MHB Graphing Functions With Asymptotes and Discontinuity

[ardentmed]

Hey guys,

I have a couple more questions about this problem set I've been working on. I'm doubting some of my answers and I'd appreciate some help.

Question:

Sorry guys, I can't upload my sketched graph. But I can describe it:

Given the four conditions, I got a dicontinuous graph that decreases on all intervals except for [2,4] in which case it increases.

for 1b, the function is not continuous because $\lim_{{x}\to{a}}$ DNE since different values are approached for the left-hand and right-hand limits.

As for 2, I got a vertical asymptote at x=-2 since "x+2" crosses out in both the nominator and denominator after factoring the function to f(x) = (2x-3)(x+2) / (x+2)(x+1)

And the domain is just (-infinity, infinity)

For 2c, I got x=/1 and y=/2 for horizontal and vertical asymptotes respectively.
Thanks in advance.

Moreover, for 2d, seeing as to how x =/ -1 because the limits approach different values from both sides, I computed the following domain:

(-infinity, -1) u (-1, infinity)

Thanks a ton for the help. You guys are the best.

 

[ardentmed]

Can anyone help me out with this question?

Thanks in advance.

 

[Dethrone]

I think the massive block of text and/or questions makes people less inclined to answer your question, but I will do so now.

Your description of the graph seems accurate. 1b) is not correct, check x = 2 and x = 4 and tell me what values they approach as you approach them from the left and right side.

 

[Dethrone]

2) If you cross a factor from the numerator and denominator, you don't get an asymptote, you get a hole. You have an asymptote (vertical) only at x = -1. The domain is slightly wrong, because you did not include where the domain is undefined or doesn't exist. 2c) Your asymptotes are wrong, 2d) You left out the point discontinuity (hole). The domain is continuous for all x-values such that it doesn't equal x = -1 and 2.

 

[ardentmed]

I think the massive block of text and/or questions makes people less inclined to answer your question, but I will do so now.

Your description of the graph seems accurate. 1b) is not correct, check x = 2 and x = 4 and tell me what values they approach as you approach them from the left and right side.

x=2 approaches 0 for both sides and f(2) = 0.

Also, x=4 approaches 2 for both sides and f(4) = 2.

Wouldn't that constitute continuity, or am I missing something crucial here?

Thanks for the help, guys.

- - - Updated - - -

2) If you cross a factor from the numerator and denominator, you don't get an asymptote, you get a hole. You have an asymptote (vertical) only at x = -1. The domain is slightly wrong, because you did not include where the domain is undefined or doesn't exist. 2c) Your asymptotes are wrong, 2d) You left out the point discontinuity (hole). The domain is continuous for all x-values such that it doesn't equal x = -1 and 2.

Alright, so for the domain, I computed:

(-$\infty$,-2)u(-2,-1)u(-1,$\infty$)

Am I on the right track?

 

[Dethrone]

x=2 approaches 0 for both sides and f(2) = 0.

Also, x=4 approaches 2 for both sides and f(4) = 2.

Wouldn't that constitute continuity, or am I missing something crucial here?

Thanks for the help, guys.

Yes, that constitutes continuity, and it was what I was trying to point out. In your original post, you said that the function wasn't continuous. I was merely pointing out that it is indeed continuous at 2 and 4.

Alright, so for the domain, I computed:

(-$\infty$,-2)u(-2,-1)u(-1,$\infty$)

Am I on the right track?

Yes, that is correct.

 

[ardentmed]

Yes, that constitutes continuity, and it was what I was trying to point out. In your original post, you said that the function wasn't continuous. I was merely pointing out that it is indeed continuous at 2 and 4. Yes, that is correct.

Here is my graph for 1a, by the way. Sorry, the webcam photos come out as inverted.

Thanks. View attachment 2909

 

[Dethrone]

Almost! Remember to use an open circle when the endpoint is not included in the interval. If you have two dots filled-in with one directly above the other, then your graph would fail the vertical line test.

 

[ardentmed]

Also, here is the (inverted) graph for 2.

As for asymptotes, I computed a vertical asymptote at x= -1, and horizonta asymptote at y=2.

The intercepts are: (3/2, 0) for the x intercept and (0, -3) for the y intercept. I rejected x = -2 as a possibility since that is where the point of discontinuity is situated at.

Thanks in advance. View attachment 2910

Edit: Oh, so for the first one, there should be a hole for x=0, right?

 

[Dethrone]

Also, here is the (inverted) graph for 2.

As for asymptotes, I computed a vertical asymptote at x= -1, and horizonta asymptote at y=2.

The intercepts are: (3/2, 0) for the x intercept and (0, -3) for the y intercept. I rejected x = -2 as a possibility since that is where the point of discontinuity is situated at.

Thanks in advance. View attachment 2910

Edit: Oh, so for the first one, there should be a hole for x=0, right?

Yup, and your graph looks correct.