Question 1050568
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Since *[tex \Large f(2)\ =\ -3], you know that when *[tex \Large x\ =\ 2], the value of the function is -3, which is to say *[tex \Large y\ =\ -3].  Hence, the point *[tex \Large (2,-3)] is on the graph of your function.  Similarly, the point *[tex \Large (-4,7)] is also on the graph of your function.  Use the two-point form of an equation of a straight line (straight line because you are trying to derive a <i>linear</i> function), to derive an equation of the line.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ y_1\ =\ \left(\frac{y_1\ -\ y_2}{x_1\ -\ x_2}\right)(x\ -\ x_1) ]


where *[tex \Large \left(x_1,y_1\right)] and *[tex \Large \left(x_2,y_2\right)] are the coordinates of the given points.


Plug in your values, simplify and then replace *[tex \Large y] with *[tex \Large f(x)] or *[tex \Large g(x)] or *[tex \Large y(x)] or TheArtistThatWasFormerlyKnownAsPrince(x)...you get the idea.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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