log10(2)
x = 2.3 hours
Rewrite as log base 10
Evaluate right side
E.g.#2 A rabbit population triples every 4 years. If there were 435 rabbits
to start with, when will it reach a population of 6781 rabbits?
y = a*bx/Δx
a = 435
to start with
b=3
triples
Δx = 4
every 4 years
y = 6781
future population
Substituting we have:
6781 = 435(3)x/4
Divide by 435 both sides
x/4
6781 = 435(3)
Reduce both sides
435 435
15.59 = (3)x/4
Rewrite as log base 10
x/4 = log10(15.59)
Evaluate right side
log10(3)
x/4 = 2.5
Multiply both sides by 4
x = 10 years
Asn(G5) Handout Q 1,3,4,5 Advanced 15,16, 18 to 26 p. 182,183
Test #7
Math 112 Trigonometry
Book 2 Mathematical Modelling (old version-blue text) required
Note (90) Trigonometry involves the study of angles and sides in relation to
triangles in particular. A review of terms is essential
1. complementary angles- angles that add up to 900.
2. supplementary angles- angles that add up to 1800.
3. acute angle- an angle that is less than 900.
4. obtuse angle- an angle that is between 900 and 1800.
5. right angle- a angle with measure exactly 900.
6. straight angle- an angle with measure exactly 1800.
7. isosceles triangle- a triangle with two sides equal.
8. equilateral triangle- a triangle with all sides equal.
9. right angle triangle- a triangle with one angle of measure 900.
10. acute triangle- a triangle with all angles less than 900
11. obtuse triangle- a triangle with one angle greater than 900
12. oblique triangle- any type of triangle that does not have a 900
(both the acute and obtuse triangles are oblique).
Note(91) Right angle triangles are special triangles in that the sides are
related in a special manner.
Pythagorean Property:
c2 = a2 + b2
A
Or
b
C
c
a
B
H2 = S12 + S22
Math 112 Trigonometry
Note(92) Right angle triangles are special triangles in that the angles are
related in a special manner.
 sum = 1800
A
A +B +C= 1800.
b
Because C is 900 we have a special
relationship between A and B as follows:
c
A +B = 900
C
a
B
Note(93) Some triangles cannot be solved using the Pythagorean Theorem
or the angle sum of a triangle. Primary trigonometric ratios can be used
instead.
E.G. solve for x in the figure below:
A
12
350
B
x
C
In such a case, trig ratios can be used for solving for the unknown.
The primary trig ratios are:
Sin() = Opp
Hyp
Cos() = Adj
Hyp
Tan() = Opp
Adj
We use an acronym for assisting in the recall
of these ratios:
SOH
CAH
TOA
Math 112 Trigonometry
Note(94) We may use the primary trig ratios in solving for the sides of a
right triangle as illustrated below:
E.g. Solve for x in the figure below:
12
x
300
Sin() = Opp
Hyp
0
Sin(30 ) = x
12
0
12Sin(30 ) = x
1) Shade the angle given
2) Using the angle as a point of reference
label the side x as opposite and 12 as the
hypotenuse.
3) Choose the trig ratio that fits the situation.
Write the primary trig ratio that matches
Substitute for  and Hyp
Multiply both sides by 12. (cross multiply)
x=6
E.g. Solve for x in the figure below:
x
10
450
Sin() = Opp
Hyp
0
Sin(45 ) = 10
x
0
x * sin(45 ) = 10
x * (.707106..) = 10
x = 14.1 cm
Write the primary trig ratio that matches
Substitute for  and Opp.
Cross multiply to move x to left side.
Divide both sides by (.707106…)
Asn(H1) Question involving primary trig ratio and sides. (In class)
Math 112 Trigonometry
Note(95) We may use the primary trig ratios in solving for the angles of a
right triangle as illustrated below:
E.g. Solve for 
1) Shade the indicated angle
2) Label the sides relative to the shaded angle
3) Choose the trig ratio that matches the situation
4

5
Tan() = Opp
Adj
Tan() = 4
5
Tan() = .8000
Write the primary trig ratio that matches
Evaluate the right side to obtain a decimal.
Use 2nd function tan-1 to find angle 
 = tan-1 (.8000)
 = 38.70
Asn(H2) Questions involving primary trig ratios and angles. (In class)
Math 112 Trigonometry
Note(96) There are 6 parts that make up a right triangle. (3 sides and 3
angles). We may solve for the unknown parts in a variety of ways:
Case #1 the triangle has 2 angles given and one side.
Suggestion: Start by solving for the 3rd angle

