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Algebra 1 Unit 6 Review

Total questions: 12

Worksheet time: 23mins

Name
Class
Date
1.

What is the equation of the line of symmetry for the parabola represented by the equation, y=−2x2+20x−44y=-2x^2+20x-44

Hint: x = −b2a\frac{-b}{2a}

a)

x = -10

b)

x = -5

c)

x = 5

d)

x = 10

2.
a)

b)

c)

d)

3.

What are the zeros of the parabola?

a)

b)

c)

d)

4.

What is the equation of the parabola?

a)

b)

c)

d)

5.

Which of the quadratic equations have no real solutions?

Select all that apply (2)

(HINT: Use the discriminant)

a)

x2+x+4=0x^2+x+4=0

b)

3x2−2x−7=03x^2−2x-7=0

c)

14x2−17x−6=014x^2−17x−6=0

d)

2x2−4x+5=−32x^2−4x+5=−3

6.

Use the quadratic formula to solve the quadratic equation, 2x2+4x−1=02x^2+4x-1=0 . What are the solutions to the equation?

a)

−1±62\frac{-1\pm\sqrt[]{6}}{2}

b)

2±62\frac{2\pm\sqrt[]{6}}{2}

c)

1±62\frac{1\pm\sqrt[]{6}}{2}

d)

−2±62\frac{-2\pm\sqrt[]{6}}{2}

7.
a)

6

b)

12

c)

23

d)

37

8.

Oscar throws an object to his friend from a point that is above ground level, with an initial velocity. Unfortunately, his friend cannot catch the object, and it hits the ground. The quadratic equation  h=−16t2+35t+4h=-16t^2+35t+4 models the height of the object, where h represents its height and t represents the number of seconds the object is in the air. The graph of this equation follows.

Select all that apply (2).

a)

The y-intercept shows that it takes the object about 4 seconds to hit the ground after it is thrown.

b)

The zero, or the x-intercept, shows that it takes the object about 2.3 seconds to hit the ground after it is thrown.

c)

The y-coordinate of the vertex shows that the maximum height the object reaches while airborne is 23.14 feet. 

d)

The x-coordinate of the vertex shows that the maximum height the object reaches while airborne is 1.09 feet. 

9.

Which graph correctly represents the quadratic equation y=0.0012x2−0.50x+25.98y=0.0012x^2−0.50x+25.98 ?

HINT: Use Desmos Graphing Calculator

a)

b)

c)

d)

10.

Renate launched an object vertically from a point that is 58.9 meters above ground level with an initial velocity of 21.6 meters per second. This situation can be represented by the equation h=−4.9t2+21.6t+58.9h=-4.9t^2+21.6t+58.9 , where h is the height of the object in meters and t is the time in seconds after the object is launched. What is the maximum height of the object?

HINT: Use Desmos Graphing Calculator

a)

2.2 meters

b)

6.31 meters

c)

58.9 meters

d)

82.7 meters

11.

What is the solution for the system of equations?

HINT: Use Desmos Graphing Calculator

a)

(-2.6, 3.7)

and

(3.1, 6.5)

b)

no solution

c)

(-1.7, 0)

and

(1.7, 0)

d)

(2.6, -3.7)

and

(-3.1, -6.5)

12.

What is the solution to the system of equations?

HINT: Use Desmos Graphing Calculator

a)

(2, 0)

and

(-1, 3)

b)

(-2, 0)

and

(1, -3)

c)

(2, 0)

and

(1, -3)

d)

(-2, 0)

and

(-1, 3)