Question

find possible formulas for each rational function drawn below.

Explanation:

Step1: Recall rational - function form

A rational function is of the form $y=\frac{f(x)}{g(x)}$, where $g(x)
eq0$. Vertical asymptotes occur at the values of $x$ that make $g(x) = 0$, and $x$ - intercepts occur at the values of $x$ that make $f(x)=0$.

Step2: Analyze problem 13

The vertical asymptotes are $x=-2$ and $x = 3$. So the denominator is $(x + 2)(x - 3)$. Since there are no $x$ - intercepts (the graph does not cross the $x$ - axis), the numerator can be a non - zero constant. Let the numerator be $1$. So the function is $y=\frac{1}{(x + 2)(x - 3)}=\frac{1}{x^{2}-x - 6}$.

Step3: Analyze problem 14

The $x$ - intercepts are $x=-3$ and $x = 1$, so the numerator is $(x + 3)(x - 1)=x^{2}+2x - 3$. The vertical asymptote is $x=-2$, so the denominator is $a(x + 2)$. Using the point $(0,-0.75)$:
Substitute $x = 0$ and $y=-0.75$ into $y=\frac{(x + 3)(x - 1)}{a(x + 2)}$.
$-0.75=\frac{(0 + 3)(0 - 1)}{a(0+2)}$,
$-0.75=\frac{- 3}{2a}$,
$-1.5a=-3$,
$a = 2$. So the function is $y=\frac{(x + 3)(x - 1)}{2(x + 2)}=\frac{x^{2}+2x - 3}{2x + 4}$.

Step4: Analyze problem 15

The $x$ - intercepts are $x=-2$ and $x = 3$, so the numerator is $(x + 2)(x - 3)=x^{2}-x - 6$. The vertical asymptotes are $x=-1$ and $x = 2$, so the denominator is $a(x + 1)(x - 2)=a(x^{2}-x - 2)$. Using the point $(0,-3)$:
Substitute $x = 0$ and $y=-3$ into $y=\frac{(x + 2)(x - 3)}{a(x + 1)(x - 2)}$.
$-3=\frac{(0 + 2)(0 - 3)}{a(0 + 1)(0 - 2)}$,
$-3=\frac{-6}{-2a}$,
$-3=\frac{3}{a}$,
$a=-1$. So the function is $y=\frac{(x + 2)(x - 3)}{-(x + 1)(x - 2)}=\frac{x^{2}-x - 6}{-x^{2}+x + 2}$.

Step5: Analyze problem 16

The $x$ - intercepts are $x=-2$ and $x = 3$, so the numerator is $(x + 2)(x - 3)=x^{2}-x - 6$. The vertical asymptotes are $x=-1$ and $x = 2$, so the denominator is $a(x + 1)(x - 2)=a(x^{2}-x - 2)$. Using the point $(0,3)$:
Substitute $x = 0$ and $y = 3$ into $y=\frac{(x + 2)(x - 3)}{a(x + 1)(x - 2)}$.
$3=\frac{(0 + 2)(0 - 3)}{a(0 + 1)(0 - 2)}$,
$3=\frac{-6}{-2a}$,
$3=\frac{3}{a}$,
$a = 1$. So the function is $y=\frac{(x + 2)(x - 3)}{(x + 1)(x - 2)}=\frac{x^{2}-x - 6}{x^{2}-x - 2}$.

Step6: Analyze problem 17

The $x$ - intercept is $x=-1$, so the numerator is $a(x + 1)$. The vertical asymptote is $x = 2$, so the denominator is $(x - 2)$. The horizontal asymptote is $y=-1$, so the leading - coefficient of the numerator and denominator must be in a ratio of $-1$. Let $a=-1$. The function is $y=\frac{-(x + 1)}{x - 2}=\frac{-x - 1}{x - 2}$. Using the point $(0,0.5)$:
Substitute $x = 0$ and $y = 0.5$ into $y=\frac{-(x + 1)}{x - 2}$, $0.5=\frac{-(0 + 1)}{0 - 2}=\frac{-1}{-2}=0.5$. So the function is $y=\frac{-(x + 1)}{x - 2}$.

Step7: Analyze problem 18

The $x$ - intercept is $x=-1$, so the numerator is $a(x + 1)$. The vertical asymptotes are $x = 1$, and the horizontal asymptote is $y = 1$, so the function is of the form $y=\frac{(x + 1)}{x - 1}$.

Answer:

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