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Writing Quadratic Functions

Writing Quadratic Functions

Assessment

Presentation

Mathematics

9th - 12th Grade

Hard

Created by

Joseph Anderson

FREE Resource

5 Slides • 10 Questions

1

Interactive Lesson

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2

Writing Quadratic Equations in Vertex Form

Have your notebook nearby to show your work!

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3

Forms of Quadratic Functions

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4

Multiple Choice

What form of a quadratic is given?

f(x)=2x2+3x4f\left(x\right)=2x^2+3x-4  

1

Vertex Form

2

Standard Form

3

Root Form

5

Multiple Choice

What form of a quadratic is given?

f(x)=3(x2)2+4f\left(x\right)=3\left(x-2\right)^2+4  

1

Vertex Form

2

Standard Form

3

Root Form

6

7

Fill in the Blank

Given the equation

f(x)=5(x2)24f\left(x\right)=-5\left(x-2\right)^2-4 , what is the vertex for the parabola? 
Write your answer as a coordinate point. (x, y)

8

Multiple Choice

Lets do another problem just to make sure we've got this.



Find the equation of a parabola that has a vertex at (2, -7) and goes through the point (3, -4).

1

y=325(x+2)27y=\frac{3}{25}\left(x+2\right)^2-7

2

y=3(x2)27y=3\left(x-2\right)^2-7

3

y=3(x+2)27y=3\left(x+2\right)^2-7

4

y=(x2)27y=\left(x-2\right)^2-7

9

Multiple Choice

Find the equation for a parabola that has a vertex at (3, 4) and goes through the point (2, 2).

1

y=(x3)2+4y=\left(x-3\right)^2+4

2

y=225(x+3)2+4y=-\frac{2}{25}\left(x+3\right)^2+4

3

y=2(x3)2+4y=-2\left(x-3\right)^2+4

4

y=2(x+3)24y=-2\left(x+3\right)^2-4

10

Multiple Choice

Find the equation for a parabola that has a vertex at (2,3) and goes through the point (4,5).

1

y=(x4)2+5y=\left(x-4\right)^2+5

2

y=(x2)2+3y=\left(x-2\right)^2+3

3

y=(x4)25y=\left(x-4\right)^2-5

4

y=(x2)23y=\left(x-2\right)^2-3

11

Multiple Choice

Find the equation for a parabola that has a vertex at (-1,2) and goes through the point (3,4).

1

y=18(x3)2+4y=\frac{1}{8}\left(x-3\right)^2+4

2

y=18(x+1)2+2y=\frac{1}{8}\left(x+1\right)^2+2

3

y=14(x+1)2+2y=\frac{1}{4}\left(x+1\right)^2+2

4

y=18(x1)2+2y=\frac{1}{8}\left(x-1\right)^2+2

12

13

Multiple Choice

Lets do another problem just to make sure we've got this.



Find the equation of a parabola that has x-intercepts at -4 and 3 and passes through the point (2,7)

1

y=76(x4)(x3)y=-\frac{7}{6}\left(x-4\right)\left(x-3\right)

2

y=76(x+4)(x3)y=-\frac{7}{6}\left(x+4\right)\left(x-3\right)

3

y=72(x4)(x3)y=\frac{7}{2}\left(x-4\right)\left(x-3\right)

4

y=76(x+4)(x3)y=\frac{7}{6}\left(x+4\right)\left(x-3\right)

14

Multiple Choice


Find the equation of a parabola that has x-intercepts at -3 and 5 and passes through the point (3,-6)

1

y=12(x+3)(x5)y=-\frac{1}{2}\left(x+3\right)\left(x-5\right)

2

y=12(x+3)(x5)y=\frac{1}{2}\left(x+3\right)\left(x-5\right)

3

y=12(x3)(x5)y=\frac{1}{2}\left(x-3\right)\left(x-5\right)

4

y=2(x+3)(x5)y=2\left(x+3\right)\left(x-5\right)

15

Multiple Choice


Find the equation of a parabola that has x-intercepts at 3 and 5 and passes through the point (2,-3)

1

y=(x3)(x5)y=\left(x-3\right)\left(x-5\right)

2

y=(x3)(x5)y=-\left(x-3\right)\left(x-5\right)

3

y=(x+3)(x+5)y=-\left(x+3\right)\left(x+5\right)

4

y=13(x3)(x5)y=\frac{1}{3}\left(x-3\right)\left(x-5\right)

Interactive Lesson

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