Create Rational Function crossing x-axis at 2, touching at

[Choppers]

One of my questions for homework is:

create a rational function that has the following characteristics: crosses at x-axis at 2; touches x-axis at -1; one vertical assymptote at x=5 and another at at x=6; and one horizontal assymptote y=3.

How would you do this problem

 

[wjm11]

Re: Rational Function

create a rational function that has the following characteristics: crosses at x-axis at 2; touches x-axis at -1; one vertical assymptote at x=5 and another at at x=6; and one horizontal assymptote y=3.

What have you tried? What are your thoughts? Did you read your book and example problems?

Here’s a start, with some simplified explanations:

Write the equation in factored form.

Factors in the denominator will define vertical asymptotes (as long as similar factors do not exist in the numerator and cancel them out to create “holes” (simplified explanation)): (x-5) and (x-6).

Factors in the numerator will define zeroes of the function: (x-2) and (x+1)^2. Note: the squared factor indicates a “bounce” by the function; the function will bounce off the x-axis rather than pass through it.

If the numerator and denominator are of the same degree, the leading coefficients of the numerator and denominator form a ratio that defines a horizontal asymptote.

Hope that helps get you started.

 

[Choppers]

so the answer would be?

x^3-3x-2
________

x^2-11x+30

i appreciate your quick response. hopefully this is right. if not can you elaborate more on what i did wrong and what i need to do differently

 

[wjm11]

so the answer would be?

x^3-3x-2
________

x^2-11x+30

So, did you try graphing this? If you do, I think you'll find you have not achieved the horizontal asymptote you desire. I'll leave that for you to figure out.