I got a quiz back and I didn't understand why I missed a question. The teacher explained it, something like "we read range from the bottom to the top," but I still don't understand it.

Here's the equation: f(x)=(2x^{3}+5x^{2}-3x)/(x^{2}-x)

I got the range to be (-inf, 3.343] U [14.657, inf) using my graphing calculator.

She said that the range was (-inf, 3) U (3, 3.343] U [14.657, inf).

Now, don't get me wrong, I know there's a hole at (0, 3), but I thought that since there was another point where it crossed the y-axis, being (-1, 3), that 3 would be a valid value of the range.

Who's right and who's wrong? If my teacher is right can someone give me an explanation or a helpful video or something?

Thanks!

RandomEpicGamer Sep 25, 2019

#1**+2 **

(2x^3 + 5x^2 - 3x)

f(x) = ______________ =

x^2 - x

x (2x^2 + 5x - 3)

______________ =

x (x - 1)

2x^2 + 5x - 3

___________ =

x - 1

(2x - 1) ( x + 3)

____________

x - 1

We have a "hole" at x = 0 and a vertical asymptote at x = 1

We also have a "slant" asymptote determined as follows :

2x + 7

x^2 - x [ 2x^3 + 5x^2 - 3x ]

2x^3 - 2x^2

________________

7x^2 - 3x

7x^2 - 7x

__________

So.....the slant asymptote is y = 2x + 7

The range is (-inf, 3.343] U [ 14.657, inf)

Note that the graph crosses the line y = 3.....so......it definitely exists when y = 3

See the graph here : https://www.desmos.com/calculator/e5di4axkpw

CPhill Sep 25, 2019