what are the vertical asymptotes of $f(x)=\frac{10}{x^{2}-1}$\n$x = - 1$\n$x = 0$\n$x = 1$\n$x = 10$\ndone
what are the vertical asymptotes of $f(x)=\frac{10}{x^{2}-1}$\n$x = - 1$\n$x = 0$\n$x = 1$\n$x = 10$\ndone
Answer
Explanation:
Step1: Recall vertical - asymptote condition
Vertical asymptotes occur where the denominator of a rational function is zero and the numerator is non - zero. Set the denominator equal to zero: $x^{2}-1 = 0$.
Step2: Solve the equation for x
Factor the left - hand side: $(x + 1)(x - 1)=0$. Then, using the zero - product property, if $ab = 0$, then $a = 0$ or $b = 0$. So $x+1 = 0$ gives $x=-1$ and $x - 1=0$ gives $x = 1$. The numerator is 10 (non - zero for all real x).
Answer:
A. $x=-1$, C. $x = 1$