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Semester Exam Review
Notes
(Put in slope-intercept form first.
2y = 3x + 8
y = 3/2 x + 4
2y = -x – 8
y = -1/2 x – 4
Steps for Solving by Elimination
• *Align like terms in each equation vertically into
the same columns (or order) before starting.
• 1. Choose a variable to eliminate.
• 2. Eliminate that variable by adding or subtracting
one equation from the other. (Sometimes you have
to multiply first.)
• 3. Solve the new equation.
• 4. Plug in your answer to find the other variable.
(Or repeat the elimination process for the other variable.)
• 5. Check your answer.
Elimination Example
5( 5x
+ 4y = -28 )
-2(3x + 10y = -13 )
 If we multiply equation (1) by 5 and equation (2) by -2, we be able to eliminate y using 20/-20.
25x + 20y = -140
-6x – 20y = 26
19x = -114
Add the equations 
x = -6
Substitute x = -6 into equation (1)
5x + 4y = -28
5(-6) + 4y = -28
-30 + 4y = -28
4y = -28 +30
4y = 2
2
y=
4
y=
1
2
Check your answer x = -6 and
equation (2)
y = ½ into
3(-6) + 10(½) = -13
-18 + 5 = -13
-13 = -13
Therefore, the solution is (-6, ½)
Solving Systems of Equations
using Substitution
Steps:
1. Solve one equation for one variable (y= ; x= ; a=)
2. Substitute the expression from step one into the
other equation.
3. Simplify and solve the equation.
4. Substitute back into either original equation to find
the value of the other variable.
5. Check the solution in both equations of the system.
Substitution Example:
x + y = 10
5x – y = 2
Step 1: Solve one equation for one variable.
x + y = 10
y = -x +10
Step 4: Substitute back into either original
equation to find the value of the other
Step 2: Substitute the expression
variable.
from step one into the other equation.
x + y = 10
5x - y = 2
2 + y = 10
5x -(-x +10) = 2
y=8
Step 3: Simplify and solve the equation.
5x + x -10 = 2
6x -10 = 2
6x = 12
x=2
Solution to the system is (2,8).
Step 5: Check the solution in both equations.
5x – y = 2
x + y =10
5(2) - (8) = 2
2 + 8 =10
10 – 8 = 2
10 =10
2=2
VERIFIED!!!
Solving
Systems by
Matrices
Writing a System of Equations
A woman owns 21 pets. Each of her pets is either a cat or a bird. If the pets have a total of 76 legs, and
assuming that none of the bird's legs are protruding from any of the cats' jaws, how many cats and how
many birds does the woman own?
There are two unknown quantities here: the number of cats the lady owns, and the number of birds the lady
owns.
The problem has given us two pieces of information: if we add the number of cats the lady owns and the number
of birds the lady owns, we have 21, and if we add the number of cat legs and the number of bird legs, we have
76.
Let's replace the unknown quantities with variables. Let x be the number of cats the lady owns, and y be the
number of birds the lady owns.
Now we can replace the pieces of information with equations. Instead of saying "if we add the number of cats the
lady owns and the number of birds the lady owns, we get 21, " we can say:
x + y = 21
What about the second piece of information: "if we add the number of cat legs and the number of bird legs, we
get 76"? Since a cat has 4 legs, if the lady owns x cats there are 4x cat legs. Since a bird has 2 legs, if the lady
owns y birds there are 2y bird legs. This means we can replace this second piece of information with an equation:
4x + 2y = 76
If x is the number of cats and y is the number of birds, the word problem is described by this system of equations:
Graphing Systems of Linear
Inequalities
1. Put each equation in slope-intercept form
(Remember that if you divide by a negative
number, you must flip the inequality sign).
2. Graph each inequality, remembering your shading
rules.
< , > Dashed Line
≤ , ≥ Solid Line
< , ≤ Shade Below
> , ≥ Shade Above
< Less Than
> Greater Than
Domain/Range of Ordered Pairs
• In the set of ordered pairs {(-2, 0), (0, 6), (2, 12), (4, 18)},
the domain is the set of the first number in every pair
(those are the x-coordinates): {-2, 0, 2, 4}. The range is
the set of the second number of all the pairs (those are
the y-coordinates): {0, 6, 12, 18}.
