Question

1. let $r(x)=\frac{x^{2}+x - 6}{x^{3}-5x^{2}+6x}$. which of the following values of $x$ are zeros on the graph of $r$? (a) $x = 2$ only (b) $x=-3$ only (c) $x = 0$ and $x = 3$ only (d) $x = 0,x = 2$, and $x = 3$

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{3}-5x^{2}+6x=x(x^{2}-5x + 6)=x(x - 2)(x - 3)$. The denominator $x^{2}+x - 6=(x + 3)(x - 2)$. So $r(x)=\frac{x(x - 2)(x - 3)}{(x + 3)(x - 2)}$.

Step2: Simplify the rational - function

Cancel out the common factor $(x - 2)$ (for $x
eq2$), we get $r(x)=\frac{x(x - 3)}{x + 3},x
eq2$.

Step3: Find the zeros

The zeros of a rational function occur when the numerator is equal to zero and the denominator is not equal to zero. Set the numerator $x(x - 3)=0$. Solving $x(x - 3)=0$ gives $x = 0$ or $x=3$. When $x = 0$, the denominator $0 + 3=3
eq0$; when $x = 3$, the denominator $3+3 = 6
eq0$. Also, when we consider the original function before simplification, we need to check if $x = 2$ makes the numerator zero. Substituting $x = 2$ into the original numerator $2^{3}-5\times2^{2}+6\times2=8-20 + 12=0$.

Answer:

C. $x = 0$ and $x = 3$ only