Question

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

Answer 1

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

Answer 2

`f(x) = (x + 1)(x - 7)`

Explanation:

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

`3 = - (b)/(2a)`

Let a=1, then we have

`3 = - (b)/(2)`

`b = - 2 * 3 = - 6`

So now our equation becomes:

`f(x) = {x}^(2) - 6x + c`

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

Let these factors be 1, and c.

Then

`c + 1 = - 6`

`c = - 6 - 1 = - 7`

Therefore the factors are :

1 and -7.

The function becomes:

`f(x) = {x}^(2) - 6x - 7`

The factored form is

`f(x) = (x + 1)(x - 7)`