Answer
See below:
Work Step by Step
We have to draw the graph of the function $f\left( x \right)={{\left( x-2 \right)}^{2}}-1$ as follows:
Step 1:
Find the vertex by comparing with standard form of the equation ${{\left( x-h \right)}^{2}}=4a\left( y-k \right)$:
Vertex is at $\left( 2,-1 \right)$.
Step 2:
Also, find the $x-\text{intercept}$ and $y-\text{intercept}$.
The $x-\text{intercept}$ is as follows:
Let $y=0$ in equation (I), then
$\begin{align}
& y={{\left( x-2 \right)}^{2}}-1 \\
& 0={{\left( x-2 \right)}^{2}}-1 \\
& \pm \sqrt{1}=x-2
\end{align}$
$x=3$ And $1$
So, intercepts are $\left( 3,0 \right)$ and $\left( 1,0 \right)$.
And the $y-\text{intercept}$ is as follows:
Place $x=0$ in equation (I):
$\begin{align}
& y={{\left( x-2 \right)}^{2}}-1 \\
& y={{\left( 0-2 \right)}^{2}}-1 \\
& y=4-1 \\
& y=3
\end{align}$
So, the intercept is $\left( 0,3 \right)$.
Step 2:
Now, use a smooth curve to join the intercept points.
And the graph of the function $f\left( x \right)={{\left( x-2 \right)}^{2}}-1$ intersects the $x-\text{axis}$ at the point $\left( 1,0 \right),\left( 3,0 \right)$ and $y-\text{axis}$ at the point $\left( 0,4 \right)$.
