Unit 3: Polynomials
Math 2 (Fall 2015)
Name: ______________________________
Date: ______________ Period _______
Test Review: Due on Test day BEFORE the test. Graded for ACCURACY! Every day late is minus 5 points.
Show all work. You will not receive full credit if you do not show your work. If I cannot read the work then it will be
counted wrong. NO Homework passes accepted for this assignment. *This is not all inclusive. Please study all quizzes
and warm ups.
You need to
know & be
able to do
Things to remember
Vertex: Point where the
graph changes direction.
Practice Problems
For the following problems find the characteristics
1) .
2) 𝑦 = (𝑥 − 2)2 + 3
Vertex:
Maximum: The
parabola opens down (if
𝒂 is negative)
Minimum: The
parabola opens up (if 𝒂
is positive)
Axis of Symmetry: The
vertical line cutting the
parabola in half
Find the
Characteristics
of a parabola
y-intercept: The point
where the parabola
crosses the y-axis
Vertex Form
𝑦 = 𝒂(𝑥 − 𝒉)2 + 𝒌
Vertex: (h, k)
AoS: x = h
y-intercept: (0, #), plug 0
in for x
Standard Form
𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
Vertex: (#, #)
𝑏
𝒙=−
2𝑎
𝒚 =plug x into equation
𝑏
AoS: 𝒙 = − 2𝑎
y-intercept: (0, #), plug 0
in for x
Max or Min:
AoS:
y-int:
Vertex:
Max or Min:
AoS:
y-int:
3) 𝑦 = 𝑥 2 − 8𝑥 + 7
Vertex:
4) 𝑦 = −3(𝑥 + 1)2 − 2
Vertex:
Max or Min:
Max or Min:
AoS:
AoS:
y-int:
y-int:
Graph a
quadratic
function
1. Find and plot the
vertex.
State the characteristics and graph the following equations.
5) 𝑦 = (𝑥 − 3)2 − 9
2. Find and graph the
axis of symmetry
(dashed vertical line)
Vertex:
3. Find the points on
the each side of the
vertex and plot.
(calculator can help
you with this)
AoS:
Max or Min:
y-int:
Solutions:
6)
𝑦 = −𝑥 2 + 6𝑥 − 5
Vertex:
Max or Min:
AoS:
y-int:
Solutions:
Solve a
Quadratic by
Factoring
Get in standard Form
Solve by factoring
7) 9𝑥 2 − 25 = 0
8) 𝑥 2 = 9𝑥 − 14
9) 𝑥 2 = −4𝑥 + 12
10) 2𝑥 2 − 20 = 6𝑥
Factor
Set each factor equal to
zero and solve
Find the discriminate and state the number and type of
solutions.
11) 2𝑥 2 + 7𝑥 + 3 = 0
12) 𝑥 2 + 4𝑥 − 1 = −5
𝑥=
−𝑏 ± √𝑏 2 − 4𝑎𝑐
2𝑎
2
Solve a
Quadratic by
Quadratic
Formula
Discriminate: 𝑏 − 4𝑎𝑐
 Positive answer
2 real roots.
 Zero, one real root
(two equal real
roots).
 Negative answer 2
imaginary roots.
To Solve:
1st: Put into standard
form.
nd
2 : Plug a, b, and c into
the quadratic formula
3rd: type the
discriminate in the
calculator
4th: Simplify the radical
5th: simplify the
numbers
13. −4𝑥 2 + 4𝑥 = 5
14. −𝑥 2 = 4𝑥
Find the solutions by using the quadratic formula.
13. 3𝑥 2 = −10𝑥 − 3
14. 3𝑥 2 + 3𝑥 = 1
15. 𝑥 2 + 4𝑥 − 2 = 0
16. 7𝑥 2 = 14 + 2𝑥
Solve the equation using the square root method.
17. 𝑥 2 = 20
18. 𝑥 2 − 16 = 0
19. 3𝑥 2 − 12 = 0
20. 9𝑥 2 − 25 = 0
Isolate the square.
Solve a
Quadratic by
Taking Square
Root
Take the square root of
both sides.
Don’t forget the ±
Simplify.
Transform a
quadratic
functions
Parent Function
𝑦 = 𝑥2
Describe the transformations
21. 𝑦 = 𝑥 2 − 3
22. 𝑦 = (𝑥 − 3)2
Transformations
𝑦 = 𝑥 2 + 𝑘 vertical shift
+k up and –k down
𝑦 = (𝑥 + ℎ)2 horizontal
shift +h left and –h right
𝑦 = −𝑥 2 reflect graph
over the x-axis
23. 𝑦 = 3 𝑥 2
1
24. 𝑦 = −𝑥 2
25. 𝑦 = 4(𝑥 + 2)2 − 5
26. 𝑦 = − 2 𝑥 2 + 4
3
Write the equation for the following transformations
𝑦 = 𝑎𝑥 2
|𝑎| > 1 compressed
(wide)
0 < |𝑎| < 1 stretched
(wide)
27. The graph goes down 7
and right 3
Convert from
Vertex Form to
Standard Form
1. Foil or the square
term.
2. Distribute the “a”
value.
3. Combine like terms.
Write in standard form
5. 𝑦 = (𝑥 − 3)2 + 1
Convert from
Standard Form
to Vertex Form
Write in vertex form
1. Find the vertex
2. Identify the “a” value. 7. 𝑦 = 2𝑥 2 − 4𝑥 − 1
3. Use these three
values and substitute
into the form:
𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘
28. The parabola is flipped and
compressed by a factor of 1/4. Its
position moves left 4 and up 8
6. 𝑦 = −3(𝑥 + 2)2 − 4
8. 𝑦 = −𝑥 2 − 8𝑥 + 4