Tools to use for solving triangles:
Tool #1 Angle sum of triangle
Tool #2 Trig ratios
Tool #3 Pythagorean Theorem
x
2
350
y
Tool#1  sum of triangle:
 + 350 + 900 = 1800.
 = 550
Tool#2 Trig ratios:
Tan(350)
= Opp
Adj
0
Tan(35 )
= 2
y
Write the trig ratio that matches the question
Substitute
y*tan(350) = 2
y
= 2/tan(350)
y
= 2.85
Multiply both sides by y. (Cross multiply)
Divide both sides by tan(350)
Evaluate
Tool#2 Trig ratios:
Sin(350)
= Opp
Hyp
Write the trig ratio that matches the question
Sin(350)
Substitute
= 2
x
0
x*sin(35 ) = 2
x
= 2/sin(350)
x
= 3.48
Multiply both sides by x. (Cross multiply)
Divide both sides by sin(350)
Evaluate
Math 112 Trigonometry
Case #2 the triangle has 2 sides given and one angle.
Suggestion: Start by solving for the 3rd side

Tools to use for solving triangles:
Tool #1 Angle sum of triangle
Tool #2 Trig ratios
Tool #3 Pythagorean Theorem
14
x

5
Tool#3 Pythagorean Theorem:
c2
= a2 + b2
(14)2 = (x)2 + (5)2
196 = x2 + 25
x2
= 196-25
2
x
= 171
x
= 171
Tool#2 Trig ratios:
Cos()
= Adj
Hyp
Cos()
= 5
14
Cos()
= .35714…

= cos-1(.35714..)

= 690
Tool#2 Trig ratios:
Sin()
= Opp
Hyp
Sin()
= 5
14
Sin()
= .35714…

= sin-1(.35714..)

= 210
Substitute for c and b
Square terms
Isolate x
Simplify
Take sq rt of 171
Write the trig ratio that matches the question
Substitute for adj and hyp
Evaluate right hand side
Use 2nd function cos-1
Write the trig ratio that matches the question
Substitute for opp and hyp
Evaluate right hand side
Use 2nd function sin-1
Asn(H3) Handout to do at home.
Math 112 Trigonometry
Note(97) Application of trigonometry to real life situations requires an
understanding of common terms.
Angle of elevation- the angle between the ground and the line of sight
looking upwards.
Angle of depression- the angle between a line horizontal to the ground and
the line of sight looking downwards
Special note: the angle of depression equals the angle of elevation.
E.g. The angle of depression from the top of a cliff to a sailboat in the bay
below is 250. If the sailboat is 100 m from the shore how high is the cliff.
250
x
angle of depression 250
angle of elevation is also 250
angle of elevation
sailboat
100 m
The angle of elevation is equal to the angle of depression since they
are alternate angles between parallel lines.
Tan(250) = x
100
x = 46.6 m
Math 112 Trigonometry
Note(98) The area of a triangle is given by the formula: A = ½ b*h.
The area may also be found by using the formula A = ½ bcSin(A) where
 A is the included angle.
E.g. Find the area of the Δ ABC given below:
A
c = 5.4
h
B
18.40
D
C
a = 9.3
The area of Δ ABC can be found using the formula:
A = ½ b*h
A = ½ 9.3*h
Using the triangle Δ ABD and sine trig ratio we find h as follows:
Sin(18.4) = Opp = h
Hyp 5.4
h = 5.4 Sin(18.4)
h = 1.7
Substituting h = 1.7 into the area formula above we have the area as:
A= ½ (9.3)(1.7)
A = 7.93
The area of Δ ABC can also be found using the formula:
A = ½ ac*Sin(B)
= ½ (9.3 )(5.4 ) sin(18.4)
A = 7.93
Conclusion: The area formula using the Sine trig ratio is given by:
A = ½ acSin(B) where  B is the included angle between sides a and c.
Asn(H4) Handout on area problems.
Math 112 Trigonometry
Note(99) The sine law can be used for solving oblique triangles. We solve
for either an unknown side or an unknown angle.
E.g. #1 Solve for side “a” in the figure given below.
A
400
b = 12
350
B
a=?
C
The sine law for sides is given by:
a =
b
=
c
Sin(A)
Sin(B)
Sin(C)
Substituting we have:
a =
Sin(40)
(12 ) =
c
Sin(35)
Sin(C)
Evaluate:
a =
(12 )
Sin(40)
Sin(35)
a
= 12 Sin(40)
Sin(35)
a
= 13.4
Since c and Sin(C ) are not known
we do not use the last ratio for
evaluation.
Cross multiply to isolate side “a”
Evaluate
Math 112 Trigonometry
E.g. #2 Solve for angle “B” in the figure given below.
A
950
c = 16
b = 13

