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# Help with range?

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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)=(2x3+5x2-3x)/(x2-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!

Sep 25, 2019

#1
+106539
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(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

Sep 25, 2019
edited by CPhill  Sep 25, 2019
#2
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Thank you so much! :)

You are the best!

RandomEpicGamer  Sep 25, 2019
#3
+106539
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I made a slight mistake on the first posting.....there should be brackets next to 3.343  and 14.657

CORRECTED!!!

CPhill  Sep 25, 2019
#4
+15
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Okay! Thanks again!

RandomEpicGamer  Sep 25, 2019
#5
+6046
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OMG IT'S CORNER FOR YOU!!!!

:D :D :D

Rom  Sep 25, 2019