Ramon222
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Explain how you could write a quadratic function in factored form that would have a vertex with an x-coordinate of 3 and two distinct roots

Respuesta :

kudzordzifrancis

Answer:

[tex]f(x) = (x + 1)(x - 7)[/tex]

Step-by-step explanation:

For a quadratic function to have a vertex with an x-coordinate of 3, then

[tex]3 = - \frac{b}{2a} [/tex]

Let a=1, then we have

[tex]3 = - \frac{b}{2} [/tex]

[tex]b = - 2 \times 3 = - 6[/tex]

So now our equation becomes:

[tex] f(x) = {x}^{2} - 6x + c[/tex]

We now find two factors of c that add up to -6.

Let these factors be 1, and c.

Then

[tex]c + 1 = - 6[/tex]

[tex]c = - 6 - 1 = - 7[/tex]

Therefore the factors are :

1 and -7.

The function becomes:

[tex]f(x) = {x}^{2} - 6x - 7[/tex]

The factored form is

[tex]f(x) = (x + 1)(x - 7)[/tex]

jordynxxz

Answer:

The vertex lies on the axis of symmetry, so the axis of symmetry is x = 3. Find any two x-intercepts that are equal distance from the axis of symmetry. Use those  x-intercepts to write factors of the function by subtracting their values from x. For example, 2 and 4 are each 1 unit from x = 3, so f(x) =  (x – 2)(x – 4) is a possible function.

Step-by-step explanation: Took test on Edg

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