Question

This graph represents a quadratic function. What is the function’s equation written in factored form and in vertex form?

Graph shows upward parabola plotted on a coordinate plane. The parabola has vertex at (2, minus 8) with the left slope at (0, 0) and the right slope at (4, 0).

f(x) = x
(x −
)

f(x) =
(x −
)2 +

Answer 1

x intercepts=0,4

So the equation

• y=a(x-b)(x-c)
• y=ax(x-4)

Now put vertex and find a

• -8=a(2)(2-4)
• -8=2a(-2)
• a=-8/-4
• a=2

So equation

• y=2x(x-4)

Answer 2

`\textsf{Factored form}: \quad f(x)=2x(x-4)`

`\textsf{Vertex form}: \quad f(x)=2(x-2)^2-8`

Explanation:

Factored form of a quadratic function

`f(x)=a(x-p)(x-q)`

where:

• p and q are the x-intercepts
• a is some constant

Given x-intercepts:

• (0, 0)
• (4, 0)

Therefore:

`\implies f(x)=a(x-0)(x-4)`

`\implies f(x)=ax(x-4)`

To find a, substitute the given vertex (2, -8) into the equation and solve for a:

`\implies 2a(2-4)=-8`

`\implies -4a=-8`

`\implies a=2`

Therefore, the function's equation in factored form is:

`f(x)=2x(x-4)`

Vertex form of a quadratic equation

`f(x)=a(x-h)^2+k`

where:

• (h, k) is the vertex
• a is some constant

Given:

• vertex = (2, -8)

Therefore:

`\implies f(x)=a(x-2)^2-8`

To find the constant a, substitute one of the x-intercepts into the equation and solve for a:

`\implies a(0-2)^2-8=0`

`\implies 4a-8=0`

`\implies a=2`

Therefore, the function's equation in vertex form is:

`f(x)=2(x-2)^2-8`