Answer
The graph is shown below.
Range $[-1,\infty)$.
Work Step by Step
The given quadratic function is
$f(x)=(x-4)^2-1$
The standard form of the parabola is
$f(x)=a(x-h)^2+k$
Compare with the original given function.
$a=1,h=4$ and $k=-1$
Step 1:- Parabola opens.
$a>0$, The parabola opens upward.
Step 2:- Vertex.
We have $h=4$ and $k=-1$. The vertex is $(h,k)=(4,-1)$.
Step 3:- $x-$intercepts.
Replace $f(x)$ with $0$.
$\Rightarrow 0=(x-4)^2-1$
Add $1$ to both sides.
$\Rightarrow 0+1=(x-4)^2-1+1$
Simplify.
$\Rightarrow 1=(x-4)^2$
Apply the square root property.
$\Rightarrow x-4=\sqrt{1}$ or $x-4=-\sqrt{1}$
Simplify.
$\Rightarrow x-4=1$ or $x-4=-1$
Add $4$ to both sides in each equation.
$\Rightarrow x-4+4=1+4$ or $x-4+4=-1+4$
Simplify.
$\Rightarrow x=5$ or $x=3$
Hence, the $x-$intercepts are $5$ and $3$. The parabola passes through $(5,0)$ and $(3,0)$.
Step 4:- $y-$intercept.
Replace $x$ with $0$.
$\Rightarrow f(0)=(0-4)^2-1$
Simplify.
$\Rightarrow f(0)=16-1$
Simplify.
$\Rightarrow f(0)=15$
Hence, the $y-$intercept is $15$. The parabola passes through $(0,15)$.
Step 5:- Graph the parabola
Use the points: vertex, $x-$intercepts and $y-$intercept to draw a parabola.
The axis of symmetry is $x=h=4$.
$A=(4,-1)$
$B=(5,0)$
$C=(3,0)$
$D=(0,15)$.
From the graph we determine the range of the function:
$[-1,\infty)$.
