Question

The axis of symmetry for the function f(x) = −x2 − 10x + 16 is x = −5. What are the coordinates of the vertex of the graph?

Answer 1

Just plugin -5 into the equation f(x) = −x^2 − 10x + 16
f(x) = −(-5)^2 − 10(-5) + 16 solve and the answer will be f(x) = 41

so  (-5, 41) is the answer. 

Answer 2

we know that

The equation of a vertical parabola in vertex form is equal to

$y=a(x-h)^{2} +k$

where

(h,k) is the vertex

if $a > 0$ -----> the parabola open upward (vertex is a minimum)

if $a < 0$ -----> the parabola open downward (vertex is a maximum)

In a vertical parabola the axis of symmetry is equal to

$x=h$

In this problem we have

$f(x)=-x^{2}-10x+16$

Complete the square

$f(x)-16=-(x^{2}+10x)$

$f(x)-16-25=-(x^{2}+10x+25)$

$f(x)-41=-(x+5)^{2}$

$f(x)=-(x+5)^{2}+41$

The parabola open downward

the vertex is the point $(-5,41)$ (is a maximum)

therefore

the answer is

$(-5,41)$