Question 465060
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A couple of facts to consider.  The equation is linear, meaning the graph is a straight line.  Recall from geometry that you only need 2 points to define a line.  Also recall that any line consists of an infinite number of points.  The instructions in "course compass" (whatever that is) that tell you to select three values to substitute for *[tex \Large x] is overkill -- you only absolutely have to have 2 points (However, remember that it is academically career limiting to ignore instructor or textbook instructions -- just sayin').  Furthermore, in any given situation,  it doesn't matter what values you choose for *[tex \Large x], so long as you do the arithmetic correctly.  Generally, it is easier to do the arithmetic if you choose small integers -- like the ones selected by your "compass" thing.


As for using zeros to replace the variables, that is a slightly different technique for graphing a straight line.  This takes advantage of the characteristics of a couple of special points on a line in a Cartesian plane.  The point where the line crosses (i.e. intersects or intercepts) the *[tex \Large y]-axis is called the *[tex \Large y]-intercept and has the characteristic that the *[tex \Large x]-coordinate of the ordered pair is equal to zero.  The upshot is if you replace the *[tex \Large x] in your equation with zero, then you will get some value for *[tex \Large y], let's call that value *[tex \Large b] and then the ordered pair for the *[tex \Large y]-intercept is *[tex \Large (0,b)].  Likewise, if you replace the *[tex \Large y] in the equation with zero, you discover the value of the *[tex \Large x]-coordinate of the *[tex \Large x]-intercept, and the point is then *[tex \Large (a,0)].  The problem with using this method on the problem you posed in this posting is that the graph goes through the origin, *[tex \Large (0,0)], which is to say that the *[tex \Large x] and *[tex \Large y] intercepts are the same point in this case.  Since you need two distinct points to define a line, using the intercepts in this case won't work.


The thing to remember is that there is nothing sacred about any of the numbers anyone suggests as values to substitute.  Do what is convenient for you and makes the arithmetic easier for you, unless you have specific instructions to do something differently.  Capisce?


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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