Download Chapter 9 Quadratic Equations and Functions

Unit 2
Quadratic Equations and Functions
• Solve quadratic equations by finding square
roots and quadratic formula
• Graph quadratic equations
• Use the discriminant
• Solve using x-intercepts
• Choose a model of best fit
5.1 Graphing quadratic functions
• Sketch the graph of a quadratic function
• Use models in applied problems
Quadratic function
Written in 3 forms
f(x) = ax2 + bx + c, where a ≠ 0.
Standard Form
f(x) = a(x-h)2 + k where a ≠ 0.
Vertex Form
f(x) = a(x - x1)(x - x2) where a ≠ 0.
Factored Form
To find the vertex from standard form:
b
1. x value 
2a
2. substitute x value into equation and solve for y value
To find the vertex from vertex form:
Vertex: (h, k)
To graph a quadratic function in standard form
1. Find the x-coordinate of the vertex
2. Make a table of values, using x-values to the left and right
of the vertex. Find y intercepts and use symmetry to help.
3. Plot the points and connect them with a smooth curve
The graph of a quadratic function is a parabola
Axis of symmetry
vertex
a is positive
a is negative
b
8
x coordinate of vertex :

1
2a 2(4)
y intercept of (0, 1)
f (1)  3
so...verte x is at (1,3)
and axis of symmetry
x 1
x
-1
0
y
13
1
1
2
3
-3
1
13
Use symmetry and a t-chart
to fill additional points
b
8
x coordinate of vertex :

2
2a 2(2)
y intercept of (0, - 3)
f (2)  5
so...verte x is at (2,5)
and axis of symmetry
x2
x
-1
0
y
-13
-3
2
4
5
5
-3
-13
Use symmetry and a t-chart
to fill additional points
Graphing in Vertex Form
1. Place the vertex in the table.
2. Place several x values above and below the vertex in table.
3. Put the x-values in formula and solve for y to get
additional points.
4. Continue with this until you form a graph.
Graph f(x) = 2(x - 3)2 - 4
x
1
2
3
y
4
-2
-4
4
5
-2
4
Graph f(x) = -2(x + 1)2 - 2
x
-3
-2
-1
0
1
y
-10
-4
-2
-4
-10
Graphing in Factored Form
1. Set each of the parenthesis equal to 0 and solve. These are your
x-intercepts. This gives you 2 points, plot them now.
2. Vertex - to find (x1 + x2)/2 gives you the x value of the vertex
3. Put the x-value in formula and solve for the y value of the vertex
4. Write this in table and graph with 3 points.
Graph: f(x) = (x - 6)(x - 2)
x
0
2
4
6
8
y
12
0
-4
0
12
Graph: f(x) = -(x + 1)(x - 2)
x
-3
y
10
-1
.5
0
2.25
2
4
0
10
5.2 Solving quadratic equations
graphically
• Solve quadratics
using x-intercepts
• Use quadratics in
applied problems
Standard quadratic function:
y  ax 2  bx  c
Opens up
y  ax  bx  c
2
Basic graphs
Opens down
max
y-int:
vertex
y-int:
x-int:
x-int:
min
vertex: (3,-3)
x-int: (1,0) & (5,0)
y-int: (0,5)
domain: all real numbers (, )
range: [3, )
vertex: (2,9)
x-int: (-1,0) & (5,0)
y-int: (0,4)
domain: all real numbers
(, )
range: [,8)
Quadratic Equations are quadratic functions set to a
specific value.
Solutions of quadratic equation are called a roots.
These are the x-intercepts when you graph the
function.
It can have 1 real solution, 2 real solutions or none.
2 solutions
0 solutions
1 solution
Find the x and y intercepts & the max/min f(x)= 2x2 - 12x + 7
y-intercept: (0,7)
x-intercepts,
zeros, roots:
(.655,0) &
(5.345,0)
Domain : (, )
vertex and min:
(3,-11)
Range : (11, )
Find the x and y intercepts & the max/min f(x)= x2 - 8x + 2
y-intercept: (0,2)
x-intercepts,
zeros, roots:
(-8.243,0) &
(.243,0)
Domain : (, )
vertex and max:
(-4,18)
Range : (,18)
Find the x and y intercepts & the max/min f(x)= -2(x – 3)2 - 1
y-intercept: (0,-19)
x-intercepts,
zeros, roots:
none
Domain : (, )
vertex and max:
(3,-1)
Range : (,1)
State the domain and range
a) f(x) = x2 -2x + 2 Domain :  ,   opens up Range : 1,  
b
2
x coordinate of vertex :

