Youssef H.

asked • 11/23/16

How to create the function?

create a rational function with a slant asymptote of y=x+8 And vertical asymptotes x=2 and x=-1

1 Expert Answer

By:

Mark M. answered • 11/23/16

Tutor
5.0 (243)

Mathematics Teacher - NCLB Highly Qualified

Youssef H.

So would it just be (x+8)( 2x+2)(2x-4)?
 
Report

11/23/16

Mark M.

No, it is a fraction:  x + 8)(x + 1)(x - 2)
                                (x - 2)(x + 1)
Report

11/23/16

Michael J.

If you graph this, you only get a straight line.  There is no curvature shape in the graph.  Since the graph has vertical asymptotes, the graph should be undefined at those specific points.  In other words, the graph should not result in a straight line if you graph it.
Report

11/23/16

Mark M.

I respectively disagree. The graph of f(x) = (x3 + 8x2 - 10x - 16) / (x2 - x - 2) has the specified asymptotes.
You might try it at wolframalpha.com.
Report

11/23/16

Michael J.

When I graphed on Desmos, it showed me a straight line.  But when I graphed it again, I did get a curvature graph. 
 
You do have the correct the function and I do understand your approach to getting your solution to the tea.  The graphing tool I used was being a jerk.  But let me assure, Desmos has never failed me when I had to check graphs.
Report

11/23/16

Michael J.

Hold on mark.  I saw an error in your calculation of the numerator part.
 
(x + 8)(x2 - x - 2) =
 
x3 - x2 - 2x + (8x2 - 8x - 16) =
 
x3 + 7x2 - 10x - 16
 
 
The coefficient of x2 in the expansion of the numerator should not be 8.
Report

11/23/16

Mark M.

Thank you for correcting my arithmetic error!
Report

11/23/16

Mark M.

Michael:
If the numerator is multiple of the denominator the resulting graph shall reduce to a straight line (the slant asymptote).
To avoid this the numerator must be altered (add or subtract something). So instead of 
x3 + 7x2 - 10x - 16, make the numerator x2 + 7x2 - 10x - 10.
The three asymptotes are correct
Report

11/23/16

Michael J.

I think the issue with this question is that it lacked one more constraint.  Perhaps a fixed point on the graph.  I replaced the -16 with -70, and it worked out.
Report

11/23/16

Mark M.

Lacking a constraint is not the issue.
In this case f(x) = P(x)/Q(x) such that P(x)/Q(x) = x + 8 plus a remainder (so the factors do not reduce).
Your substitution of -70 for the -16 created the needed remainder.
Great working with you!
Report

11/23/16

Still looking for help? Get the right answer, fast.

Ask a question for free

Get a free answer to a quick problem.
Most questions answered within 4 hours.

OR

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.