Re: curve sketching!! help!

Quote:

Originally Posted by

**hellokitty999** g(x)= (2x-3.6)/(2x-12.96)^2

please help me find the following:

where is the function increasing and where is it decreasing?

where is it concave upwards and when is it concave downwards?

what is the relative minimum or the relative maximum?

what is the inflection point ?

the line x= ------- is a (vertical/horizontal) asymptote for the graph of g.

the line y= ------- is a (vertical/horizontal ) asymptote for the graph of g.

find the y-intercept and x-intercept.

Hi hellokitty999! :)

Can you determine the derivative of g(x)?

Re: curve sketching!! help!

Hi hellokitty999

The first step is to find the derivative of g(x).

Find the domain of g(x)

Then set g'(x) = 0 and solve for x

also set g'(x) = Does not exist, then solve for x, those will be your critical points.

Then using those critical points make an interval table using the domain and the critical points.

Then plug in the test values from the intervals into your g'(x).

Then if the value is positive then that part of interval is increasing.

If the value is negative then that part of interval is decreasing.

Your critical points will either be your relative max or relative min.

Find g''(x)

Then set g''(x) = 0 and solve for x

also set g''(x) = Does not exist, then solve for x, those will be your inflection points.

Then using those inflection points make an interval table using the domain and the inflection points.

Then plug in the test values from the intervals into your g''(x).

Then if the value is positive then that part of interval is concave upwards.

If the value is negative then that part of interval is concave downwards.

To find vertical asymptotes, you set g(x) = does not exist

To find horizontal asymptotes, you set the lim g(x) when x approaches to +inf and do another lim and set x approaches to -inf

To find x-int, set y=0 and solve for x

To find y-int, set x=0 and solve for y