 1  f (1)  1  vertex: (1,1)
2a 2(1)
1
b) 𝑓 𝑥 = − 𝑥 2 + 2𝑥 − 3
3
b
2
vertex :

 3  f (3)  0  vertex: (3,0)
2a
 1
2  
Domain :  ,  
 3
opens down
Range :  ,0 
c) f(x) = 0.25x2 + 2x
b
2
vertex :

 4  f (4)  4  vertex: (-4, -4)
2a 2.25
opens up Range :  4,  
Domain :  ,  
Rate of Change/Applications
• Find the rate of change for a quadratic function.
• Write and solve application problems using quadratic
functions.
Vertical motion problems talk about an
object falling to earth.
h(t )  16t 2  ho
Formula for feet
h(t )  16t 2  vt  ho
h(t )  4.9t 2  h0
Formula for meters
h(t )  4.9t 2  vt  h0
h = height at time t
v = initial velocity
s = starting height
A ball is thrown vertically upward with an initial
velocity of 48 feet per second with an initial height of
8 feet off the ground.
a) Write a function that models this scenario.
h(t )  16t 2  vt  ho h(t )  16t 2  48t  8
b) What is the maximum height of the ball?
44 feet
c) When did this occur?
1.5 seconds
d) State the domain and range for the function.
Domain : (0, 3.16) seconds
Range : (0, 44) feet
An object is launch at 20 meters per second
from a 60 meter tall platform.
a) Write a function that models this
scenario.
h(t )  4.9t 2  vt  ho h(t )  4.9t 2  20t  60
b) How high is the object at 3 seconds?
75.9 feet
c) What is the maximum height of the
ball?
80.4 feet
d) When did this occur?
2.04 seconds
e) State the domain and range for the
function Domain : (0, 6.09) seconds
Range : (0, 80.4) feet
Average rate of change
The Average Rate of Change between x =a & x = b
of a function f(x) is:
f ( x) f  b   f  a 

x
ba
This is equal to the SLOPE of the secant line
between x = a & x = b:
y
Secant
Line
f(x)
x
COOL
Determine the rate of change
on the given intervals
a) Find the rate of change for
x = 1 to x = 2
f (2)  3
f (1)  0
30
m
 3
2 1
b) Find the rate of change for
x = 0 to x = 2
f (2)  3
f (0)  5
m
35
 4
20
c) Which is the greater rate
of change?
Since -3 is further right on the number line than -4,
the rate of change from x = 1 to x = 2 is greater
Given: f ( x)  x 2  x  6
a) Find the rate of change for x = 1 to x = 2
f (2)  0
f (1)  4
0  (4)
m
4
2 1
b) Find the rate of change for x = 0 to x = 3
f (3)  6
f (0)  6
6  (6)
m
4
30
c) Which is the greater rate of change?
They are equal.
Factoring Map
Pull out the GCF and factor.
a) 3x + 9 3(x + 3)
b) 15 x  25 x  5 x(3  5 x)
2
c) x  7 x  x 2 ( x 2  7 )
4
2
d) 9 x  18 x  9 x ( x  2)
3
2
2
e) 2 x  4 x  8 x  2 x( x 2  2 x  4)
3
2
Difference of squares
x  64  ( x  8)( x  8)
2
x  144  ( x  12)( x  12)
2
 (4 x  9)( 4 x  9)
16 x  81
2
100 x  9 y
2
2
 (10 x  3)(10 x  3)
Guess and Check Factoring:
If it is a quadratic with a leading coefficient of one:
• Find two terms which multiply to the third term
and add to the second term
x  8 x  15  ( x  5)( x  3)
2
x  11x  30  ( x  5)( x  6)
2
x  8 x  9  ( x  9)( x  1)
2
x  3x  18
2
 ( x  6)( x  3)
“BOTTOMS UP FACTORING”
1.
Factor out GCF
2. Multiply first and last
numbers
3. Find two numbers that
multiply to the product and
add to the middle term
4. Write two factors with
fractions
5. Reduce fractions
6. Kick denominator to front
(“bottoms up”)
6 x  11x  4
2
8
24
11
3
8 
3

