Download 1.01 Write equivalent forms of algebraic expressions to solve

1.01 Write equivalent forms of algebraic
expressions to solve problems.
a) Apply the laws of exponents.
1
Dividing Powers
(same base)
For all numbers x (not zero) and all integers m and n ,
There are a few rules that simplify our dealings with
exponents. Given the same base, there are ways that
we can simplify various expressions. For instance:
Multiplying Powers
(same base)
Simplify (x
•
3
)(x4)
Copyright © Elizabeth Stapel 2006-
2008 All Rights Reserved
when you are dividing, and the bases are the
same, you SUBTRACT the exponents."
Example
The bases are the same, so the
exponents are subtracted.
The numbers in front of the bases
are divided.
Think in terms of what the exponents mean:
(x3)(x4) = (xxx)(xxxx)
= xxxxxxx
= x7
(3+4)
...which also equals x
. This demonstrates a basic
exponent rule: Whenever you multiply two terms with the
same base, you can add the exponents:
(x
m
n
)(x )=x
A negative exponent just means that the base is on the
wrong side of the fraction line, so you need to flip the
–2
base to the other side. For instance, "x " just means
2
2
"x , but underneath, as in 1/(x )".
Write x
(m+n)
–4
using only positive exponents.
Power to a Power
Simplify (x
2 4
)
1.01 b) Operate with polynomials.
Again, think in terms of what the exponents mean:
2 4
2
2
2
2
(x ) = (x )(x )(x )(x )
= (xx)(xx)(xx)(xx)
= xxxxxxxx
= x8
Adding polynomials is just a matter of combining
like terms, with some order of operations considerations
thrown in. As long as you're careful with the minus signs,
and don't confuse addition and multiplication, you should
do fine.
Simplify (3x
3
+ 3x2 – 4x + 5) + (x3 – 2x2 + x – 4)
( 2×4 )
...which also equals x
. This demonstrates another
rule: Whenever you have an exponent expression that is
raised to a power, you can multiply the exponent and
power:
m
n
(x ) =x
mn
Horizontally:
(3x3 + 3x2 – 4x + 5) + (x3 – 2x2 + x – 4)
= 3x3 + 3x2 – 4x + 5 + x3 – 2x2 + x – 4
= 3x3 + x3 + 3x2 – 2x2 – 4x + x + 5 – 4
= 4x3 + 1x2 – 3x + 1
Vertically:
Rights
Stapel 2006-2008 All
Reserved
2
.
Subtracting polynomials is quite similar to
adding polynomials, but you have that pesky minus sign
to deal with. Here are some examples, done both
horizontally and vertically:
Simplify (x
3
+ 3x2 + 5x – 4) – (3x3 – 8x2 – 5x + 6)
The first thing I have to do is take that negative through
the parentheses. Some students find it helpful to put a
"1" in front of the parentheses, to help them keep track
of the minus sign:
Horizontally:
(x3 + 3x2 + 5x – 4) – (3x3 – 8x2 – 5x + 6)
That is, FOIL tells you to multiply the first terms in each
of the parentheses, then multiply the two terms that are
on the "outside" (furthest from each other), then the two
terms that are on the "inside" (closest to each other),
and then the last terms in each of the parentheses. In
other words, using the previous example:
•
Use FOIL to simplify (x
+ 3)(x + 2)
2
= (x3 + 3x2 + 5x – 4) – 1(3x3 – 8x2 – 5x + 6)
= x3 + 3x2 + 5x – 4 – 3x3 + 8x2 + 5x – 6
= x3 – 3x3 + 3x2 + 8x2 + 5x + 5x – 4 – 6
= –2x3 + 11x2 + 10x –10
"first": (x)(x) = x
"outer": (x)(2) = 2x
"inner": (3)(x) = 3x
"last": (3)(2) = 6
So: Copyright © Elizabeth Stapel 2006-2008 All Rights
Reserved
Vertically:
Stapel 2006-2008 All Rights Reserved
(x + 3)(x + 2) = x2 + 2x + 3x + 6 = x2
+ 5x + 6
•
In the horizontal case, you may have noticed that
running the negative through the parentheses changed
the sign on each term inside the parentheses. The
shortcut here is to not bother writing in the subtaction
sign or the parentheses; instead, I'll change all the signs
in the second row and add down:
– 4)(x – 3) ( Elementary School
Multiplication)
Simplify (x
So the answer is: x
Either way, I get the answer:
–2x3 + 11x2 + 10x – 10
polynomial multiplication
There is also a special method, useful ONLY for a twoterm polynomial times another two-term polynomial. The
method is called "FOIL". The letters F-O-I-L come from
the words "first", "outer", "inner", "last", and are a
memory device for helping you remember how to
multiply horizontally, without having to write out the
distribution like I did, and without dropping any terms.
