Question

select the x - intercept(s) and y - intercept of the function, if one exists.
$f(x)=\frac{x^{2}-2x}{x^{2}-9x + 8}$
x - int: (2, 0)
y - int: (0, $\frac{1}{4}$)
x - int: none
y - int: none
y - int: (0, 0)
x - ints: (0, 0) and (2, 0)

Explanation:

Step1: Find x - intercepts

Set $y = f(x)=0$. So we set the numerator equal to 0 since $\frac{a}{b}=0$ when $a = 0$ and $b
eq0$. Given $f(x)=\frac{x^{2}-2x}{x^{2}-9x + 8}$, set $x^{2}-2x=0$. Factor out $x$: $x(x - 2)=0$. Solving $x(x - 2)=0$ gives $x=0$ or $x = 2$. When $x = 0$, $x^{2}-9x + 8=8
eq0$; when $x = 2$, $x^{2}-9x + 8=4-18 + 8=-6
eq0$. So the x - intercepts are $(0,0)$ and $(2,0)$.

Step2: Find y - intercept

Set $x = 0$. Substitute $x = 0$ into $f(x)=\frac{x^{2}-2x}{x^{2}-9x + 8}$. Then $f(0)=\frac{0^{2}-2\times0}{0^{2}-9\times0 + 8}=0$. So the y - intercept is $(0,0)$.

Answer:

x - ints: $(0,0)$ and $(2,0)$
y - int: $(0,0)$