Question

Find the vertex of f(x) = 3(x-7)(x+5)

Please show your work.

^^ I’ve been having a bit of trouble with this question :( although it’s simple quadratics is not my strong suit

Answer 1

vertex (1, -108)

Explanation:

First find the zeros

f(x) = 3(x-7)(x+5)

0 = 3(x-7)(x+5)

Using the zero product property

x-7 = 0 x+5 = 0

x = 7 x = -5

The x coordinate of the vertex is the average of the zeros

(7+-5)/2 = 2/2 =1

To find the y coordinate, substitute the x coordinate into the equation

y = 3(1-7)(1+5) = 3(-6)(6) = -108

Answer 2

The vertex is at (1, -108).

Explanation:

We have the function:

`f(x)=3(x-7)(x+5)`

And we want to find its vertex point.

Note that this is in factored form. Hence, our roots/zeros are x = 7 and x = -5.

Since a parabola is symmetric along its vertex, the x-coordinate of the vertex is halfway between the two zeros. Hence:

`\displaystyle x=(7+(-5))/(2)=(2)/(2)=1`

To find the y-coordinate, substitute this back into the function. Hence:

`f(1)=3((1)-7)((1)+5)=3(-6)(6)=-108`

Therefore, our vertex is at (1, -108).