Answer
See graph
Work Step by Step
The function that we want to graph is shown below. Follow the steps outline to graph the function.$$
\begin{aligned}
f(x) & =(x-5)^2-32.
\end{aligned}
$$ Step 1: We first determine the direction in which the function opens: The given function opens upward because the constant $a= 1$ is positive.
Step 2: Determine the vertex and the equation for the axis of symmetry. The vertex is $(5,-32)$ and the equation for the axis of symmetry is $x = -32$.
Step 3: Find the $y$ and $x$ intercepts. To find the $y$ intercept, set $x= 0$ to find the value of $y$. Similarly, to find the $x$ intercept, set $ y= 0$ to find the values of $x$. $$
\begin{aligned}
f(0)& =(0-5)^2-32\\
&= -7\\
\textbf{y-intercept: (0,-7)}.
\end{aligned}
$$ Set $ y = 0$ and solve. $$
\begin{aligned}
(x-5)^2-32 & =0\\
(x-5)^2&= 32 \\
x-5& = \pm\sqrt{32}\\
x&= 5+\pm\sqrt{32}.
\end{aligned}
$$Find the two separate solutions:
$$
\begin{array}{cl}
x= 5+\sqrt{32} &\text { or } x= 5-\sqrt{32} \\
x= 10.66 & \text { or } x=-0.66\\
\textbf{x-intercept: (10.66,0), (-0.66,0)}.
\end{array}
$$ Step 4: Find the domain and range of the function and sketch it.
Domain: All real numbers,
Range: $[-32, \infty) $.
