Question #15993

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

Expert's answer

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

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

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