identify the vertical asymptote and the hole on the graph of the function $f(x)=\frac{x^{2}+x - 6}{x^{2}-6x…

identify the vertical asymptote and the hole on the graph of the function $f(x)=\frac{x^{2}+x - 6}{x^{2}-6x + 8}$. there is an asymptote at $x = square$. there is a hole at $(square,square)$.

identify the vertical asymptote and the hole on the graph of the function $f(x)=\frac{x^{2}+x - 6}{x^{2}-6x + 8}$. there is an asymptote at $x = square$. there is a hole at $(square,square)$.

Answer

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{2}+x - 6=(x + 3)(x - 2)$. The denominator $x^{2}-6x + 8=(x - 2)(x - 4)$. So $f(x)=\frac{(x + 3)(x - 2)}{(x - 2)(x - 4)}$.

Step2: Find the hole

A hole occurs when a factor can be canceled out. Since we can cancel out $(x - 2)$ (where $x\neq2$), to find the $y$-coordinate of the hole, substitute $x = 2$ into the simplified function. The simplified function is $y=\frac{x + 3}{x - 4}$ (for $x\neq2$). When $x = 2$, $y=\frac{2+3}{2 - 4}=-\frac{5}{2}$. So the hole is at $(2,-\frac{5}{2})$.

Step3: Find the vertical asymptote

The vertical asymptote occurs when the denominator of the simplified - function is zero. Set $x-4 = 0$, then $x = 4$.

Answer:

There is an asymptote at $x = 4$. There is a hole at $(2,-\frac{5}{2})$.