the vertical asymptote and horizontal asymptote
[tex]\begin{gathered} f(x)=\frac{x^2-1}{x^2+3x+2} \\ f(x)=\frac{(x-1)(x+1)}{(x+1)(x+2)} \\ f(x)=\frac{x-1}{x+2} \end{gathered}[/tex]
The vertical asymptote is x+2 = 0, x = -2
The horizontal asymptote is x - 1 = 0, x = 1
the x-intercept and the y-intercept
The intercept with x is calculated as 0 f(x)
[tex]\begin{gathered} f(x)=0 \\ \frac{x-1}{x+2}=0 \\ (x-1)=0 \\ x=1 \\ \end{gathered}[/tex]
The intercept with x is (1,0)
The intercept with y is calculated for x = 0
[tex]\begin{gathered} f(0) \\ \frac{x-1}{x+2} \\ \frac{0-1}{0+2} \\ \frac{-1}{2} \\ \end{gathered}[/tex]
The intercept with y is (0,-1/2)
