Question

Solve this identifying holes, vertical asymptotes, and horizontal asymptotes for

f(x)=x²+7x+12/-x²-3x+4

Answer 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)=(x^2+7x+12)/(-x^2-3x+4)=((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:}`

`\lim\limits_(x\to\pm\infty)(x^2+7x+12)/(-x^2-3x+4)=\lim\limits_(x\to\pm\infty)(x^2\left(1+(7)/(x)+(12)/(x^2)\right))/(x^2\left(-1-(3)/(x)+(4)/(x^2)\right))=(1)/(-1)=-1\\\\\boxed{y=-1}`