Question 1165348
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I presume that you know how to use the Two-Point form of a line to create an equation of a line that passes through two given points, namely:


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


Where *[tex \Large (x_1,y_1)] and *[tex \Large (x_2,y_2)] are the given points


and then putting the equation in Slope-Intercept form, *[tex \Large y\ =\ mx\ +\ b], which is simply expressing the set of points comprising a line in the form of *[tex \Large y] as a function of *[tex \Large x] which can be expressed using function notation as *[tex \Large y(x)\ =\ mx\ +\ b].


Note that any point on this line can be expressed as *[tex \Large \(x,y(x)\)].


So, the two given points on your line are: *[tex \Large \(1,h(1)\)] and *[tex \Large \(-2,h(-2)\)], and since you are given the values of *[tex \Large h(1)] and *[tex \Large h(-2)], you know that your two points are *[tex \Large (1,4)] and *[tex \Large (-2,13)]


I'll let you take it from there.

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