Answer:
The expression is given below as
[tex]a(x)=\frac{x^2+2x-3}{x^2-1}[/tex]
The horizontal intercepts will be at y=0
[tex]\begin{gathered} a(x)=\frac{x^2+2x-3}{x^2-1} \\ \frac{x^2+2x-3}{x^2-1}=0 \\ x^2+2x-3=0 \\ x^2+3x-x-3=0 \\ x(x+3)-1(x+3)=0 \\ (x-1)(x+3)=0 \\ x-1=0,x+3=0 \\ x=1,x=-3 \end{gathered}[/tex][tex]\begin{gathered} x^2-1=0 \\ x^2=1 \\ x=\pm1 \\ x=1,x=-1 \end{gathered}[/tex]
Hence,
The horizontal intercepts is at x = -3
The vertical intercept is at x=0
[tex]\begin{gathered} a(x)=\frac{x^2+2x-3}{x^2-1} \\ y=\frac{0^2+2(0)-3}{0^2-1} \\ y=\frac{-3}{-1} \\ y=3 \end{gathered}[/tex]
Hence,
The vertical intercept is at y=3
A vertical asymptote is a vertical line that guides the graph of the function but is not part of it. It can never be crossed by the graph because it occurs at the x-value that is not in the domain of the function. A function may have more than one vertical asymptote.
[tex]\begin{gathered} a(x)=\frac{x^2+2x-3}{x^2-1} \\ a(x)=\frac{(x-1)(x+3)}{(x-1)(x+1)_{}} \\ \text{hence, the vertical aymspote will be at} \\ x+1=0 \\ x=-1 \end{gathered}[/tex]
Hence,
The vertical asymptotes is at x= -1
The horizontal asymptotes will be calculated using the image below
[tex]\begin{gathered} a(x)=\frac{x^2+2x-3}{x^2-1} \\ a=1,b=1,n=m=1 \\ y=\frac{a}{b} \\ y=\frac{1}{1} \\ y=1 \end{gathered}[/tex]
Hence,Do you have any questions about the steps to solve your question?
The horizontal asymptotes is y=1
The graph is represented below as
