## Precalculus (10th Edition)

$(-3,3)$
We are given the inequality: $x^2-9<0$ $f(x)=x^2-9$ Determine the intercepts: $y$-intercept: $f(0)=0^2-9=-9$ $x$-intercepts: $x^2-9=0$ $(x+3)(x-3)=0$ $x+3=0$ or $x-3=0$ $x=-3$ or $x=3$ The $y$-intercept is -9; the $x$-intercepts are -3 and 3. Determine the vertex: $x_V=-\dfrac{b}{2a}=-\dfrac{0}{2(1)}=0$ $y_V=f(x_V)=0^2-9=-9$ The vertex is $V(0,-9)$. Graph the function. The graph is below the $x$-axis for: \$-3
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