How do you graph #y=(x^2-2x)/(2x+3)# using asymptotes, intercepts, end behavior?

1 Answer
Aug 24, 2016

Answer:

graph{(x^2-2x)/(2x+3) [-20, 20, -10, 10]}

Explanation:

#y=(x^2-2x)/(2x+3)#

a)
You can already tell that it has a vertical asymptote. Just let the denominator equal to zero and solve for #x#.
#2x+3=0# so #x=-3/2# (evident on the graph above).

b)
#y# shows zero values when #x=0# and #x=2#

c)
Next, if #x->+-oo#, #y->+-oo#
Roughly speaking;
#y=(oo^2)/(oo)=oo# and #y=oo^2/-oo=-oo#
*Mathematicians hate seeing the infinity symbol used like this. I think it is because infinity cannot simply be seen as a number but more of a concept.

d)
Finally, find the turning points;
#dy/dx=2/(2x+3)^2(x^2+3x-3)=0#
In other words; #x^2+3x-3=0#
Using #(-b+-sqrt(b^2-4ac))/2# you will find #x=(-3+-sqrt(21))/2#.

So far, you have found these important points:
-Asymptote #x=-3/2#
-Points #(0,0)# and #(2,0)#
-#(x->oo,y->oo)# and #(x->-oo,y->-oo)#
-Turning points at #x=(-3+-sqrt(21))/2#

These should be enough to help you see the shape of your graph.
Cheers.