Answer
x-intercepts: $-3$ and $3$
y-intercept: $-9$
Work Step by Step
RECALL:
(1) To find the x-intercept/s of an equation, set $y=0$ then solve for $x$.
(2) To find the y-intercept/s of an equation, set $x=0$ then solve for $y$.
Set $y=0$ then solve for $x$ to find the x-intercept/s:
$y=x^2-9
\\0=x^2-9
\\0+9=x^2
\\9=x^2$
Take the square root of both sides:
$\pm \sqrt9 = \sqrt{x^2}
\pm 3 = x$
Thus, the x-intercepts are $-3$ and $3$.
Set $x=0$ then solve for $y$ to find the y-intercept/s:
$y=x^2-9
\\y=0^2-9
\\y=-9$
Thus, the y-intercept is $-9$.