Here is what FOIL stands for:
"Box"
Method
2
– 7x + 12
This is a "table" version of
the distributive method.
This style applies to all
polynomial multiplications.
To multiply by the grid method, place one binomial
at the top of a 2x2 grid (for binomials) and the
second binomial on the side of the grid. Place the
terms such that each term with its sign lines up with
a row or column of the grid. Multiply the rows and
columns of the grid to complete the interior of the
grid. Finish by adding together the entries inside
the grid.
3
x2 +2x + 3x + 6
Answer: x2 + 5x + 6
If c is negative, then the factors you're looking for are of
alternating signs;
that is, one is negative and one is positive.
If b is positive, then the larger factor is positive.
If b is negative, then the larger factor is negative.
In either case, you're looking for factors that are b units
apart.
To factor a "hard" quadratic, we have to handle all three
coefficients, not just the two we handled above. In this
case, we first need to multiply "a" and "c", and then find
factors of the product "ac" that add up to "b". For
instance:
Square Roots
In general, how do you figure out what can "come out" of
a square root? Factor the innards, and any factor that
occurs in pairs can come out. For example:
Simplify:
If c is positive, then the factors you're looking for are
either both positive or else both negative.
If b is positive, then the factors are positive
If b is negative, then the factors are negative.
In either case, you're looking for factors that add to b.
.
•
Factor 2x
2
+ x – 6.
Looking at this quadratic, we have a = 2, b = 1, and c =
–6, so ac = (2)(–6) = –12. Then we need to find factors
of –12 that add up to +1. The pairs of factors for 12 are
1 and 12, 2 and 6, and 3 and 4. Since –12 is negative,
we need one factor to be positive and one to be negative
(positve times negative is negative). Then we want to
use the pair "3 and 4", and we want the 3 to be negative,
because –3 + 4 = +1
1.01 c) Factor polynomials.
If your text or teacher has you factoring "by grouping",
you'll find that it is very easy to make mistakes with the
signs. You'll still have to find the numbers that add to the
coefficient in the middle, but your steps would look like
this:
2x2 + x – 6
•
Factor x
2
– 5x + 6.
The constant term is 6, but the middle coefficient this
time is negative. Since we multiplied to a positive six,
then the factors must have the same sign. (Remember
that two negatives multiply to a positive.) Since we're
adding to a negative (–5), then both factors must be
negative. So rather than using 2 and 3, as in the first
example, this time we will use –2 and –3:
x2 – 5x + 6 = (x – 2)(x – 3)
Note that you can use clues from the signs to determine
which factors to use, as we did in this last example
above:
= 2x2 + 4x – 3x – 6
= 2x(x + 2) – 3(x + 2) =
(x + 2)(2x – 3)
1.02 Use formulas and algebraic
expressions, including iterative and
recursive forms, to model and solve
problems.
4
This next one involves a little "trick" to solve it. See if you
can see what it is:
Solve Q
= 3a + 5ac for a
At first glance, these problems appear to be much worse
than your usual equation solving, but they really aren't
that bad. You pretty much do what you've done all along
for solving linear equations, except that, due to all the
variables, you won't necessarily be able to simplify your
answers as much as you're used to doing. Here's how
"solving literal equations" works:
1.03 Model and solve problems using
direct variation.