• {(-2, 0), (0, 6), (2, 12), (4, 18)}
• Domain: {-2, 0, 2, 4}
• Range: {0, 6, 12, 18}
This is a function because
none of the x-values
repeat with a different yvalue.
Also use this method if you have a scatterplot graph.
Domain/Range of a Graph
• For the domain: Look at the values of x that are covered by the
graph.
• From left to right, the graph goes from -4 to 4.
• So our values of x are greater than or equal to -4 and less than 4.
• In set notation we write: 𝑥𝑥 −4 ≤ 𝑥𝑥 < 4 .
• In algebraic notation we write: −4 ≤ 𝑥𝑥 < 4.
• In interval notation we write: [−4, 4).
• For the range: Look at the values of y that are covered by the
graph.
• From bottom to top, the graph goes from -3 to 4.
• So our values of x are greater than or equal to -3 and less than
or equal to 4.
• In set notation we write: 𝑦𝑦 −3 ≤ 𝑦𝑦 ≤ 4 .
• In algebraic notation we write:−3 ≤ 𝑦𝑦 ≤ 4.
• In interval notation we write: [−3, 4].
Parent Functions
Refer to your
parent function
packet from the
beginning of the
year. Make your
graphs exact…
don’t just sketch
lines!
𝑙𝑙𝑙𝑙𝑙𝑙𝑏𝑏 𝑥𝑥
Transformations Rules
The Rules
𝑓𝑓 𝑥𝑥
𝑎𝑎𝑎𝑎 𝑥𝑥
What They Mean
the function
vertical
stretch/compression
−𝑓𝑓 𝑥𝑥
reflection across x-axis
𝑓𝑓 𝑥𝑥 + ℎ
horizontal shift left
𝑓𝑓 𝑥𝑥 − 𝑘𝑘
vertical shift down
𝑓𝑓 𝑥𝑥 − ℎ
horizontal shift right
𝑓𝑓 𝑥𝑥 + 𝑘𝑘
vertical shift up
PARENT FUNCTION
TRANSFORMATION FORMS
• Quadratic: y = a(x – h)2 + k
• Absolute Value: y = a│x – h│+ k
• Cubic: y = a(x – h)3 + k
• Square Root: y = a (𝐱𝐱 – 𝐡𝐡) + k
• Cube Root: y = a 𝟑𝟑 (𝐱𝐱 – 𝐡𝐡) + k
• Exponential: y = a∙b(x – h) + k
• Logarithmic: y = a∙logb(x – h) + k
• Reciprocal: y =
𝒂𝒂
𝒙𝒙−𝒉𝒉
+k
• Linear: y = a(x – h) + k
Steps for solving absolute value equations:
1)
2)
3)
4)
5)
**Need to isolate the absolute value expression**
Undo addition or subtraction outside of absolute value.
Undo multiplication or division outside of absolute value.
Set expression inside absolute value equal to the given value and to its opposite.
Solve for variable using steps for solving equations.
Check your solutions!
Ex1:
|3x – 6| – 5 = -7
+5 +5
Ex 2: |x + 2| = 7
x+2=7
-2 -2
x=5
or
x + 2 = -7
-2 -2
x = -9
or
Check:
|x + 2| = 7
|(5) + 2| = 7
|7| = 7
7=7
|3x – 6| = -2
Already isolated
3x – 6 = -2
+6 +6
3x = 4
x=4/3
or
or
3x – 6 = 2
+6 +6
3x = 8
x = 8/3
Check:
|3(8/3) – 6| – 5 = -7
|x + 2| = 7 |3(4/3) – 6| – 5 = -7
|8 - 6| = -2
|4 - 6| = -2
|(-9) + 2| = 7
|2| = -2
|-2| = -2
|-7| = 7
2 = -2
2 = -2
7=7
{-9, 5}
No Solution (remember that an absolute value cannot be equal
to a negative number – that can save you some work!)
