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)?

Question

Julia Adams · Accepted Answer

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#

Nathan Axel · Answer

undefined The equation for the rational function is f(x) = (x-2)(x-3)/(x(x-2)(x-3)).