  x   x  
6 
6

4 
1

  x   x  
3 
2

 3x  42x  1
Factor:
5 x  17 x  12
2
60
 12  5
 17
12 
5

 x   x  
5 
5

12 

 x   x  1
5

5x 12x 1
2x  x 1
2
2
2
1
1
2 
1

 x   x  
2 
2

1

x  1 x  
2

x  12 x  1
5.3 Solving equations by factoring
• Solve equations by factoring
Zero-product property: If the product of two
numbers is zero, then one of them must be zero.
If ab  0, then a  0 or b  0
So if x( x  4)  0 then x  0 OR x  4  0
x = 0 OR x = 4
a ) x  x  12
2
x  x  12  0
( x  4)( x  3)  0
2
x  4  0 OR x  3  0
x  4 OR x  3
b) 25 x 2  16
16
2
x 
25
c) 3x  2 x  21  0
2
 63
9
7
2
16
x
25
9 
7

 x   x    0
3 
3

4
x
5
7

x  3 x    0
3

7
x  3 OR x 
3
Factor by grouping
• Learn to factor by
grouping
To factor by grouping:
1) Factor out GCF (if any)
2) Group the first two terms, and the last two
terms. Be sure to group the signs as well
and add them together.
3) Factor the GCF out of both groups
4) Remaining binomial should be the same for
both groups. Factor the common binomial.


x  4 x  4 x  16  x  4 x   4 x  16
3
2
3
2


 x x  4  4x  4  x  4  x  4
2
2
Factor:


b) x 3  6 x 2  2 x  12  x 3  6 x 2   2 x  12 


 x x  6  2 x  6  x 2  2  x  6 
2
c) 4 xy  3 y  20 x  15  4 xy  3 y    20 x  15
 y 4 x  3  54 x  3   y  54 x  3
Factoring Map
Solve.
c) x  4 x  45  0
2
x  9x  5  0
x  9 OR
x5
d)  3 x  10 x  8  0
2
3 x 2  10 x  8  0
 24
2
12
12 
2

 x   x    0
 10
3 
3

2

x  4 x    0
3

x  4 OR
2
x
3
5.4 Complex Numbers
• Use the imaginary unit i to write
complex numbers
• Add, subtract, and multiply complex
numbers
• Use complex conjugates to divide complex
numbers
• Find complex solutions of quadratic
equations
Some equations have no real solutions:
x2  1  0
i  1
i 2  1
3
i  i
So we create a number system with the
imaginary unit i
After the fourth power, the cycle repeats.
i.e.  i  i
5
i 1
4
5  i 5
a)  121 b)  28 c) (3i )(2i ) d) (2i )(4i )(5i )
2
3
 6i
 40i
 i 121  i 28
 11i
 2i 7
e) i  i  1
14
2
 40i
6
f) i  i  i
27
3
CAREFUL!!!
You must convert to standard form before multiplying
5  5 is NOT 25!
  
5 5  i 5 i 5
 i 2 25
 5i  5
2
Note the difference from (5)(5) which does
equal 25
  2 
 i 12 i 2 
g)

 12
 i 2 24  2 6
Complex numbers: The standard form of complex numbers is a + bi,
where a is the real number part and bi is the imaginary part.
3  7i
3 is the real number part
7i is the imaginary number part
Example: Perform the operation
a) (-2 + 5i) + (1 - 7i)
b) (4 + 6i) - (-1 + 2i)
 1 2i
5  4i
18  6i  36i  12i
18  30i  12(1)
2
30  30i
3  6i  12i  24i 2
3  6i  24(1)
27  6i
Complex conjugates: (a  bi ) and (a  bi ) are conjugates
Conjugates are an example of how the product of two
complex numbers can be a real number.
2  3i
4  2i
2  3i  4  2i 



4  2i  4  2i 
2
8  4i  12i  6i
8  6  16i 2  16i
1 4


2



i
16  4
16  4i
20
10 5
3  5i
1  2i
2  i 1 i 
Evaluate:


