## Precalculus (10th Edition)

$(-\infty,-4)\cup(3,\infty)$
We are given the inequality: $x^2+x>12$ Rewrite the inequality: $x^2+x-12>0$ $f(x)=x^2+x-12$ Determine the intercepts: $y$-intercept: $f(0)=0^2+0-12=-12$ $x$-intercepts: $x^2+x-12=0$ $(x+4)(x-3)=0$ $x+4=0$ or $x-3=0$ $x=-4$ or $x=3$ The $y$-intercept is -12; the $x$-intercepts are -4 and 3. Determine the vertex: $x_V=-\dfrac{b}{2a}=-\dfrac{1}{2(1)}=-0.5$ $y_V=f(x_V)=(-0.5)^2+(-0.5)-12=-12.25$ The vertex is $V(-0.5,-12.25)$ Graph the function. The graph is above the $x$-axis for: $x\in (-\infty,-4)\cup(3,\infty)$ The solution set is: $(-\infty,-4)\cup(3,\infty)$