## College Algebra (10th Edition)

x-intercepts: $-3$ and $3$ y-intercept: $-9$
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 $-9$.