write a rational function that has a vertical asymptote at x=1, a horizontal asymptote at y=2, and a hole at x=-1.

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embizze | High School Teacher | (Level 2) Educator Emeritus

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A vertical asymptote will occur in a rational function when a factor of the denominator causes the denominator to be zero at a point, and that factor does not also occur in the numerator. A "hole" will occur if there is a common factor in the numerator and denominator that is zero for some value of x.

If the degree of the numerator and the denominator is the same, there will be a horizontal asymptote at the value of the quotient of the leading coefficients, assuming the numerator and denominator are written in standard form.

(1) To achieve the "hole" let x+1 be a factor of the numerator and denominator.

(2) To achieve the vertical asymptote, let x-1 be a factor of the denominator, but not the numerator.

(3) To achieve a horizontal asymptote of y=2, we make the numerator have the same degree as the denominator with the quotient of the leading coefficients=2.


Then `y=((2x+1)(x+1))/((x+1)(x-1))` ==> `y=(2x^2+3x+1)/(x^2-1)` is a rational function with the desired properties.


The graph:

** Note that most graphing utilities will not display the "hole". On a graphing calculator you can look at the table to see error at x=-1.**

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