Answer
$(-\infty,-4)\cup(3,\infty)$
Work Step by Step
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:
$(-\infty,-4)\cup(3,\infty)$
