Question

Write a rational function that has the following attributes:Vertical asymptote: x=-1, x=2Horizontal asymptote: y=-2x-intercept: x=3, x=-2(Write function as if you were typing it into your calculator)

Answer 1

You know that the Rational Function must have these attributes:

- Vertical asymptote:

`\begin{gathered} x=-1 \\ x=2 \end{gathered}`

- Horizontal asymptote:

`y=-2`

- The x-intercepts must be:

`\begin{gathered} x=3 \\ \\ x=-2 \end{gathered}`

Then, you need to remember that, by definition, Rational Functions have this form:

`(p(x))/(q(x))`

Where the numerator and the denominator are polynomials, and:

`q(x)\\e0`

Notice that the denominator cannot be zero.

• If the function has the form:

`y=(a)/(x-b)+c`

Then the Vertical asymptote is:

`x=b`

And the Horizontal asymptote is:

`y=c`

Therefore, knowing the Vertical asymptotes given in the exercise, you can determine that the denominator of the Rational Function is:

`q(x)=(x+1)(x-2)`

Simplifying it, you get:

`\begin{gathered} q(x)=(x)(x)-(x)(2)+(x)(1)-(1)(2) \\ q(x)=x^2-2x+x-2 \\ q(x)=x^2-x-2 \end{gathered}`

• Knowing that the Horizontal asymptote is:

`y=-2`

By definition:

-When the degree of the numerator is less than the degree of the denominator:

`y=0`

- When the degree of the numerator is greater than the degree of the denominator, the function does not have a Horizontal asymptote.

- When the degrees are equal, the Horizontal asymtopte is equal to the ratio of the leading coefficient of the numerator and the leading coefficient of the denominator.

Therefore, in this case, you know that the degree of the numerator is:

`degree\colon2`

And, since the leading coefficient of the denominator is 1, the leading coefficient of the numerator is:

`a=-2`

By definition, the x-intercepts of a Rational Function can be found when the numerator is equal to zero. Therefore, you can set up the following expression for the numerator:

`p(x)=-2(x-3)(x+2)`

Simplifying, you get:

`\begin{gathered} =-2\lbrack(x)(x)+(x)(2)-(3)(x)-(3)(2)\rbrack \\ =-2\lbrack x^2+2x-3x-6\rbrack \\ =-2\lbrack x^2-x-6\rbrack \\ =-2x^2+2x+12 \end{gathered}`

• Therefore, you can set up the following function:

`y=(-2x^2+2x+12)/(x^2-x-2)`

Hence, the answer is:

`y=(-2x^2+2x+12)/(x^2-x-2)`