1 i 1 i 
 2  2i  i  i 2 

 
2

1 i


 2  3i  1 


 11 
1  3i 1 3

  i
2
2 2
Solve each equation
5 x 2  20  0
5 x 2  20
x 2  4
x   4
x  i 4
x  2i
4 x 2  100  0
4 x 2  100
x 2  25
x    25
x  i 25
x  5i
Find the values for x and y that make the equations true
(set real parts equal to each other and the imaginary parts equal
to each other) 3x - 5 + (y - 3)i = 7 + 6i
3x  5  7
3x  12
Real parts
x4
y 3  6
y 9
y 9
(5x + 1) + (3 + 2y)i = 2x - 2 + (y - 6)i
5x 1  2x  2
3x  3
x  1
3 2y  y  6
y  9
imaginary parts
5.5 Completing the square
• Solve a quadratic
equation by
completing the square
• Choose a method for
solving a quadratic
Solve: x 2  5 x  14  0
( x  7)( x  2)  0
x  7  0 OR x  2  0
x  7 OR x  2
Sometimes the quadratic does not factor. We can make it
“factorable” by the process of completing the square.
x 2  16 x  9
Step 4 Factor the left, simplify the right
Step 5 Take the square root of both sides, and solve the equation
x 2  16 x  9
x 2  6 x  25
x2 + 16x + 64 = 9 + 64
 x  8
2
 73
 x  8  
73
x  8  73
x2 - 6x + 9 = -25 + 9
x  3
2
 16
 x  3  
 16
x  3  i 16
x  3 4i
1) The constant must be on the other side of the equals sign
2) The coefficient of the squared term must be divided out so that
it is equal to one.
3x 2  5 x  7  0
 3 x 2  5 x  7 Move constant
5
7
2
x  x
Divide out squared term
3
3
2
2
5
7 5
5
2
x  x    
Complete the square
3
6
3
6
 
 
2
5  109

x  
6
36

Factor left side and simplify right side
5
109
x 
Take the square root of both sides
6
36
 109 5
 109  5
x
 
Simplify and solve
6
6
6
5.6 Quadratic Formula
• Use the quadratic formula to solve a quadratic
equation
• Use vertical motion models to solve applied
problems
Felix the cat jumps off of a rooftop 15 feet high
with an initial upward speed of 3 ft/sec. How
long will it take Felix to hit the ground?
So far we have learned to solve quadratic
equations by taking the square root and
by graphing. Now we will learn to solve
any quadratic equation with the quadratic
formula.
The solutions of the equation ax  bx  c  0 are
2
b  b  4ac
2
x
when a  0 and b  4ac  0
2a
2
Solve:
2
3x  5 x  7  0
2
2x  3  2x
 b  b 2  (4)(a )(c)
x
2a
 5  25  (4)(3)(7)
x
6
 5  25  84
x
6
 5  109
x
6
1 i 5

2
Vertical motion problems talk about an
object falling to earth.
h(t )  16t 2  ho
Formula for feet
h(t )  16t 2  vt  ho
h(t )  4.9t 2  h0
Formula for meters
h(t )  4.9t 2  vt  h0
h = height at time t
v = initial velocity
s = starting height
Back to the cat problem:
Felix the cat jumps off of a rooftop 15 feet high
with an initial upward speed of 3 ft/sec. How
long will it take Felix to hit the ground?
h  16t  vt  ho
2
0  16t  3t  15
2
3  32  4(16)(15)
x
2(16)
3  969
x
32
One answer
approximates
to a negative
time…not
possible…the
other is about
1.07 seconds
In the quadratic formula, the expression
inside the radical is called the discriminant.
b  b2  4ac
x
2a
b 2  4ac is the discriminant and is
used to find the number of solutions
b2  4ac  0 then the equation has two solutions
b2  4ac  0 then the equation has one solution
b2  4ac  0 then the equation has no real solution
The number of x-intercepts is equal to the
number of solutions of an equation
Use the discriminant to tell if each
equation has two solutions, one solution,
or no real solutions
a.) x  2 x  4  0
2
no real solutions
b.)  3 x  5 x  1  0
two solutions
c.)  x  10 x  25  0
one solution
2
2
Ways to solve quadratics:
1) Find square roots
2) Graphing
3) Use the quadratic formula
4) Factoring
5) Complete the square
Method
Can be Used When to Use
Graphing
sometimes
Factoring
sometimes
Square Root
Property
sometimes
Completing the
Square
always
when b is an even
number x + 6x - 14
Quadractic
Formula
always
when other methods
fail or are too tedious
when exact answer
not needed
If c = 0 or factors
easily found
when equation is a
perfect square
VERY COOL
Derive the quadratic formula from the general
equation by completing the square: ax  bx  c  0
2
5.7 Transformations of quadratics
• Find the vertex form
of a quadratic and
graph using
transformations
Let’s look at the basic quadratic function of
yx
2
A quadratic equation
in vertex form
y = a(x - h)2 + k
Horizontal Translation - h
moves to the left - if h is added to x
moves to the right - if h is subtracted from x
Vertical Translation - k
moves up - if k is positive
moves down - if k is negative
a - reflection and dilation
a is positive and opens up or negative and opens down
a > 1 stretched vertically or 0 < a < 1 is compressed vertically
Graph and state the transformation from
yx
2
:
y = (x - 5)2 + 3
Original
yx
2
Go 5 units to the
right, and 3 units up
Graph and state the transformation from
yx
2
:
y = -2(x - 5)2
Original
yx
2
Go 5 units to the right, reflect
over x axis, vertical stretch of 2
A quadratic function is a function that can written in
the general form:
2
y  ax  bx  c
The standard form or vertex form of a quadratic
function is given by:
2
f ( x )  a ( x  h)  k
The vertex is at the point (h, k) and the axis of
symmetry is the vertical line x = h.
Complete the square to write the function in
vertex form.
Write in vertex form

