Question

question
consider the graph of the function $f(x)=\frac{x^{3}-12x^{2}+27x}{x^{2}-9x}$. which is a removable discontinuity for the graph? select all that apply.
select all that apply:
$x = - 9$
$x=-3$
$x = 0$
$x = 3$
$x = 9$

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{3}-12x^{2}+27x=x(x^{2}-12x + 27)=x(x - 3)(x - 9)$. The denominator $x^{2}-9x=x(x - 9)$. So $f(x)=\frac{x(x - 3)(x - 9)}{x(x - 9)}$, $x
eq0,9$.

Step2: Simplify the function

After canceling out the common factors $x$ and $(x - 9)$ (for $x
eq0,9$), we get $f(x)=x - 3$ for $x
eq0,9$. A removable - discontinuity occurs when a factor in the denominator can be canceled out. The values of $x$ that make the original denominator zero are $x = 0$ and $x=9$. Since we can cancel out the factors corresponding to these $x$ - values in the rational - function, they are removable discontinuities.

Answer:

C. $x = 0$, E. $x = 9$