B
a = 14
C
The sine law for angles is given by:
Sin(A) = Sin(B)
a
b
= Sin(C)
c
The ratios are reciprocated
for easier evaluation.
Substituting we have:
Sin(95) = Sin(B) = Sin(C)
(14)
(13)
(16)
Since we are not solving for C we
do not use the last ratio to evaluate.
Evaluate:
Sin(95) = Sin(B)
(14)
(13)
Cross multiply to isolate Sin(B)
13Sin(95) = Sin(B)
14
Evaluate left side
0.9250 = Sin(B)
Apply 2nd function Sin-1
Sin-1(0.9250) = B
Evaluate
 B = 67.70
Asn(H5) Q 1 to 3 p. 259
Math 112 Trigonometry
Special Note: We use the Sine Law if the angles and sides are opposites.
We look for the “butterfly- pattern” as illustrated below;
A
950
c = 16
b = 13

B
a = 14
C
Special Note: Occasionally a required angle is missing in which case we can
utilize the angle sum of a triangle to solve for the missing angle.
Solve for “b”
A
300
c = 12
b=?
500
?
B
Missing angle ( sum gives 1000)
C
We cannot use the Sine Law until we find the size of the unknown (?). To
do this we use the angle sum of a triangle:
 sum = 1800 so:
A +B +C = 1800
30 + 50 + C = 180
 C = 100
We can now use the Sine Law to solve for side b.
Asn(H6) Q 4a) b) 5 p. 259
Asn(H7) Handout Q 5,6,7 Sine Law Problems
Math 112 Trigonometry
Note(100) The cosine law can also be used for solving oblique triangles
where the sine law is not helpful.
E.g.#1 Solve for side “a”
C
a=?
B
b =12
620 A
c = 10
In the case above , we are lacking information that is required for the
sine law, that is a second angle.
We use another law designated as the cosine law given by:
a2 = b2 + c2 –2bcCos(A)
b2 = a2 + c2 –2acCos(B)
c2 = a2 + b2 –2abCos(C)
where A is included between two sides a,b
where B is included between two sides a,c
where C is included between two sides b,c
Using the first of these from above we have:
a2 = b2 + c2 –2bcCos(A)
Substitute
2
2
2
a = (12) + (10) –2(12)(10)Cos(62) Enter into graphing calculator
a2 = 131.33
a = 131.33
Apply square root to both sides
a = 11.5
Math 112 Trigonometry
Note(100) The cosine law cont’d
E.g.#2 Solve for angle “A”
The following rearranged cosine laws are useful in solving for angles:
1) Cos(A) = b2 + c2 - a2
2bc
2
2) Cos(B) = a + c2 - b2
2ac
2
3) Cos(C) = a + b2 - c2
2ab
C
b = 15
A
a = 12

B
c = 16
Cos(A) = b2 + c2 - a2
2bc
Cos(A) = (15)2 + (16)2 - (12)2
2(15)(16)
Cos(A) = .70208….
A = cos-1(.70208….)
A = 45.40
Substitute
Evaluate
Apply second function cosine
Asn(H8) Q 8,9,10,12,13,15 p. 261,262 Cosine Law problems
Asn(H9) Handout cosine law practice sheet. (Optional)
Asn(H10) Handout- problem solving. Sine, Cosine and Area questions.
Advanced Q 11, 14 p. 262 Q 21 p 264 q.24, 25. p. 265
Math 112 Trigonometry
Note(101) The ambiguous case for the sine law.
Some geometric constructions allow for 2 possible constructions from
the given data. Two correct results are possible.
E.g. Construct 2 unique triangles using the following data:
A = 270 b = 12.3 cm
a = 9.6 cm
Construction #1
C
b = 12.3
a = 9.6
A 270
B
In the above diagram B = 35.60 if measured with a protractor.
 B is acute.
Construciton #2
C
b = 12.3
a = 9.6
270
A
B
In the above diagram B = 144.40 if measured with a protractor.
B is obtuse.
Since both diagrams are correct then there are 2 possible answers for
B as follows:
B = 35.60
B = 144.40
It is noteworthy that the measures 35.60 and 144.40 are supplementary.
Math 112 Trigonometry
Note(102) For the sine function we notice that the sine ratios for
supplementary angles are identical.
E.g. Sine(35.6) = .5821229 = Sine(144.4)
This is true for all supplementary angles:
Sine(10)
= .1736..
= Sine(170)
Sine(30)
= .5000
= Sine(150)
In general:
Sine() = Sine(180 - )
Note(103) If given an obtuse angled triangle, it is possible to obtain a result
that does not match the diagram. In such a case we suggest that the
supplementary angle is the correct result.
E.g. Solve for B in the figure below:
b = 14
a = 10
30
A
Solve for B:
0
B
Sin(B) = Sin(A)
b
a
Sin(B) = 14Sin(30)
10
Sin(B) = .7000
B = sin-1(.7000)
B = 44.40 Because this does not match the appearance of the
diagram we suggest that the supplement to this angle is the solution so that:
B = (1800 - 44.40)= 135.60
Asn(H11) Q 19,17 p. 264
Test # 8