Answer
$\{x|x2\}=(-\infty,-5)\cup(2,\infty)$
Work Step by Step
We are given the inequality:
$x^2+3x-10>0$
$f(x)=x^2+3x-10$
Determine the intercepts:
$y$-intercept: $f(0)=0^2+3(0)-10=-10$
$x$-intercepts: $x^2+3x-10=0$
$(x+5)(x-2)=0$
$x+5=0$ or $x-2=0$
$x=-5$ or $x=2$
The $y$-intercept is -10; the $x$-intercepts are -5 and 2.
Determine the vertex:
$x_V=-\dfrac{b}{2a}=-\dfrac{3}{2(1)}=-1.5$
$y_V=f(x_V)=(-1.5)^2+3(-1.5)-10=-12.25$
The vertex is $V(-1.5,-12.25)$
Graph the function.
The graph is above the $x$-axis for:
$x>2$
The solution set is:
$\{x|x>2\}=(-\infty,-5)\cup(2,\infty)$
