I have a question about the rational function I recently posted. Would the range still be (-∞,2)U(2,∞) if the rational function is what Reiny posted "(2x^2-18)/(x^2+3x-10)." The y-intercept confuses me because its (0,1.8) and when I look at the line it passes through the horizontal asymptote.

Question

Reiny · Answer

Here is a wonderful webpage that let's your graph just about any curve

http://rechneronline.de/function-graphs/

enter your function with brackets in the form
(2x^2-18)/(x^2+3x-10)

set: "range x-axis from" -20 to 20
set: "range y-axis from" -20 to 20

you will see your y-intercept correct at (0,1.8)
and the two vertical asymptotes of x = -5 and x = 2
starting to show.

Horizontal asymtotes begin to show up only when x approaches ± infinity, so if you look to the far right, the curve approaches y = 2 from the bottom up, and if you look far to the left, the curve approaches y - 2 from the top down

It is very common for the curve to intersect the horizontal asymptote for reasonable small values of x.

If we set our function equal to 2,
(2x^2-18)/(x^2+3x-10) = 2 , and cross-multiply we get
2x^2 - 18x = 2x^2 + 6x - 20
-24x =-20
x = 20/24 = 5/6 , which is shown on the graph

But as x --> ±∞ , the function will never again reach the value of 2

(try it on your calculator, set x = 500 and evaluate
then let x = -500 and evaluate,
in the first case you should get 1.988... a bit below 2
in the 2nd case you should get 2.012... a bit above 2
the larger you make your x, the closer you will get to 2, but you will never reach it, and that is your concept of an asympote )

Enlia · Answer

Thank you very much!

Reiny · Answer

you are welcome, does it make sense now?

Enlia · Answer

Yes it does ^^