take+home+assignment

IF YOU HAVE NOT SOLVED PART i) OF THE ASSIGNMENT WE GOT ON SEPTEMBER 30th 2008, DO NOT LOOK AT THIS FILE.

It's just an eyeballed approximation using Geometer's Sketchpad and it might not be right, but the numbers look pretty good. Just as a reference for your own personal visualizations. Part ii) to follow.



Here is an algebraic solution to i) that corresponds to the sketchpad solution:

PART ii)

Geometer's sketchpad rough sketch of the functions/possible tangents: Algebraic solution to part ii)

(I am not sure if this is correct, this is my attempt, please give feedback so we can all be sure of how to solve the problem :) thanks!)

Heyy guys. Here's a full out explanation for the for the first part. Alright, what I did was to find the equations for both of the tangents for x^2+2x+3, using the same method we were using in class. The equations for those tangents are: - y=2x+3 - y=14x+39 After graphing both of those against both parabolas/paraboli (I don't remember the correct name) I found that the line y=2x+3 is a tangent for both. It touches the parabola x^2+2x+3 at the point (0,3) and the parabola -x^2+4x+2 at the point (1,5). So therefore the equation for the tangent of both parabolas is y=2x+3. night:) Debbie

Ladies and Gentlemen! (ok well that was probably not important enough to have an exclamation point... but I'm sure you liked it anyways...) I don't know if you will like this but tried part ii) from a different way than you might have expected... I completed the square on both of the quadratics to find the vertecies and found that f(x) had a vertex of (-1,2) and g(x) had a vertex of (2,4). I showed that said tangent is not tangent to g(x) by saying that if a line tangent to f(x) were to pass through the point (-5,2), one would be horizontal, y=2 (the y values for the point and f(x) are both 2) and because we know that the vertex of the fuction g(x) is (2,4) and we know it opens down, a horizontal line, y=2, has to cross it twice making it not tangent to the function g(x). As for the other tangent, I have not found a different way than the ones above...