How do you graph #f(x)=(x-2) / (x+2)#?

1 Answer
Nov 6, 2015

Just to run through some general points:

#f(x) = (x-2) / (x+2)#
What is f(x) when x = 0?
#(0-2)/(0+2) = -2/2 = -1#
Thus our first point on the graph is (0,-1).
What is x at 1?
#(1-2)/(1+2) = -1/3#
Thus our second point on the graph is (1, -1/3).
We continue in this manner until you have enough points on the graph.

Explanation:

One thing to note: when #x=-2#, we have #4/0# (we can't divide by zero!) so our graph will not reach x=-2. Thus we can see an asymptote in our graph at x=-2.

As we can see on the y axis, when x is positive it never reaches #f(x) = 1# (as we will always be dividing x by the larger number x+2). The opposite is true when x is negative, as the nominator x will always be larger than the denominator until x=0.

graph{(x-2)/(x+2) [-10, 10, -5, 5]}