identify the vertical asymptote(s) of the function.\n$f(x)=\\frac{x + 2}{x^{2}-3x - 4}$\n$x=-4$\n$x=-2$\n$x=…

identify the vertical asymptote(s) of the function.\n$f(x)=\\frac{x + 2}{x^{2}-3x - 4}$\n$x=-4$\n$x=-2$\n$x=-1$\n$x=1$\n$x=2$\n$x=4$\ndone

identify the vertical asymptote(s) of the function.\n$f(x)=\\frac{x + 2}{x^{2}-3x - 4}$\n$x=-4$\n$x=-2$\n$x=-1$\n$x=1$\n$x=2$\n$x=4$\ndone

Answer

Answer:

C. ( x = - 1 ), F. ( x = 4 )

Explanation:

Step1: Factor the denominator

For the quadratic (x^{2}-3x - 4), we use the formula (ax^{2}+bx + c=a(x - x_1)(x - x_2)), where (x_{1,2}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}). Here (a = 1), (b=-3), (c=-4). Then (x=\frac{3\pm\sqrt{9+16}}{2}=\frac{3\pm5}{2}), so (x^{2}-3x - 4=(x + 1)(x - 4)). The function becomes (f(x)=\frac{x + 2}{(x + 1)(x - 4)}).

Step2: Find the vertical asymptotes

Vertical asymptotes occur where the denominator is zero (and the numerator is non - zero). Set the denominator ((x + 1)(x - 4)=0). Solving (x+1 = 0) gives (x=-1), and solving (x - 4=0) gives (x = 4). When (x=-1), the numerator (x + 2=-1 + 2=1\neq0); when (x = 4), the numerator (x+2=4 + 2=6\neq0).