I'm not quite sure where you got the 0 and 13 from for the range??
Okay, firstly; you need to know the general shape of
g(x)=x1 graph and the fact that this has one horizontal asymptote, and one vertical asymptote. You can even see that this function goes over ALL x and y values EXCEPT those asymptote ones. Here is a graph of your function and
g(x) for comparison:
Your function of
f(x)=x−32x+5 is a similar to
g(x) as you still have linear (function) over linear as the fraction therefore there will still essentially be ONE horizontal and ONE vertical asymptote as well. You are told the vertical one,
x=3, as
f(3) is not valid due to division by 0, this is the straight forward part.
However for the horizontal asymptote, it is trickier to determine it because we need to observe what happens to
f(x) as
x tends to positive/negative infinity, therefore I'd suggest dividing (using long division or otherwise)
2x+5 by
x−3 and expressing it in the form
A+x−3B which you'd find to be
f(x)=2+x−311 then consider what happens as x→±∞. Of course, you can see that the second term goes to 0 therefore you are left with 2; hence the function TENDS to 2 therefore
f(x)→2 as
x→±∞.
It never reaches 2, therefore it cannot be equal to it. Hence where the
f(x)∈R,
f(x)=2 comes from.
EXTRA NOTE: When it comes to finding the horizontal asymptote for this question (and any alike) you may notice that you can divide (just as you can multiply) the top and bottom of the fraction by something. In this case, you can divide top and bottom by
x.
You get left with
f(x)=1−x32+x5.
Now if you take the limit as
x→±∞ then you can see that
x5 AND
x3 both go to 0, and your overall fraction is
f(x)→12⇒f(x)→2 as we had before.
Hope this helps! I didn't want to just plain out do the question for you, but I'd find this difficult to explain otherwise without raising even more confusion so take it with you because it will definitely help on similar questions as you're now aware of how to approach them with a solid method. This should clear up any confusion you may have about the question.
As an example, find the range and domain for
h(x)=2x+13x−5