Question

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 $\left( {0,1} \right)$?

Answer 1

Hint: We are the horizontal and vertical asymptotes and a y-intercept. We have to find an equation of rational function using the asymptotes and y-intercept. First, find the factors at the denominator using vertical asymptotes. Then, multiply the factors to obtain a polynomial. Then, use the horizontal asymptote to obtain the degree of the polynomial at the numerator. Then, substitute the value of the y-intercept to the numerator and denominator. Then, determine the value of a, by substituting values of vertical asymptotes in the expression at the numerator which must be not equal to 0.

Complete step by step solution:
Given that the vertical asymptotes are at $x = 2$ and $x = 3$ which means the function has $\left( {x - 2} \right)$ and $\left( {x - 3} \right)$ as factors in the denominator.

Now, we will multiply the factors at the denominator to obtain the polynomial.

$ \Rightarrow \left( {x - 2} \right)\left( {x - 3} \right) = {x^2} - 2x - 3x + 6$

On combining like terms, we get:

$ \Rightarrow \left( {x - 2} \right)\left( {x - 3} \right) = {x^2} - 5x + 6$

Now, we are given the horizontal asymptote at $y = 0$ which means the highest degree of the numerator is equal to the highest degree of denominator that is 2 and the ratio of leading coefficients is one.

Now, we will write the polynomial at the numerator with highest degree 2.

$ \Rightarrow {x^2} + ax + b$

Write the rational function by substituting the values of numerator and denominator.

$ \Rightarrow y = \dfrac{{{x^2} + ax + b}}{{{x^2} - 5x + 6}}$

Now, y-intercept of the function is $\left( {0,1} \right)$. Thus, substitute $x = 0$ and $y = 1$ into the rational function.

$ \Rightarrow \dfrac{{{{\left( 0 \right)}^2} + a\left( 0 \right) + b}}{{{{\left( 0 \right)}^2} - 5\left( 0 \right) + 6}} = 1$

On simplifying the expression, we get:

$ \Rightarrow \dfrac{b}{6} = 1$

$ \Rightarrow b = 6$

Now, write the rational function by substituting the value of b.

$ \Rightarrow y = \dfrac{{{x^2} + ax + 6}}{{{x^2} - 5x + 6}}$

Now, the value of $a$ cannot be equal to $ - 5$ so that $\left( {x - 2} \right)$ and $\left( {x - 3} \right)$ cannot be factors of the numerator.

The rational function is $\dfrac{{{x^2} + ax + 6}}{{{x^2} - 5x + 6}}$ where $a \ne - 5$

Note: The students please note in the vertical asymptotes of rational functions, $f\left( x \right) = \dfrac{{p\left( x \right)}}{{q\left( x \right)}}$ the $p\left( x \right)$ and $q\left( x \right)$ have no factors common. The vertical asymptotes of the rational function obtained at points where the denominator is equal to zero. The students must also remember the properties of rational function such as:
If the horizontal asymptote is equal to the x-axis, then in that case the degree of the numerator is less than or equal to the degree of the denominator. If the rational function has no asymptote then the degree of the numerator is greater than degree of denominator.