Answer
See graph
Work Step by Step
We are given the function:
$f(x)=-\sqrt{64-16x^2}$
Determine the domain:
$64-16x^2\geq 0$
$16(4-x^2)\geq 0$
$4-x^2\geq 0$
$x^2\leq 4$
$x\in[-2,2]$
Determine the $x$-intercepts:
$y=0$
$64-16x^2=0$
$16x^2=64$
$x^2=4$
$x=\pm 2$
Determine the $y$-intercept:
$x=0$
$y=-\sqrt{64-16(0^2)}=-\sqrt {64}=-8$
We can write:
$y=-\sqrt{64-16x^2}$
$y^2=64-16x^2$
$16x^2+y^2=64$
$\dfrac{16x^2}{64}+\dfrac{y^2}{64}=1$
$\dfrac{x^2}{4}+\dfrac{y^2}{64}=1$
As $y=-\sqrt{64-16x^2}\leq 0$, the function's graph is the lower half of the above ellipse.
Graph the function:
