Answer
Vertex: $(1,4)$
$x$-intercepts: $-1$; $3$;
$y$-intercept: $-3$
Work Step by Step
Bring the function to the vertex form $f(x)=a(x-h)^2+k$:
$$\begin{align*}
f(x)&=(x^2-2x+1)-4\\
&=(x^2-2x+1)-4\\
&=(x-1)^2-4.
\end{align*}$$
Identify the constants $a$, $h$, $k$:
$$\begin{align*}
a&=1\\
h&=1\\
k&=-4.
\end{align*}$$
Determine the vertex of the function:
$$(h,k)=(1,-4).$$
The axis of symmetry is $x=1$. It opens upward because $a>0$.
Determine the $x$-intercepts by solving the equation $f(x)=0$:
$$\begin{align*}
(x-1)^2-4&=0\\
(x-1)^2&=4\\
x-1&=\pm 2\\
x-1=-2&\text{ or }x-1=2\\
x=-1&\text{ or }x=3.
\end{align*}$$
The $x$-intercepts are $-1$ and $3$.
Determine the $y$-intercept:
$$f(0)=(0-1)^2-4=-3.$$
The $y$-intercept is $-3$.
We plot the vertex $(1,-4)$ and the intercept points $(-1,0)$,
$(3,0)$ and $(0,-3)$, then join them to sketch the graph of the function:
