Algebra: 9.1.1 (Day 2) Solving Quadratics
Bell Work 1/29
Solutions
Name _________________________
Block ____ Date ___________
Solve each quadratic equation:
a. 0 = (𝑥 − 1)2 − 4
b.
4 = (𝑥 − 1)2
2 = (𝑥 + 3)2
±2 = 𝑥 − 1
±√2 = 𝑥 + 3
1±2= 𝑥
𝟑=𝒙
0 = (𝑥 + 3)2 − 2
−𝟑 ± √𝟐 = 𝒙
−𝟏= 𝒙
Write each equation in standard form and then in graphing form by using generic rectangles, and then
y
graph each equation. Label the roots and vertex.
1.
y = (x – 2)(x – 3)
−2 −2𝑥
6
𝑥 𝑥 2 −3𝑥
𝑥
2.
−3
Standard Form:
𝒚 = 𝒙𝟐 − 𝟓𝟓 + 𝟔
y = (x – 2)(x + 2)
+2 +2𝑥 −4
𝑥 𝑥 2 −2𝑥
𝑥
−2
Standard Form:
𝒚 = 𝒙𝟐 − 𝟒
3.
−2.5 −2.5𝑥 6.25
+6
𝑥
𝑥2
−2.5𝑥
𝑥
-6.25
2
3
(2.5, -0.25)
−2.5
Graphing Form:
𝒚 = (𝒙 − 𝟐. 𝟓)𝟐 − 𝟎. 𝟐𝟐
−2.5 0𝑥
𝑥 𝑥2
𝑥
0
x
y
-4
0𝑥
2
-2
−2.5
Graphing Form:
𝒚 = 𝒙𝟐 − 𝟒
x
(0, -4)
y
y = x(x + 4)
𝑥 𝑥 2 4𝑥
𝑥 +4
Standard Form:
𝒚 = 𝒙𝟐 + 𝟒𝟒
2 2𝑥
𝑥 𝑥2
𝑥
4
-4
2𝑥
+2
Graphing Form:
𝒚 = (𝒙 + 𝟐)𝟐 − 𝟒
-4
0
(-2, -4)
x
Standard Form
A
y = 5x + 5x
2
Factored Form
𝒚 = 𝟓𝟓(𝒙 + 𝟏)
5𝑥 5𝑥 2 5𝑥
𝑥
B
𝟐
𝒚 = 𝟐𝟐 + 𝟓𝟓 + 𝟐
+1 1𝑥
+1
y = (2 x + 1)( x + 2)
𝒚 = 𝒙 − 𝟒𝟒
−2 −2𝑥
4
𝑥 𝑥 2 −2𝑥
4
𝑥
−2
-4
𝒚 = 𝒙(𝒙 − 𝟒)
𝑥 𝑥2
𝑥
−4𝑥
𝑥 𝑥 2 . 5𝑥
-1
0
x
– 0.25
𝑥 +.5
𝑦 = 2(𝑥 2 + 2.5𝑥) + 2
y
= 2(𝑥 + 1.25)2 + 2 + 2(−1.5625)
𝑥2
𝑥
1.25𝑥
-0.5
-2
+2
x
– 1.5635
+1.25
y
y = ( x − 2) − 𝟒
2
4
0
x
y
𝑦 = 𝑥 2 − 2𝑥 − 8
𝒚 = 𝒙𝟐 − 𝟐𝟐 − 𝟑
+1 1𝑥 −3
𝒚 = (𝒙 + 𝟐)(𝒙 − 𝟒)
+2 2𝑥 −8
𝑥
−4
𝑦 = (𝑥 − 3)(𝑥 + 1)
𝑥 −3
𝒚 = 𝟒𝟒𝟐 − 𝟏
𝟐
𝒚 = (𝒙 − 𝟏) − 𝟗
−1 −𝑥 +1 −1
-2
4
x
𝑥 𝑥 2 −𝑥 −8
𝑥 −1
𝒚 = (𝒙 − 𝟏)𝟐 − 𝟒
−1 −𝑥 +1 −1
y
-1
3
x
𝑥 𝑥 2 −𝑥 −3
𝑥 𝑥 2 −3𝑥
6
. 5 . 5𝑥 . 25
−4
𝑥 𝑥 2 −4𝑥
5
𝒚 = 𝟓(𝒙 + 𝟎. 𝟓)𝟐 − 𝟏. 𝟐𝟐
𝑥
𝟐
y
𝑦 = 5(𝑥 + 0.5)2 + 5(−0.25)
1.25 1.25𝑥 1.5625
𝑥 +2
Graph with roots
𝑦 = 5(𝑥 2 + 𝑥)
𝒚 = 𝟐(𝒙 + 𝟏. 𝟐𝟓)𝟐 − 𝟏. 𝟏𝟐𝟐
2
2𝑥 2𝑥 2 4𝑥
C
Graphing Form
𝒚 = (𝟐𝟐 + 𝟏)(𝟐𝟐 − 𝟏)
𝑥 −1
y
y = 4x − 1
2
-0.5
0.5
x
Use the equations to find the x-intercepts for the graphs. Describe how the graphs for each of the following
compares to the graph of the parent y = x2, then graph. Label the vertex, roots, and axis of symmetry.
