Question

What is the y-value of the vertex of the function f(x)=-(x+3)(x-14)?

Answer 1

If we expand the product, we have

$f(x)=-(x+3)(x-14)=-x^2+11x+42$

The x-coordinate of the vertex is given by

$V_x=-\dfrac{b}{2a}=-\dfrac{11}{-2}=\dfrac{11}{2}$

Plug this value in the equation to find the correspondant y-value:

$f\left(\dfrac{11}{2}\right)=-\left(\dfrac{11}{2}\right)^2+11\left(\dfrac{11}{2}\right)+42=-\dfrac{121}{4}+\dfrac{121}{2}+42$