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