Mathematics, 27.06.2019 00:30 sarah957
Solve this identifying holes, vertical asymptotes, and horizontal asymptotes for f(x)=x²+7x+12/-x²-3x+4
Answers: 1
![x^2+7x+12=x^2+4x+3x+12=x(x+4)+3(x+4)=(x+4)(x+3)\\\\-x^2-3x+4=-(x^2+3x-4)=-(x^2+4x-x-4)\\\\=-[x(x+4)-1(x+4)]=-(x+4)(x-1)\\\\f(x)=\dfrac{x^2+7x+12}{-x^2-3x+4}=\dfrac{(x+4)(x+3)}{-(x+4)(x-1)}\\\\\text{Vertical asymptotes:}\\\\(x+4)(x-1)=0\iff x+4=0\ \vee\ x-1=0\\\\\boxed{x=-4\ and\ x=1}\\\\\text{Horizontal asymptotes:}](/tpl/images/0021/3905/6144d.png)
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