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#