## Thursday, November 16, 2006

### 4.5 Summary of Curve Sketching

To start things off, I'd like to remind Kyle to do the next blog.

this lesson is (as its title suggests) a summary of what we have learned thus far. I don't know about you, but for almost everything I learn, I make little devices or words to remember them. For this section, the things you need to remember when sketching a graph, given the equation are:
1. remember any Asymptotes
2. find the Intercepts of the graph -for y, plug in 0 for x, vice versa for x-intercepts
3. keep in mind the Domain
4. For time-saving's sake, see if you can apply Symmetry
5. use the derivative to find Intervals of increase/decrease on the graph
6. find Max's andmin's-- using the first derivative test, if f ' changes from positive to negative, it is a local max.
7. f '' =0 is a pt. of Inflection and where this f '' is greater than 0, it is Concave up.
8. Sketch the graph

You can rearrange the bold letters any way that helps you remember, but I personally like to think : I SAID MISC , or IM AIDS SIC
for any other creative ideas to remember these guidelines, comment on this blog....

for the graph of y = (5x^3)/(x+1)
asymptotes: when x is -1, there will be a vertical asymptote
intercepts: when x=0, y= 0.......... when y=0, x also must =0 place a dot at the origin
domain: when x gets increasingly negative, what happens? a negative over a negative equals a postive. so while the graph has an asymptote at x=-1, the graph will resume on the left of the x-axis and get increasingly large.
symmetry: there is no symmetry because the function is neither odd nor even.
Intervals: the der. is (using the quotient rule) [(x+1)(15x^2)-5x^3] / (x+1)^2

so this derivative starts out negative, but gets closer to zero, where it does some funky stuff, then becomes positive when x is positive. using this info, the graph will start out with a negative slope, then because of the asymptote become positive.

local max, min: the der. is 0 at -1.5 and 0 and undefined at -1
concavity: do some "simplifying" to get [(x+1)^2 (30xx+30x) - (10xxx+15xx)(2x+2)]/ (x+1)^4

the graph of f" is negative only between -1 and 0 so f(x) should be concave up most of the time

pts. of inflection: when f" = 0 x= 0

finally attempt to sketch the curve. You can check your graph on your calculator.

http://www.mathgraphs.com/mg_clc7e.html
advice for next time: use an easier equation : )

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