Question 637114
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*[tex \LARGE f(x)] means the value of the function *[tex \LARGE f] at *[tex \LARGE x].


Your function is in slope-intercept form, namely *[tex \LARGE y\ =\ mx\ +\ b], where *[tex \LARGE f(x)\ =\ y], *[tex \LARGE m] is the slope, and *[tex \LARGE b] is the *[tex \LARGE y]-coordinate of the *[tex \LARGE y]-intercept.  To find the *[tex \LARGE x]-coordinate of the *[tex \LARGE x]-intercept, substitute *[tex \LARGE 0] for *[tex \LARGE y] and then solve the result for *[tex \LARGE x].  Note:  The *[tex \LARGE x]-coordinate of the *[tex \LARGE y]-intercept is *[tex \LARGE 0], and the *[tex \LARGE y]-coordinate of the *[tex \LARGE x]-intercept is *[tex \LARGE 0].


Given two points on the line you can calculate the slope thusly:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m\ =\ \frac{y_1\ -\ y_2}{x_1\ -\ x_2} ]


or you can derive an equation of the line using the two-point form:


*[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 in either case.


Given the slope and a point, you can derive an equation of the line using the point-slope form thusly:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ y_1\ =\ m(x\ -\ x_1) ]


where *[tex \Large \left(x_1,y_1\right)] are the coordinates of the given point and *[tex \Large m] is the given/calculated slope.


If you are given a point and a line either parallel or perpendicular, you can use either:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  L_1\ \parallel\ L_2 \ \ \Leftrightarrow\ \ m_1\ =\ m_2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  L_1\ \perp\ L_2 \ \ \Leftrightarrow\ \ m_1\ =\ -\frac{1}{m_2}\ \text{ and } m_1,\, m_2\, \neq\, 0]


and the point-slope form to derive an equation of the desired line.


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