Question

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

Answer 1

Answer:

The vertex is at (1, -108).

Step-by-step 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:

$\displaystyle x=\frac{7+(-5)}{2}=\frac{2}{2}=1$

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).

Answer 2

Answer:

vertex (1, -108)

Step-by-step 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