Answer to Question #15993 in Algebra for rick

Question

1. solve x^2+3x-4>0 show work
2. graph the function f(x)=-e^-2-3 show work

Answer 1

x^2+3x-4>0
First of all, let's solve x^2+3x-4=0
using Vieta's theorem we can find that x1=-4, x2=1
So x^2+3x-4=(x+4)(x-1)>0
Using interval notation:

__+___x=-4______-_____x=1___+____
So the answer is: x<-4, x>1