How do you graph #y=1/[x(x-2)]#?

Step 1. Find the #y#-intercepts.

#y = f(x) = 1/(x(x-2))#

#f(0) = 1/(0(0-2)) = 1/0#

There is no #y#-intercept.

Step 2. Find the #x#-intercepts.

#0 = 1/(x(x-2))#

#0 = 0#

There is no #x#-intercept.

Step 3. Find the vertical asymptotes.

Set the denominator equal to zero and solve for #x#.

# x(x-2) = 0#

#x = 0# or #x-2 = 0#

#x = 0# or #x=2#

There are vertical asymptotes at #x = 0# and #x = 2#.

Step 4. Find the horizontal asymptote.

The degree of the denominator is greater than the degree of the numerator, so

The horizontal asymptote is at #y=0# (the #x#-axis).

Step 5. Draw your axes and the asymptotes.

The vertical asymptotes divide the graph into three regions of #x#s.

(a) The left hand region has the #x#- and #y#-axes as asymptotes.

#f(-1) = 1/(-1)(-1-2) = 1/(-1)(-3)) = 1/(3) ≈ 0.33#.

The point at (#-1,0.33#) is in the second quadrant, so we have a "hyperbola" above the horizontal asymptote.

(b) The right hand region has #x =2# and the #x#-axis as asymptotes.

#f(3) = 1/(3(3-2)) = 1/(3×1) = 1/3 ≈ 0.33#.

The point at (#3,0.33#) is in the second quadrant, so we have a "hyperbola" above the horizontal asymptote.

So we have a mirror-image hyperbola in the first quadrant.

(c) In the middle region, we have

#f(1) = 1/(1(1-2)) = 1/(1(-1)) = 1/(-1) = -1# and

#f(1.5) = 1/(1.5(1.5-2)) = 1/(1.5(-0.5)) = 1/(-0.75) ≈ -1.33#

#f(0.5) = 1/(0.5(0.5-2)) = 1/(0.5(-1.5)) = 1/(-0.75) ≈-1.33#

The points at (#0.5, -1.33#), (#1,-1#), and (#1.5,-1.33#) are all below the #y#-intercept, so we have an "inverted parabola" between the vertical asymptotes.

And we have our graph.