How do you graph #y=(34x-2)/(16x+4)# using asymptotes, intercepts, end behavior?

1 Answer
Nov 23, 2016

Answer:

I have managed to present the rectangular hyperbola represented by your equation.. Vertical asymptote: #x =-1/4#; the horizontal one: #y = 17/2#. See explanation, for more.

Explanation:

graph{(y-17/2) (x+1/4)+9/4=0 [-40, 40, -20, 20]}

By actual division and reorganization,

#(y-17/2)(x+1/4)=-9/4#

This represents athe rectangular hyperbola with asymptotes #y

=17/2 and x = -1/4#.

The center is at the common point #(-1/4, 17/2)#..

The graph would clarify these statements.

As # y to 17/2, x to +-oo#.

As #x to -1/4, y to +-oo#.

Not easy to see in the graph, the x-intercept is 1/68 and y-intercept

is #-#1/2.

Note that the general equation of a rectangular hyperbola is of the

form

#(y-mx+a)(x+my+b) = c#,

enabling us to read the equations of the asymptotes as

#y -mx + a =0=x+my+b#