Question

find possible formulas for each rational function drawn below.
13.
14.

Explanation:

Step1: Recall the general form of a rational function

A rational function is of the form $y = \frac{f(x)}{g(x)}$, where vertical - asymptotes occur at the values of $x$ for which $g(x)=0$, and $x$ - intercepts occur at the values of $x$ for which $f(x) = 0$.

Step2: Analyze problem 13

The vertical asymptotes are $x=-2$ and $x = 3$. So the denominator $g(x)=(x + 2)(x - 3)=x^{2}-x - 6$. There is no $x$ - intercept shown, so we can assume the numerator is a non - zero constant. Let the numerator $f(x)=k$. Since the function passes through the origin $(0,0)$ (by symmetry, we can assume a simple case), substituting $x = 0$ and $y=0$ into $y=\frac{k}{(x + 2)(x - 3)}$, we get $0=\frac{k}{(0 + 2)(0 - 3)}$, which is not possible. Let's assume the function has a horizontal asymptote $y = 0$ (degree of numerator less than degree of denominator). A possible function is $y=\frac{1}{(x + 2)(x - 3)}=\frac{1}{x^{2}-x - 6}$.

Step3: Analyze problem 14

The vertical asymptote is $x=-2$, so the denominator $g(x)=(x + 2)$. The $x$ - intercepts are $x=-3$ and $x = 1$, so the numerator $f(x)=a(x + 3)(x - 1)=a(x^{2}+2x-3)$. The horizontal asymptote is $y = 1$, which means the degree of the numerator and denominator are the same and the ratio of the leading coefficients is 1. So $a = 1$. Also, we can use the point $(0,-0.75)$ to check. Substitute $x = 0$ into $y=\frac{(x + 3)(x - 1)}{x + 2}$, we have $y=\frac{(0 + 3)(0 - 1)}{0+2}=-\frac{3}{2}=-1.5$. But if we consider $y=\frac{(x + 3)(x - 1)}{2(x + 2)}=\frac{x^{2}+2x-3}{2x + 4}$, when $x = 0$, $y=\frac{-3}{4}=-0.75$.

Answer:

1. $y=\frac{1}{(x + 2)(x - 3)}$
2. $y=\frac{(x + 3)(x - 1)}{2(x + 2)}$