Solving Absolute Value Inequalities
≤
≥
Less than sets up as “between”
Greater than sets up as “outside” or “beyond” the boundary values
1− 2x − 3 ≥ 3
3x − 1 ≤ 5
− 5 ≤ 3x − 1 ≤ 5
+1
-4
+1 +1
6
− 4 ≤ 3x ≤ 6
3
3
3
4
− ≤x≤2
3
+3 +3
6
1− 2x ≥ 6
1 − 2 x ≤ −6 or 1 − 2 x ≥ 6
− 2 x ≤ −7
-2
-2
7
or
x≥
2
− 2x ≥ 5
-2
-2
5
x≤−
2
Graphing Absolute Value Functions
Graph 𝒇𝒇 𝒙𝒙 = − 𝒙𝒙 + 𝟐𝟐 + 𝟑𝟑
• 1. Vertex = (-2, 3)
• 2. Apply the movement
a=-1 from the vertex
(down 1, right 1)
• 3. Use the line of
symmetry x=-2 to plot the
3rd point (reflect the
point).
• 4. Complete the graph.
• If a>0, the parabola opens upward.
If a<0, the parabola opens downward.
• The axis of symmetry is the line 𝒙𝒙 =
• The x-coordinate (h) of the vertex is
−𝒃𝒃
.
𝟐𝟐𝟐𝟐
−𝑏𝑏
.
2𝑎𝑎
Quadratic Standard Form:
𝒇𝒇 𝒙𝒙 = 𝒂𝒂𝒙𝒙𝟐𝟐 + 𝒃𝒃𝒃𝒃 + 𝒄𝒄
The y-coordinate (k) of the vertex is the y-value of the function when 𝑥𝑥 =
or 𝑦𝑦 = 𝑓𝑓
−𝑏𝑏
2𝑎𝑎
.
• The y-intercept is (0, c).
QUADRATIC FACTS
−𝑏𝑏
2𝑎𝑎
,
• The MAX or MIN will be found at the “k” of the Vertex
Example: Find the vertex, axis of symmetry, the maximum or
minimum value, and the range of 𝑓𝑓 𝑥𝑥 = 2𝑥𝑥 2 + 8𝑥𝑥 − 2.
• Vertex: (−2, −10)
−𝑏𝑏
−8
−8
=
=
= −2 (x-value)
2𝑎𝑎
2(2)
4
Plug in: 2(−2)2 +8 −2 − 2 = −10
• Axis of Symmetry: 𝑥𝑥 = −2
• The graph opens upward so it will have a minimum value at −10.
• The range is 𝑦𝑦 ≥ −10.
Converting Standard Form to Vertex Form
• Find the vertex.
• Use the vertex (h, k) and the “a” value to write the vertex form.
Example: Convert
𝑓𝑓 𝑥𝑥 = 2𝑥𝑥 2 + 20𝑥𝑥 + 7 to Vertex Form
• Step 1: Find the vertex.
•
−𝑏𝑏
2𝑎𝑎
=
−20
2(2)
=
−20
4
= −5
(h)
• 𝑓𝑓 −5 = 2(−5)2 +20 −5 + 7 = −43 (k)
• Vertex: −5, −43
(h, k)
• Step 2: Plug in a, h, and k (a=2)
• 𝒇𝒇 𝒙𝒙 = 𝟐𝟐(𝒙𝒙 + 𝟓𝟓)𝟐𝟐 −𝟒𝟒𝟒𝟒
Attributes from a Quadratic Graph
Axis of Symmetry: x=-1
y-intercept: (0, -1)
Vertex: (-1, -3)
Minimum value is -3 found at (-1, -3)
Quadratic Regression (Finding the
equation from 3 points)
Quadratic Regression: Calculator Steps
STAT
1: EDIT ENTER
Type in data from “x” into L1.
Type in data from “y” into L2.
