Question

1. write a function that has zeros at x = 2 and x = 3. it has vertical asymptotes x = 5. it has a horizontal asymptote of y = -3.

Explanation:

Step1: Construct the numerator for zeros

Since the function has zeros at $x = 2$ and $x=3$, the numerator of the rational - function can be written as $a(x - 2)(x - 3)=a(x^{2}-5x + 6)$, where $a$ is a non - zero constant.

Step2: Construct the denominator for vertical asymptotes

Since the function has a vertical asymptote at $x = 5$, the denominator of the rational function can be written as $(x - 5)$.

Step3: Determine the value of the constant for horizontal asymptote

For a rational function $y=\frac{a(x^{2}-5x + 6)}{x - 5}=\frac{ax^{2}-5ax+6a}{x - 5}$, when the degree of the numerator is one more than the degree of the denominator, we can use long - division or the fact that for $y=\frac{ax^{2}+bx + c}{mx + n}$, the oblique asymptote can be found. But if we want a horizontal asymptote $y=-3$, we consider the general form of a rational function $y=\frac{a(x - 2)(x - 3)}{(x - 5)}$. As $x\to\pm\infty$, for a rational function $\frac{f(x)}{g(x)}$ where $\text{deg}(f)=\text{deg}(g)$, the horizontal asymptote is $y=\frac{\text{leading coefficient of }f}{\text{leading coefficient of }g}$. We want $y=-3$, so $a=-3$.

Answer:

$y=\frac{-3(x - 2)(x - 3)}{x - 5}=\frac{-3(x^{2}-5x + 6)}{x - 5}=\frac{-3x^{2}+15x - 18}{x - 5}$