Question 767085
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All functions of an independent variable *[tex \LARGE x] in *[tex \LARGE \mathbb{R}^2] can be graphed using the same process.


Step 1:  Pick a value for *[tex \LARGE x] and substitute it into the function in place of *[tex \LARGE x] wherever it occurs.


Step 2:  Do the indicated arithmetic to determine the value of the function, *[tex \LARGE y], that results.


Step 3:  Create an ordered pair *[tex \LARGE (x,\,y)] using the value of *[tex \LARGE x] you selected in step 1 and the value of *[tex \LARGE y] you calculated in step 2.


Step 4:  Plot the point on an appropriately scaled coordinate plane.


Step 5:  Repeat steps 1 through 4 using a different value of *[tex \LARGE x] each time, a sufficient number times to develop a general idea of the shape of the curve.  For your problem, I would use the following set of values: -3, -2, -1, 0, 1, 2, 3, 4.  Don't forget that a negative exponent means the reciprocal, that is *[tex \LARGE a^{-n}\ =\ \frac{1}{a^n}]


If you do it correctly, you should have a curve that takes off quickly in a positive direction as *[tex \LARGE x] gets larger in a positive direction, and is asymptotic to the *[tex \LARGE x]-axis as *[tex \LARGE x] gets small.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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