How do you write an equation for a rational function that has a vertical asymptote at x=2 and x=3, a horizontal asymptote at y=0, and a y-intercept at (0,1)?

Jan 17, 2017

The desired rational function is $y = \frac{{x}^{2} + a x + 6}{{x}^{2} - 5 x + 6}$, where $a \ne - 5$

Explanation:

As we have vertical asymptote at $x = 2$ and $x = 3$,

we have in denominator $\left(x - 2\right) \left(x - 3\right)$ or ${x}^{2} - 5 x + 6$

and as we have horizontal asymptote at $x = 0$, we have in numerator same degree as that of denominator i.e. numerator is of the type ${x}^{2} + a x + b$

i.e. $y = \frac{{x}^{2} + a x + b}{{x}^{2} - 5 x + 6}$

We also have an intercept at $\left(0 , 1\right)$ i.e., when $x = 0$, $y = 1$

therefore $\frac{{0}^{2} + a \times 0 + b}{{0}^{2} - 5 \times 0 + 6} = 1$ i.e. $b = 6$

and $y = \frac{{x}^{2} + a x + 6}{{x}^{2} - 5 x + 6}$

Now we should have $a$, so that $\left(x - 2\right)$ or $\left(x - 3\right)$ are not a factor of $\left({x}^{2} + a x + 6\right)$ i.e. they are not its zeros.

i.e. ${2}^{2} + a \times 2 + 6 \ne 0$ i.e. $2 a \ne - 10$ or $a \ne - 5$ and

${3}^{2} + a \times 3 + 6 \ne 0$ i.e. $3 a \ne - 15$ or $a \ne - 5$

Hence the desired rational function is $y = \frac{{x}^{2} + a x + 6}{{x}^{2} - 5 x + 6}$, where $a \ne - 5$