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)?

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Answer 1 · 2017-01-17T08:25:59

The desired rational function is $y=(x^2+ax+6)/(x^2-5x+6)$ , where $a!=-5$

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

we have in denominator $(x-2)(x-3)$ or $x^2-5x+6$

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 $x^2+ax+b$

i.e. $y=(x^2+ax+b)/(x^2-5x+6)$

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

therefore $(0^2+axx0+b)/(0^2-5xx0+6)=1$ i.e. $b=6$

and $y=(x^2+ax+6)/(x^2-5x+6)$

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

i.e. $2^2+axx2+6!=0$ i.e. $2a!=-10$ or $a!=-5$ and

$3^2+axx3+6!=0$ i.e. $3a!=-15$ or $a!=-5$

Hence the desired rational function is $y=(x^2+ax+6)/(x^2-5x+6)$ , where $a!=-5$