Help! How do I graph this function?

Here are a few thoughts that may help:

First, all you need is to draw something; don't worry about being able to write an equation for the function. But it does have to be a function; what does that tell you?

Also, there will be many correct answers; don't waste time trying to decide between two possibilities. Just do something, and then decide whether anything fails with that choice. Have fun with it, and be willing to correct wrong choices later.

The domain and range tell you a box that your graph has to stay within, keep that in mind as you draw.

Then I would draw the intercepts and asymptotes (limits). Generally, draw the most definite things first, and fill in the more flexible parts between them. But draw lightly, in case you make a late discovery.

Please make some attempt, and show us what you've drawn, ideally with a bit of explanation of your choices in words. Then we'll have something to discuss.
 
Dr.Peterson said:
Here are a few thoughts that may help:

First, all you need is to draw something; don't worry about being able to write an equation for the function. But it does have to be a function; what does that tell you?

Also, there will be many correct answers; don't waste time trying to decide between two possibilities. Just do something, and then decide whether anything fails with that choice. Have fun with it, and be willing to correct wrong choices later.

The domain and range tell you a box that your graph has to stay within, keep that in mind as you draw.

Then I would draw the intercepts and asymptotes (limits). Generally, draw the most definite things first, and fill in the more flexible parts between them. But draw lightly, in case you make a late discovery.

Please make some attempt, and show us what you've drawn, ideally with a bit of explanation of your choices in words. Then we'll have something to discuss.
Click to expand...
image.jpg

I did this... I tried... I don’t really know the limit when x approaching 1
 
Yes, the last two bullet points are interesting. At first glance they seem contradictory. Here is a possible answer (sorry I could not work out a way of giving you a clue, so I just give you my thoughts directly).

Have you seen notation like this before:-
\(\displaystyle
f(x)=\begin{cases}
\text{g(x)} & x\neq1\\
1, & x=1
\end{cases}
\)

You could mention this in the description, and perhaps show it like this (click) on your graph
-------

Also, your graph at x=-2 seems wrong. The graph does not have to go through the x axis at a point where f(x)=0.

Otherwise, it looks very good to me.
 
What you have is good. You say that you don't know the limit at 1, but it is given to be 4. Basically draw the graph as if f(1)=4 but remove the one point (1,4) and place a point at (1,1) to satisfy f(1)=1
You need to make sure that f(-1)=f(-3) and that f(-1)>f(-2).
You almost have it. Please try some more and if you have trouble be specific about what you can't do.
 
As others have said, the main issue is to complete the graph around x=1, by making it go through (1, 4) and marking that as a "hole". You should have seen examples of such a removable discontinuity, but it is easy to think that is too weird when the idea is new to you!

What you did at x = -1, -2, -3 is valid; -1 and -3 don't have to be zeros, but nothing says they can't be.
 
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