Math help me please

Tutoring | Bookmark Us | FREE to Join! New User? Sign Up | Sign In | Help

Students Ask Questions--Tutors Answer
TutorAnswer.com

ask.

answer.

discover.

My Profile

Home > Math & Algebra > Open Questions

eb8524

Open Questions

Math help me please

I'm graphing a rational function. f(x) = (3x^2 + x - 4) / (2x^2 - 5x). The x-intercept is -4/3. The vertical asymptotes are x=0 and 5/2. Now my problem is I need to plot points between and beyond x-intercept and vertical asymptotes. What points do I choose from the graph. Hard to find out. Please help. Thank you.

Posted 154 day ago

Report Abuse

Find Interesting

Email to Friends

Bookmark

Subscribe to Answer Alert

Answers (3)

atta

Given the vertical asymptotes x = 0, 5/2 && x-intercept(s) of -4/3, the points of this graph would be drawn between 0 and 2.5. Therefore, you would have a series of x-intercepts approaching the vertical asymptote (x = 0) from the left and a series of x-intercepts approaching the vertical asymptote (x = 2.5) from the right.

J.C

Posted 123 day ago

( 0 )

Report Abuse

samzappala

By graphing this formula, I set the horizontal to -1.33 and 2.75; verticle -60 and 18. The two verticle bars (asymptotes) are at 0,0 and 2.5,0. Choose pts ranging from .02,19 to .24,2.5 to 1.08, 0. to 1.5, -2.5 to 1.68, -2.5 to 2.3,-15 to 2.4,-44.2. The more, the better. These pts are between the asymptotes. These pts are from upper left to lower rt. Any pt on this graph intersecting the curve is what you're looking for.

Posted 153 day ago

( 0 )

Report Abuse

math_fanatic

You can find other points on the graph by choosing an x value that isn't 0, 5/2.

If the part of the function you're looking at lies between x= 0 and x = 5/2 then that means you can't choose an x value less than or equal to 0 or greater than or equal to 5/2. You also can't choose an x value that allows y to be -4/3.

So what are some numbers that lie between x=0 and x= 5/2?

There's 1/2, 1, 3/2, 2 to try. As long as these values don't yield a y value of -4/3, the coordinates will lie
on the curve of the function and meet the criteria stated in the problem.

I'll do x= 1 and x =2 for you since I don't have a calculator handy:
When x = 1, f(x) or y = 0
When x = 2, y = 10/-2 = -5

I don't think this is a continuous function though. There are clearly x values that lie outside the asymptotes for which this function would be defined. Perhaps you are only looking at a small section of it?

Posted 153 day ago

( 0 )

Report Abuse

Other Open Questions in Math & Algebra

» ABSOLUTE VALUE INEQUALITIES

» How does the square of numbers relate to the product of numbers?

» What number contains an integral factor whose sum of two numbers are identical?

» Using the Commutative rule of algebra, write a formula that generalizes the product of numbers

» How to generalize from the product of numbers?

Edit your answer. Click save, when done.
Question Title	Math help me please
Your Answer	Character Count ( Max. - 5000 ) : 314