Question

find any points of discontinuity for the two rational functions. describe any vertical or horizontal asymptotes and any holes.
y= 2x^2+3 (over) x^2+2

Answer 1

help @whpalmer4

Answer 2

\[\frac{ 2x^2+3 }{ x^2+2 }\]

Answer 3

Discontinuities will be at \(x^2+2=0\).

Answer 4

Since \(x^2\) is positive for all real numbers, there are no discontinuities with real numbers.

Answer 5

ok can you explain alittle more please

Answer 6

Vertical asymptotes are caused by division by \(0\). The horizontal ones are found by letting \(x\) go to \(\pm \infty\)

Answer 7

ok so why cant x^2 = -2

Answer 8

Well, think about it this way. \(-\sqrt{2}\times -\sqrt{2} = 2\)

Answer 9

You would have to let \(x =\pm\sqrt{2}i\)

Answer 10

ok so for this problem there are no discontinuity

Answer 11

yes

Answer 12

is it because we can not find a real number to make x^2+2 to equal 0

Answer 13

Yes.

Answer 14

are there any holes or anything else in the graph for the rest of the problem??

Answer 15

Polynomials are continuous. A polynomial divided by a polynomial is continuous so long as the bottom one is not zero.

Answer 16

There may be horizontal asymptotes.

Answer 17

ok how do we find that out?

Answer 18

are you there???

Answer 19

The horizontal asymptote we can find by dividing the first term of the numerator by the first term of the denominator. What do you get if you do that?

Answer 20

I already told you. Find them by letting x go to positive or negative infinity.

Answer 21

um not sure are we dividing 2x^2/x^2???

Answer 22

when we let \(x\rightarrow\infty\) the first term is much bigger than all of the other terms, so it essentially reduces the fraction to \[\frac{2x^2}{x^2}\]What does that equal?

Answer 23

umm 2???

Answer 24

Yes.

Answer 25

ok so the horizontal asymptote is 2??

Answer 26

\[
\frac{ 2x^2+3 }{ x^2+2 }
\cdot
\frac{x^{-2}}{x^{-2}} = \frac{2+3x^{-2}}{1+2x^{-2}}
\]

Answer 27

Yes, \(y=2\) is the horizontal asymptote.

@wio that's an elegant way of demonstrating it!

Answer 28

ok so if we use wio equation it will get the same answer??

Answer 29

Yes, his equation looks like this:
\[\frac{2+\frac{3}{x^2}}{1+\frac{2}{x^2}}\]
Hopefully you can see that as \(x\) gets large, the two little terms on the right get ever closer to 0.

Answer 30

and if they get arbitrarily close to 0, then the fraction is arbitrarily close to 2/1.

Answer 31

As you can see from this plot of @wio's fraction, even at only \(x=10\) the fraction is essential exactly 2.