Question

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.

Answer 1

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

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

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 )

Answer 2

Thank you very much!

Answer 3

you are welcome,
does it make sense now?

Answer 4

Yes it does ^^