Answer
a) See graph
b) Domain: $(-\infty,\infty)$
Range: $[-16,\infty)$
c) Decreasing: $(-\infty,0)$
Increasing: $(0,\infty)$
Work Step by Step
We are given the function:
$f(x)=\dfrac{1}{4}x^2-16$
a) The function is in the form $y=ax^2+bx+c$.
Identify $a,b,c$:
$a=\dfrac{1}{4}$
$b=0$
$c=-16$
Because $a>0$, the graph opens up.
Determine the vertex:
$x_V=-\dfrac{b}{2a}=-\dfrac{0}{2\left(\dfrac{1}{4}\right)}=0$
$y_V=\dfrac{1}{4}(0^2)-16=-16$
$V(x_V,y_V)=(0,-16)$
The axis of symmetry is:
$x=x_V\Rightarrow x=0$
Determine the intercepts:
$x=0\Rightarrow y=\dfrac{1}{4}(0^2)-16=-16$
$y=0\Rightarrow \dfrac{1}{4}x^2-16=0\Rightarrow x^2=64\Rightarrow x=\pm 8$
Graph the function.
b) The domain of the function is:
$(-\infty,\infty)$
The range is:
$[-16,\infty)$
c) The function is decreasing on the interval:
$(-\infty,0)$
The function is increasing on the interval:
$(0,\infty)$
