## Precalculus (10th Edition)

The intercepts are $(0,-9), (-3,0),$ and $(3,0)$.
1. In order to find the x-intercepts for the equation $y=x^2-9$, you will need to let $y = 0$ then solve for $x$: $0=x^2-9$ which we will then factor to get $0=(x+3)(x-3)$. 2. We then apply the Zero Factor Property by equating each factor to zero: So either $x+3=0$ or $x-3=0$. There are two solutions. To find $x$ of $x+3=0$ we let x standalone so $x=0-3$. The answer is $x=-3$. To find the $x$ of $x-3=0$ we let x standalone so $x=0+3$. The answer is $x=3$. Thus, the x-intercepts are $(-3,0)$ and $(3,0)$. 3. For the y-intercept we substitute $x$ with $0$: $y=0^2-9$ So, $y=-9$. The y-intercept is $(0,-9)$. 4. The Intercepts are $(0,-9), (-3,0)$, and $(3,0)$.