For Quiz 5.7-5.8, you must know how to do the following:
Graph a quadratic equation
Identify the vertex and write the equation of the axis of symmetry
Write a quadratic equation in vertex form (𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘)
Understand how the value of a affects the graph (opens up or down, graph is wide or
narrow)
Given the vertex and a point on a parabola, write an equation in vertex form
Solve and graph inequalities
Modern Algebra II Quiz 5.7-5.8 Review Solutions
1. −𝑥 2 − 4𝑥 + 12
a.) y intercept: 12
b.) equation of axis of symmetry: 𝑥 = −2
c.) vertex: (-2, 16)
d.) x-intercepts: ( - 6, 0) and (2, 0)
To get equation into vertex form, complete the square as follows:
−12 = −𝑥 2 − 4𝑥
−12 = −1(𝑥 2 + 4𝑥)
−12 − 4 = −1(𝑥 2 + 4𝑥 + 4)
−16 = −1(𝑥 + 2)2
𝑦 = −1(𝑥 + 2)2 + 16
Graph: (graph by making a table of values)
2. 𝑦 ≤ 4𝑥 2 − 16𝑥 + 16
a.) y intercept: 16
b.) equation of axis of symmetry: 𝑥 = 2
c.) vertex: (2, 0)
d.) x-intercepts: (2, 0)
Graph: (graph by making a table of values)
3. 𝑦 = −4𝑥 2 + 5𝑥 − 1 opens downward because a is negative.
2
4. 𝑦 = 5 𝑥 2 + 4𝑥 − 1 will open wider because it has a smaller value of a.
5. 𝑦 = 𝑥 2 − 10𝑥 − 23
To write the equation in vertex form (𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘), complete the square.
23 = 𝑥 2 − 10𝑥
23 = 𝑥 2 − 10𝑥 + 25
25 + 23 = 𝑥 2 − 10𝑥 + 25
48 = (𝑥 − 5)2
𝑦 = (𝑥 − 5)2 − 48
Move constant term to the other side of the equation
Take half of the middle term and square it
Add 25 to BOTH sides
Factor the right side
Bring constant term back to the right side of the equation
The vertex would be located at (5, -48).
6. 𝑦 =
1 2
𝑥 − 6𝑥 + 5
3
Again, to write the equation in vertex form (𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘), complete the square.
1
−5 = 3 𝑥 2 − 6𝑥
Move constant term to the other side of the equation
−5 = 3 (𝑥 2 − 18𝑥)
Factor out 3
−5 = 3 (𝑥 2 − 18𝑥 + 81)
Take half of the middle term and square it
1
1
1
1
1
27 − 5 = 3 (𝑥 2 − 18𝑥 + 81) Add 27 to both sides ( 3 ∙ 81)
1
22 = 3 (𝑥 − 9)2
Factor the right side
𝑦 = 3 (𝑥 − 9)2 − 22
Bring constant term back to the right side of the equation
1
The vertex would be located at (9, - 22).
7. 𝑦 =
1 2
𝑥 + 8𝑥
2
Again, to write the equation in vertex form (𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘), complete the square.
1
0 = 𝑥 2 + 8𝑥
Move constant term to the other side of the equation
0 = 2 (𝑥 2 + 16𝑥)
Factor out 2
0 = 2 (𝑥 2 + 16𝑥 + 64)
Take half of the middle term and square it
2
1
1
1
1
1
32 + 0 = 2 (𝑥 2 + 16𝑥 + 64) Add 32 to both sides ( 2 ∙ 64)
1
32 = 2 (𝑥 + 8)2
Factor the right side
𝑦 = 2 (𝑥 + 8)2 − 32
Bring constant term back to the right side of the equation
1
The vertex would be located at (-8, -322).
8. 𝑣𝑒𝑟𝑡𝑒𝑥 (4, 1) 𝑝𝑜𝑖𝑛𝑡 (2, −2)
To write an equation that passes through the given points, you only need to calculate the
value of a by substituting h, k, x, and y from above into the equation 𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘
as follows:
−2 = 𝑎(2 − 4)2 + 1
−2 = 𝑎(−2)2 + 1
−2 = 4𝑎 + 1
−3 = 4𝑎
3
𝑎=−
4
Then write the equation as follows:
3
𝑦 = − (𝑥 − 4)2 + 1
4
9. 𝑥 2 + 8𝑥 − 9 > 0
First, set the inequality equal to zero, factor, and solve.
𝑥 2 + 8𝑥 − 9 = 0
(𝑥 + 9)(𝑥 − 1) = 0
(𝑥 + 9) = 0 OR
(𝑥 − 1) = 0
𝑥 = −9
OR
𝑥=1
Put open circles on the number line at -9 and 1 then test and shade.
1
-9
X
10. 𝑥 2 − 9𝑥 ≤ 10
Move the constant term to the left, set the inequality equal to zero, factor, and solve.
𝑥 2 − 9𝑥 − 10 = 0
(𝑥 − 10)(𝑥 + 1) = 0
(𝑥 − 10) = 0 OR
(𝑥 + 1) = 0
𝑥 = 10
OR
𝑥 = −1
Put closed circles on the number line at – 1 and 10 then test and shade.
10
-1
X
X
11. 6𝑥 2 + 5𝑥 > 4
Move the constant term to the left, set the inequality equal to zero, factor, and solve.
6𝑥 2 + 5𝑥 − 4 = 0
(2𝑥 − 1)(3𝑥 + 4) = 0
2𝑥 − 1 = 0,
2𝑥 = 1, 𝑥 =
1
2
3𝑥 + 4 = 0, 3𝑥 = −4, 𝑥 = −
4
3
1
4
Put open circles on the number line at 2 and − 3 then test and shade.
1
2
4
-3
X
12. 𝑝𝑜𝑖𝑛𝑡𝑠 (0, 1) (2, 1) (1, 3)
First, plot the points to see which is the vertex, then use either of the other points to
calculate a.
If you plot the points above, you find that (1, 3) is the vertex. So using that and the point
(0, 1) or (2, 1), calculate the value of a as follows: (I used the point (0, 1))
1 = 𝑎(0 − 1)2 + 3
1 = 𝑎(−1)2 + 3
1 = 1𝑎 + 3
−2 = 𝑎
Then write the equation as follows:
𝑦 = −2(𝑥 − 1)2 + 3