how to find the vertex and the focus...and why do series have limiting sums? (1 Viewer)

[helz_h]

Hey, just some basic last minute clarifications before monday:

a) how do you find the vertex and focus of a parabola?

b) in the 2002 exam (I think) they asked a question about why a particular series had a limiting sum. I had no clue, I hadn't even thought of that before...and even after thinking about it, briefly haha, I still have no clue. Can you help me out? Why would a series have a limiting sum?

ALSO any tips on motion graphs and how to draw f'(x) from f(x) etc would me much appreciated. The idea seems to escape my brain fairly quickly...

Cheers, and good luck with your study

DON'T FORGET DAYLIGHT SAVINGS!!!

 

[airie]

Well, you know that for the standard parabola x²=4ay, the vertex would be at (0,0) and the focus at (0,a). Then just turn it about the origin 90 degrees at a time for the other three standard parabolas And all parabolas with horizontal or vertical axes of symmetry can be written in the form (x-h)²=4a(y-k) [vertical axis] or (y-k)²=4a(x-h) [horizontal axis] where (h,k) is the vertex, and (h,k+a) or (h+a,k) is the focus. So if you can figure out the values of h, k, a from the coordinates of points, you'd get the focus and the vertex

As for the limiting sum, it exists for a GP if and only if the common ratio is greater than -1 and less than 1, since the absolute value of each term would therefore become smaller and smaller, thus the sum would approach a limit. You can prove this result from the formula for calculating sums ie. a(1-rⁿ)/(1-r) where a is the first term, r the common ratio and n the number of terms. Therefore as n approaches infinity, rⁿ approaches zero, if and only if r is in the open interval (-1, 1). Thus, the sum approaches a/(1-r)

To draw the graph of f'(x) from f(x), just consider the gradients at various sections of the domain. If the graph is increasing, the value of f'(x) is positive, and negative if f(x) is decreasing. The value of f'(x) is zero when the gradient of f(x) is zero ie. at a max/min turning point, or a horizontal point of inflexion. And where f(x) is undefined, f'(x) is also undefined. Also, the degree of f'(x) would be one less than that of f(x), for all polynomial functions

Good luck for your exams

 

[P_Dilemma]

Crud, owned bu an 07'er. We're screwed.

-P_D

 

[SoulSearcher]

Not really, she's not doing the exam this year