STAT
CALC
5: QuadReg, ↓CALCULATE
ENTER
QuadReg 𝑦𝑦 = 𝑎𝑎𝑎𝑎 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐
a=
b=
c=
Plug in values for a, b, and c to write the equation in standard form.
Types of Equations (Hint: pay
attention to the largest exponent)
• Linear equations are equations of the type
ax + b = 0, with a ≠ 0, or any other equation in
which the terms can be operated and simplified
into an equation of the same form.
• Quadratic equations are equations of the type
ax2 + bx + c = 0, with a ≠ 0.
• Cubic equations are equations of the type
ax3 + bx2 + cx + d = 0, with a ≠ 0.
• Quartic equations are equations of the type
ax4 + bx3 + cx2 + dx + e = 0, with a ≠ 0.
Multiplying Polynomials
• Examples are worked out here:
• https://www.mathsisfun.com/algebra/polynomialsmultiplying.html
• Remember that area is just multiplying the length
and width so use the same method as above.
Multiplying polynomials is just DISTRIBUTION! There is a
distribution for each term of the first factor.
FACTORING: Difference of Squares
•If there are only 2 terms, check for difference of squares (2 terms that you can take the square
root of).
•Factor like this…
a2 – b2 = (a + b)(a – b)
*Always look for GCF first!
Example 1
Factor: x2 – 9
Is there a GCF? No.
Remember: a2 – b2 = (a + b)(a – b)
a = x and b = 3
So…factored form is… (x + 3)(x – 3)
Example 3
Example 2
Is there a GCF? No.
Factor using: a2 – b2 = (a + b)(a – b)
a = 5x and b = 7y
So…factored form is… (5x + 7y)(5x – 7y)
Factor: 16x2 – 4y2
Is there a GCF? Yes, 4. So divide both terms: 4(4x2 – y2)
Factor inside the ( ) using: a2 – b2 = (a + b)(a – b)
a = 2x and b = y
So…factored form is… 4(2x + y)(2x – y)
Factor: 25x2 – 49y2
• Factoring Trinomials
Build / Simplify / Fix
From:
Is there a GCF?
𝑎𝑎𝑎𝑎 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐
Look for a Greatest Common Factor first!
Step 1:
Identify:
“product” (𝒂𝒂 � 𝒄𝒄) number
“sum” (𝒃𝒃) number
Step 3: Build two fractions…
𝒂𝒂
and
YES…-1…so rewrite the expression as:
a= 6 b= 1 c= -2
-(6x2 + x – 2)
Step 1: Identify product (a∙c) number: -12
And sum (b) number: 1
Step 2: What 2 numbers multiply to -12 and add to 1?
Step 2: Find a factor pair that:
multiplies to get your “product” number
adds to get your “sum” number
𝑓𝑓1 and 𝑓𝑓2 (factors) so that…
𝑓𝑓1 � 𝑓𝑓2 = 𝑎𝑎 � 𝑐𝑐 AND 𝑓𝑓1 + 𝑓𝑓2 = 𝑏𝑏
𝒇𝒇𝟏𝟏
Example: Factor -6x2 - x + 2
𝒇𝒇𝟐𝟐
Factors of -12: 1, -12 -1, 12
Which ones add to 1?
Step 4: Simplify the fractions, if possible…
Step 5: “Fix” your binomials with simplified f’s and a’s…
ANSWER: (𝒂𝒂𝒂𝒂 + 𝒇𝒇𝟏𝟏 )(𝒂𝒂𝒂𝒂 + 𝒇𝒇𝟐𝟐 )
3, -4
-3, 4
So, f1=-3 and f2=4
Step 3: BUILD 2 fractions:
Step 4: SIMPLIFY:
𝒂𝒂
2, -6 -2, 6
𝑓𝑓1
𝑎𝑎
=
−3
6
and
−1
2
𝑓𝑓2
𝑎𝑎
and
=
2
3
4
6
Step 5: FIX your fractions by bringing your denominator
to the front of each factor:
Don’t forget about the -1 GCF!
−(2𝑥𝑥 − 1)(3𝑥𝑥 + 2)