How do you write a rational function that has the following properties: a zero at x= 4, a hole at x= 7, a vertical asymptote at x= -3, a horizontal asymptote at y= 2/5?

1 Answer
Shwetank Mauria
Jan 31, 2018

Rational function is #(2x^2-22x+56)/(5x^2-20x-105)#

Explanation:

A zero at #x=4# means we have #(x-4)# as a factor in numerator;

a hole at #x=7# means, we have #x-7# a factor both in numerator as well as denominator;

a vertical asymptote at #x=-3# means #x+3# a factor in denominator only

a horizontal asymptote at #y=2/5# means highesr degrees in both numerator and denominator are equal and their coefficients are in ratio of #2:5#

Hence desired rational function is #(2(x-4)(x-7))/(5(x-7)(x+3))#

i.e. #(2x^2-22x+56)/(5x^2-20x-105)#

See its graph down below. Observe vertical asymptote #x=-3# and horizontal asymptote #y=2/5#. We have a zero at #x=4# as functtion passes through #(4,0)#. Hole is not seen as #x-7# cancels out, but we know, the function is not definedat this point.

graph{(2x^2-22x+56)/(5x^2-20x-105) [-10.67, 9.33, -4.4, 5.6]} 1

Related questions
See all questions in Graphs of Rational Functions
Impact of this question
23674 views around the world
You can reuse this answer
Creative Commons License