Solve A
= bh for b
Solve d
= rt for r Copyright © Elizabeth Stapel 2006-2008 All Rights
Reserved
Solve P
= 2l + 2w for w
When two variable quantities have a constant
(unchanged) ratio,
their relationship is called a direct variation. It
is said that one variable "varies directly" as
the other.
The constant ratio is called the constant of
variation.
The formula for direct variation is y= kx,
where k is the constant of variation. "y varies
directly as x"
Solving for k:
Example
Solve Q
= (c + d)/2 for d
Solve V =
3k
/t for t
A salesman's commission varies directly as
his sales. Write the formula relating this
situation and denote the variables. If the
commission is $100 for $1000 in sales, find
the commission for $1750 in sales.
C= Commission and S = Sales
Then C=KS or C/S = K
100/1000 = C/1750
1000C = 175000
C =$ 175
2.01 Find the lengths and midpoints of
segments to solve problems
The Midpoint Formula works exactly the same way. If
you need to find the point that is exactly halfway
between two given points, just average the x-values and
the y-values. Here are some typical examples:
Find the midpoint between (–1,
2) and (3, –6).
Apply the Midpoint Formula:
So the answer is P
= (1, –2).
Technically, the Midpoint Formula is the following:
5
2.02 Use the parallelism or
perpendicularity of lines and segments to
solve problems.
Parallel lines have the
same slope.
The symbol to indicate parallel lines is two
vertical bars.
It looks
something
like the
number 11. where l1 and l2 are lines
m1 and m2 are slopes
y = 3x + 5
y = 3x - 7
y = 3x + 0.5
y = 3x
These lines are ALL parallel.
They all have the same slope
(m).
(Remember y = mx + b.)
But as long as you remember that you're averaging the
two points' x- and y-values, you'll do fine
Distance Formula:
Given the two points (x1, y1) and (x2, y2), the distance
between these points is given by the formula:
Don't let the subscripts scare you. They only indicate
that there is a "first" point and a "second" point; that is,
that you have two points. Whichever one you call "first"
or "second" is up to you.
Example : Find the distance between the points
(–2, –3) and (–4, 4).
Perpendicular lines have negative
reciprocal slopes.
The symbol to indicate perpendicular is an upside-down capital T.
where l1 and l2 are lines
m1 and m2 are slopes
Just plug them in to the Distance Formula:
These lines are
perpendicular.
Their slopes (m) are
negative reciprocals.
(Remember y = mx + b.)
Then the distance is sqrt(53), or about 7.28,
rounded to two decimal places.
6
•
For the following matrix A, find 2A and –1A.
3.01 Use matrices to display and interpret data.
3.02 Operate (addition, subtraction, scalar
multiplication) with matrices to solve
problems.
Matrix addition is fairly simple, and is done entry-wise.
For instance:
•
Add the following matrices:
Add the pairs of entries, and simplify for the
Subtracti
on works
entry-wise, too.
Copyright © Elizabeth Stapel 2006-2008 All Rights
Reserved
•
To do the scalar multiplication, I just multiply a 2 on
every entry in the matrix:
Given the following matrices, find A
A – C, or explain why you can not.
– Band
3.03 Create linear models for sets of data
to solve problems.
a) Interpret constants and coefficients in
the context of the data.
b) Check the model for goodness-of-fit
and use the model, where appropriate, to
draw conclusions or make predictions.
Graphing Calculator
Solution:
A and B are the same size, each being
so I can subtract, working entry-wise:
2 × 3 matrices,
There are two types of multiplication for matrices: scalar
multiplication and matrix multiplication. Scalar
multiplication is easy. You just take a number (called a
"scalar") and multiply it on every entry in the matrix. For
example:
1. Enter the data in
the calculator lists.
Place the data in L1
and L2.
STAT, #1Edit,
type values into the
lists
2. Prepare a
scatter plot of the
data. Set up for the
scatterplot.
2nd StatPlot choose the first
icon - choices
shown at right.
Choose ZOOM #9
ZoomStat.
7
Graph shown
below.
3. Have the
calculator
determine the line
of best fit.
