How do you write an equation with a vertical asymptote of #3#, slant asymptote of #y=x+1#, and #x# intercept at #2#?

1 Answer
Jun 22, 2016

Answer:

#f(x) = (x^2-2x)/(x-3)#

Explanation:

Let #f(x) = (n(x))/(d(x)# be the choosed fractional function.

It has a vertical asymptote at #3# so #d(x) = (x-3)#
It has a slant symptote given by #y = x+1# so #n(x)# degree must be #2#.

Now, the general polynomial of degree #2# is

#n(x) = a x^2+b x+ c# so

#f(x) = (a x^2+b x+ c)/(x-3)#

We know also that #f(2) = 0# and

#a x^2+b x+ c=(x-3) (x+1) + d# because

#f(x) = (a x^2+b x+ c)/(x-3) = (x+1)+d/(x-3)# so for big values of #abs x#

#f(x) approx (x+1)#

then

#a x^2+b x+ c = x^2-2x+d-3#

so #a = 1, b = -2, c = d-3# but from #f(2) = 0# we extract

#0 = 4a+2b+c = 4-4+c->c=0->d=3#

Finally

#f(x) = (x^2-2x)/(x-3)#

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