find the values of x where the given function has vertical asymptotes. list them in increasing order. f(x) =…

Explanation:

Step1: Factor the denominator

Factor $x^{2}-2x - 3$. We use the formula $ax^{2}+bx + c=a(x - x_1)(x - x_2)$ where for $x^{2}-2x - 3$, $a = 1$, $b=-2$, $c=-3$. We find the roots of the quadratic equation $x^{2}-2x - 3=0$ using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $x=\frac{2\pm\sqrt{(-2)^{2}-4\times1\times(-3)}}{2\times1}=\frac{2\pm\sqrt{4 + 12}}{2}=\frac{2\pm4}{2}$. The roots are $x_1=-1$ and $x_2 = 3$, so $x^{2}-2x - 3=(x + 1)(x - 3)$.

Step2: Find the vertical asymptotes

A rational function $y=\frac{f(x)}{g(x)}$ has vertical asymptotes at the values of $x$ for which $g(x)=0$ and $f(x)\neq0$. Since $f(x)=1$ and $g(x)=(x + 1)(x - 3)$, we set $g(x)=0$. Then $(x + 1)(x - 3)=0$. By the zero - product property, if $ab = 0$, then $a = 0$ or $b = 0$. So $x+1=0$ gives $x=-1$ and $x - 3=0$ gives $x = 3$.

Answer:

$x=-1,3$