Answer
The graph is shown below.
Range $[-1,\infty)$.
Work Step by Step
The given function is
$f(x)=2(x+2)^2-1$
The standard form of the parabola is
$f(x)=a(x-h)^2+k$
Compare with the original given function.
$a=2,h=-2$ and $k=-1$
Step 1:- Parabola opens.
$a>0$, The parabola opens upward.
Step 2:- Vertex.
The value of $h=-2$ and $k=-1$. The vertex is $(h,k)=(-2,-1)$.
Step 3:- $x-$intercepts.
Replace $f(x)$ with $0$.
$\Rightarrow 0=2(x+2)^2-1$
Add $1$ to both sides.
$\Rightarrow 0+1=2(x+2)^2-1+1$
Simplify.
$\Rightarrow 1=2(x+2)^2$
Divide both sides by $2$.
$\Rightarrow \frac{1}{2}=\frac{2(x+2)^2}{2}$
Simplify.
$\Rightarrow \frac{1}{2}=(x+2)^2$
Apply the square root property.
$\Rightarrow x+2=\sqrt{\frac{1}{2}}$ or $x+2=-\sqrt{\frac{1}{2}}$
Subtract $2$ from both sides in each equation.
$\Rightarrow x+2-2=\sqrt{\frac{1}{2}}-2$ or $x+2-2=-\sqrt{\frac{1}{2}}-2$
Simplify.
$\Rightarrow x=\sqrt{\frac{1}{2}}-2$ or $x=-\sqrt{\frac{1}{2}}-2$
Hence, the $x-$intercepts are $\sqrt{\frac{1}{2}}-2$ and $-\sqrt{\frac{1}{2}}-2$. The parabola passes through $(\sqrt{\frac{1}{2}}-2,0)$ and $(-\sqrt{\frac{1}{2}}-2,0)$.
Step 4:- $y-$intercept.
Replace $x$ with $0$.
$\Rightarrow f(0)=2(0+2)^2-1$
Simplify.
$\Rightarrow f(0)=8-1$
Simplify.
$\Rightarrow f(0)=7$
Hence, the $y-$intercept is $7$. The parabola passes through $(0,7)$.
Step 5:- Graph.
Use the points vertex, $x-$intercepts and $y-$intercept to draw a parabola.
The axis of symmetry is $x=h=-2$.
$A=(-2,-1)$
$B=(\sqrt{\frac{1}{2}}-2,0)$
$C=(-\sqrt{\frac{1}{2}}-2,0)$
$D=(0,7)$.
From the graph the range of the function is
$[-1,\infty)$.
