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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 +

Respuesta :

semsee45

Answer:

[tex]\textsf{Factored form}: \quad f(x)=2x(x-4)[/tex]

[tex]\textsf{Vertex form}: \quad f(x)=2(x-2)^2-8[/tex]

Step-by-step explanation:

Factored form of a quadratic function

[tex]f(x)=a(x-p)(x-q)[/tex]

where:

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

Given x-intercepts:

  • (0, 0)
  • (4, 0)

Therefore:

[tex]\implies f(x)=a(x-0)(x-4)[/tex]

[tex]\implies f(x)=ax(x-4)[/tex]

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

[tex]\implies 2a(2-4)=-8[/tex]

[tex]\implies -4a=-8[/tex]

[tex]\implies a=2[/tex]

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

[tex]f(x)=2x(x-4)[/tex]

Vertex form of a quadratic equation

[tex]f(x)=a(x-h)^2+k[/tex]

where:

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

Given:

  • vertex = (2, -8)

Therefore:

[tex]\implies f(x)=a(x-2)^2-8[/tex]

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

[tex]\implies a(0-2)^2-8=0[/tex]

[tex]\implies 4a-8=0[/tex]

[tex]\implies a=2[/tex]

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

[tex]f(x)=2(x-2)^2-8[/tex]

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)

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