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