How do you sketch the curve of #f(x) = (2x-3) / (2 x -9)^2#?

1 Answer
Sep 11, 2015

See explanation section.

Explanation:

#f(x) = (2x-3) / (2 x -9)^2#

Domain: #RR " - " {9/2}#

#x#-intercept: #3/2#
#y# intercept #-1/27#

Symmetry: None

Vertical Asymptote: #x = 9/2#
Horizontal Asymptote: #y = 0# (a.k.a. the #x#-axis)

#f'(x) = (-2(2x+3))/(2x-9)^3#

Sign analysis of #f'# shows that:
#f# is decreasing on #(-oo, -3/2)# and on #(9/2, oo)# and
#f# is increasing on #(-3/2, 9/2)#

Local Minimum #-1/24# (at #x=-3/2#)

#f''(x) = (8(2x+9))/(2x-9)^4#
Sign analysis of #f''# shows that:
#f# is concave down on #(-oo, -9/2)# and
#f# is concave up on #(-9/2, 9/2)# and on #(9/2, oo)#

#(-9/2, -1/27)# is the only inflection point.

The graph looks like this:

graph{y=(2x-3) / (2 x -9)^2 [-6.16, 11.62, -2.91, 5.985]}

(You can scroll in and out using a mouse wheel and also drag the graph around to see features more clearly.)