STAT →
CALC #4
LinReg(ax+b)
4. Have Calculator
graph it for you
Y= Vars
#5→→ EQ enter
Graph
y=11.73128088x +
193.8521475
4. To get a predicted
value within the window,
hit 2nd Trace #1,type the
desired value. The screen
below shows x = 22.
4.01 Use linear functions or inequalities to
model and solve problems; justify
results.
a) Solve using tables, graphs, and
algebraic properties.
b) Interpret constants and coefficients in
the context of the problem
Slope is a ratio and can be expressed as:
change
in y
over or
or
or
change
in x.
Zero Slope:Lines that are horizontal
have zero slope.
No Slope or Slope Undefined :Vertical
lines have no slope, or undefined slope.
Slope Intercept Form
y = mx + b
m = slope
b = y-intercept
Point Slope Form
= any point on
Use this form when
you know the slope
and the y-intercept
(where the line
crosses the y-axis).
Use this form when
you know a point on
the line and the slope
(or can determine the
slope).
the line
Standard Form
Ax + By = C
Slope = –A/B
Y-intercept = C/B
Use this form when
the coefficients are
all integers.
4.02 Graph, factor, and evaluate
quadratic functions to solve problems.
8
2
The general form of a quadratic is "y = ax + bx + c".
For graphing, the leading coefficient "a" indicates how
"fat" or "skinny" the parabola will be.
For | a | > 1 (such as a = 3 or a = –4), the parabola will
be "skinny", because it grows more quickly (three times
as fast or four times as fast, respectively, in the case of
our sample values of a). For
| a | < 1 (such as a = 1/3 or a = –1/4 ), the parabola will
be "fat", because it grows more slowly (one-third as fast
or one-fourth as fast, respectively, in the examples).
Also, if a is negative, then the parabola is upside-down.
Graph the parabola y = x2 + 6x - 1
To find the axis of symmetry, use the
formula x = -b/2a In this example, a = 1
and b = 6.
Substituting gives: x = -(6)/2(1) = -6/2 = -3
Axis of symmetry: x = -3
Since the x-coordinate of the turning point (
Point of Vertex) is -3, use this value as the
middle value for x in the chart. Include 3
values above and below -3 in the chart.
There's no magic to solving quadratic equations.
Quadratic equations can be solved by factoring
and also by graphing.
Solve for x:
Here are the steps you should follow:
1. Move all
terms to
the same
side of the
equal sign, This places the equation in
so the
standard form.
equation is
set equal
to 0.
Substitute each value of x into the quadratic
equation (the parabola), to find the
corresponding values for y and complete the
table. When the table is complete, plot the
points and draw the graph.
x
-6
-5
-4
-3
-2
-1
0
y
-1
-6
-9
-10
-9
-6
-1
2.
Factor the (x + 3) and (x + 2) are called
algebraic factors. These are factors of the
expression. expression
x2 - x - 6.
3. Set each
factor
equal to 0.
(This process
9
is called the
"zero product
property". If
the product
of two factors
equals 0,
then either
one or both
of the factors
must be 0.)
4. Solve each
resulting
equation.
x = 3 and x = -2 are called
roots. These are roots of the
equation x2 - x - 6 = 0.
2. Replace the "y"
value in the first
equation by what "y"
now equals. Grab the
"y" value and plug it
into the other equation.
3(9 - 2x) - 2x = 11
3. Solve this new
equation for "x".
(27 - 6x) - 2x = 11
27 - 6x - 2x = 11
27 - 8x = 11
-8x = -16
x=2
4. Place this new "x"
value into either of the
ORIGINAL equations
in order to solve for
"y". Pick the easier
one to work with!
y + 2x = 9 or
y = 9 - 2x
y = 9 - 2(2)
y=9-4
y=5
5. Check: substitute x
= 2 and y = 5 into
BOTH ORIGINAL
equations. If these
answers are correct,
BOTH equations will
be TRUE!
3y - 2x = 11
OR use the Quadratic Formula
The Quadratic Formula uses the "a", "b", and "c" from
2
"ax + bx + c", where "a", "b", and "c" are just numbers.