y  x

y  x  4x  6
2
2

4  6  ____
4
 4 x  ___
y  ( x  2)  2
2
Vertex is (-2, 2). Axis of symmetry is x = -2
To move from general form to standard form of an
equation, use the process of completing the
square.
f ( x)  2 x  8 x  7
2
f ( x)  2( x  4 x)  7
2
Write equation
Factor the x-terms
f ( x)  2( x  4 x  4)  7  2(4)
2
f ( x)  2( x  2)  1
2
Complete the
square
Factor and
simplify
The vertex is at the point (-2,-1). Axis of
symmetry is x = -2.
To write from a graph, all that is needed is the
2
vertex and 1 point
y  a ( x  h)  k
1. replace h,k with vertex
2. x and y with the point
3. solve for a
Write an equation in vertex form
vertex (-4,3)
point (-3,6)
y  a ( x  h) 2  k
y  a( x  4) 2  3
6  a(3  4) 2  3
y  3( x  4)  3
2
6  a(1)  3
2
3  a(1)
Write an equation in vertex form y  a( x  h) 2  k
vertex (5,4)
2
y  a( x  5)  4
point (6,1)
1  a(6  5) 2  4
y  3( x  5) 2  4
1  a(1) 2  4
 3  a (1)
Inverse Functions
Determine if inverse relations are functions from a
table and a graph
Inverse is where the x and y values get flipped.
So if the original (a,b) the inverse
would be (b, a)
yes
Find the inverse{(1,-2),(2,3)(3,-4)(4,5),(5,-6)}
{(-2,1), (3,2), (-4,3),(5,4),(-6,5)}
Is the inverse a function?
yes
x
y
0
1
5
-3
3
0
Inverse is "a reflection in the line of y = x"
Draw the line on the graph
Reflect the points across the line.
Quadratic Regressions
Use the calculator to find a
quadratic equation to model
data
The faster a car goes, the longer it takes to stop. The table
gives the stopping distances of cars at various speeds. Write a
quadratic equation that models its stopping distance.
speed
distance
37
45
51
86
66
144
81
217
96
304
y  .03x 2  .08 x  2.40
Find a quadratic function that represents the table in
standard form and vertex form using the calculator
1 - Press stat and edit
2 - Enter x's in L1 and y's in L2
3 - Press stat and calc
4 - Press 5 for Quadreg or down arrow key
y  2 x  8 x  4
2
Find a quadratic function that represents the table in
standard form and vertex form using the calculator
y  2 x  8x  5
2
Non-Linear Systems of
Equations
1) Suppose that you are aboard a spaceship. The
Earth is at the origin of the coordinate plane, and the
path of your spaceship is the graph of:
4 x 2  3 y  18
2) To ascertain at which of these points you are
located, you find that you are also on the graph of:
y  12 x  30
At what point are you located?
Graph the 2 equations to verify your answer.
4 x 2  3 y  18
y  12 x  30
4 x 2  3(12 x  30)  18
4 x  36 x  90  18
2
4 x  36 x  72  0
2
x  9 x  18  0
( x  6)( x  3)  0
2
x  6  y  42
x 3 y 6
Key Chapter points:
• Graphing quadratic
equations
• Solving quadratic equations
by graphing
• Solving by quadratic formula
• Using the discriminant
• Write quadratic equations