Answer
The axis of symmetry of the given function is $x=-1$; the domain is $\left( -\infty ,\infty \right)$ , and the range is $\left[ 4,\infty \right)$.
Work Step by Step
The graph of the given function can be drawn by following the steps as given below.
On comparing the function $f\left( x \right)={{\left( x+1 \right)}^{2}}+4$ with the standard form of a parabola $f\left( x \right)=a{{\left( x-h \right)}^{2}}+k$ , it is observed that:
$a=1,h=-1,k=4$
Since the vertex of the parabola is given by $\left( h,k \right)$ , and here, $h=-1,k=4$. So, the vertex is at $\left( -1,4 \right)$.
To find the $y$ -intercept, find $f\left( 0 \right)$ as:
$\begin{align}
& f\left( 0 \right)=-1{{\left( 0+1 \right)}^{2}}+4 \\
& =-1+4 \\
& =3
\end{align}$
Then, the y-intercept is 3.
To calculate the x-intercept, equate the function $f\left( x \right)$ to $0$ , that is put $f\left( x \right)=0$ , as follows:
$\begin{align}
& f\left( x \right)=-{{\left( x+1 \right)}^{2}}+4 \\
& 0=-{{\left( x+1 \right)}^{2}}+4 \\
& {{\left( x+1 \right)}^{2}}=4
\end{align}$
This implies, $x+1=2\text{ or }x+1=-2$.
Thus,
$\begin{align}
& x+1=2 \\
& x=2-1 \\
& x=1
\end{align}$
And
$\begin{align}
& x+1=-2 \\
& x=-2-1 \\
& x=-3
\end{align}$
Therefore, the x-intercepts are 1 and $-3$.
Lastly, since the coefficient of ${{x}^{2}}$ is positive, that is, $1$ , therefore, the parabola opens upwards.
As can be observed from above graph, the axis of symmetry of the parabola is given as $x=-1$.
Also, since the parabola opens upwards, its minimum value is 4 and its maximum value is $\infty $. Thus, the range of the function is $\left[ 4,\infty \right)$.
And, the function is defined for all real values of x, thus its domain is given by $\left( -\infty ,\infty \right)$.