y
7.
y = (x + 1)2 – 2
Left 1, down 2
0 = (𝑥 + 1)2 − 2
2 = (𝑥 + 1)2
±√2 = 𝑥 + 1
−𝟏 ± √𝟐 = 𝒙
−1 + √2 ≈ 𝟎. 𝟒
−1 + √2 ≈ −𝟐. 𝟒
8.
2
y = 2(x – 3) – 8
Twice as steep, right 3, down 8
2
0 = 2(𝑥 − 3) − 8
-2.4
.4
x
(-1, -2)
Axis: 𝒙 = −𝟏
8 = 2(𝑥 − 3)2
4 = (𝑥 − 3)2
y
1
±2 = 𝑥 − 3
5
x
𝟑±𝟐 = 𝒙
3+2= 𝟓
3−2= 𝟏
9.
y = -(x + 4)2 + 1
Inverted Graph, left 4, up 1
0 = −(𝑥 + 4)2 + 1
−1 = −(𝑥 + 4)2
1 = (𝑥 + 4)2
±1 = 𝑥 + 4
Axis: 𝒙 = 𝟑
y
(3,-8)
(-4, 1)
-3
-5
x
−𝟒 ± 𝟏 = 𝒙
−4 + 1 = −𝟑
−4 − 1 = −𝟓
10.
y = ½(x – 6)2 + 2
Half as steep, right 6, up 2
0 = 1�2 (𝑥 − 6)2 + 2
Axis: 𝒙 = −𝟒
−2 = 1�2 (𝑥 − 6)2
−4 = (𝑥 − 6)2
y
(6, 2)
±√−4 = 𝑥 − 6
𝑵𝑵 𝑹𝑹𝑹𝑹 𝑹𝑹𝑹𝑹𝑹
Axis: 𝒙 = 𝟔
x
11. A person that is 1.8 meters tall and throws a ball upward from the top of a building that is 68.6
meters tall with a velocity of 22 meters per second. h is the height of the ball after t seconds and is
given by the equation:
ℎ = −4.9𝑡 2 + 22𝑡 + 70.4
1.8 m
a. Complete the drawing and label for this situation:
68.6 m
b. Graph the function (label the axes). Label the positive root, the
vertex and the h-intercept and explain what each mean in this
problem situation:
h
h-intercept: The height when the
ball was released was 70.4 meters.
Vertex: The maximum height of the ball
is 95.1 meters after 2.2 seconds.
(2.2, 95.1)
70.4
Root: The ball
reaches the ground in
6.7 seconds.
t
9-17. Solve the following quadratic equations by factoring and using the Zero Product Property and
then solve by completing the square. (Show sufficient work including two generic rectangles for each!)
a.
x2 − 13x + 42 = 0 𝒙 = 𝟔 and 𝒙 = 𝟕
b.
0 = 3x2 + 10x – 8 𝒙 = −𝟒 and 𝒙 =
c.
2x2 − 10x = 0 𝒙 = 𝟎 and 𝒙 = 𝟓
d.
4x2 + 8x − 60 = 0 𝒙 = −𝟓and 𝒙 = 𝟑
𝟐
𝟑