The Formula is derived from the process of completing
the square, and is formally stated as:
2
For ax + bx + c =
0, the value of x is
given by:
4.03 Use systems of linear equations or
inequalities in two variables to model and solve
problems. Solve using tables, graphs, and
algebraic properties; justify results.
Using the substitution method:
Solve this
system of
equations
(and check):
3y - 2x = 11
y + 2x = 9
1. Solve one of the
equations for either "x
=" or "y =".
This example solves
the second equation for
"y =".
3y - 2x = 11
y = 9 - 2x
3(5) - 2(2) = 11
15 - 4 = 11
11 = 11 (check!)
y + 2x = 9
5 + 2(2) = 9
5+4=9
9 = 9 (check!)
Solve this system of equations using the
addition or subtraction method.
10
1. Solve this
system of
equations
and check:
x - 2y = 14
x + 3y = 9
a. First, be sure that
x - 2y = 14
x + 3y = 9
f.
x - 2y = 14
x + 3y = 9
substitute x = 12 and y
= -1 into BOTH
ORIGINAL equations.
If these answers are
correct, BOTH
equations will be
TRUE!
the variables are "lined
up" under one another.
In this problem, they
are already "lined up".
b. Decide which
variable ("x" or "y")
will be easier to
eliminate. In order to
eliminate a variable, the
numbers in front of
them (the coefficients)
must be the same or
negatives of one
another. Looks like "x"
is the easier variable to
eliminate in this
problem since the x's
already have the same
coefficients.
c. Now, in this
problem we need to
subtract to eliminate
the "x" variable.
Subtract ALL of the
sets of lined up terms.
(Remember: when you
subtract signed
numbers, you change
the signs and follow the
rules for adding signed
numbers.)
d. Solve this simple
equation.
e. Plug "y = -1" into
either of the
ORIGINAL equations
to get the value for "x".
Check:
x - 2y = 14
x - 2(-1) = 14
x + 2 = 14
x = 12
x - 2y = 14
12 - 2(-1) = 14
12 + 2 = 14
14 = 14 (check!)
x + 3y = 9
12 + 3(-1) = 9
12 - 3 = 9
9 = 9 (check!)
x - 2y = 14
-x - 3y = - 9
- 5y = 5
-5y = 5
y = -1
11
Solve
graphically:
4x - 6y = 12
2x + 2y = 6
To solve a system of equations graphically, graph
both equations and see where they intersect. The
intersection point is the solution.
First, solve each equation for "y =".
4x - 6y = 12
slope =
y-intercept =
-2
2x + 2y = 6
slope = -1
y-intercept
=3
4.04 Graph and evaluate exponential
functions to solve problems
Exponential functions
form
Notice:
The variable x is an exponent.
As such, the graphs of these functions are not
straight lines. In a straight line, the "rate of
change" is the same across the graph. In these
graphs, the "rate of change" increases or
decreases across the graphs.
Example:
when a > 0 and the b
is between
0 and 1, the graph will be
decreasing (decaying).
For this example, each time x is
increased by 1, y decreases to one
half of its previous value.
Graph
the
lines.
The slope
intercept
method of
graphing was
used in this
example.
The point of
intersection
of the two
lines, (3,0), is
the answer
Such a situation is called
Exponential Decay.
12
Example:
when a > 0 and the b
is greater
At the Algebra level, there are two functions
that can be easily used to illustrate the
concepts of growth or decay in applied
situations. When a quantity grows by a fixed
percent at regular intervals, the pattern can be
represented by the functions,
than 1, the graph will be increasing
(growing).
For this example, each time x is
increased by 1, y increases by a factor
of 2.
Growth:
Decay:
a = initial amount before measuring
growth/decay
r = growth/decay rate (often a percent)
x = number of time intervals that have passed
Example:
A bank account balance, b, for
an account starting with s dollars, earning an
annual interest rate, r, and left untouched for n
years can be calculated as
(an
exponential growth formula). Find a bank
Such a situation is called
Exponential Growth.
account balance to the nearest dollar, if the
account starts with $100, has an annual rate of
4%, and the money left in the account for 12
years.