Answer
See explanation
Work Step by Step
The quadratic function in standard form is:
$$f(x)=x^2+16x+61$$
We bring the equation to the vertex form:
$$f(x)=(x^2+16x+64)-64+61=(x+8)^2-3$$
Notice that the vertex is: $$(-8,-3)$$
The axis of symmetry is: $$x=-8$$
We determine the $x$-intercepts:
$$\begin{aligned}
(x+8)^2-3&=0\\
(x+8)^2&=3\\
x+8&=\pm\sqrt 3\\
x&=-8\pm\sqrt 3.
\end{aligned}$$
The $x$-intercepts are: $$(-8-\sqrt 3,0),(-8+\sqrt 3,0)$$
We graph the function.
