Respuesta :

gmany

[tex]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{The domain:} x\neq-4\ and\ x\neq1\\\\canceled\ (x+4)\\\\f(x)=\dfrac{x+3}{-(x-1)}=-\dfrac{x+3}{x-1}\ for\ x\neq-4\ and\ x=1[/tex]