Question 235564
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The slope intercept form is <i><b>always</i></b>


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ mx\ +\ b]


Well, what about something like


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 5x\ -\ 1]


you ask.  Well remember that *[tex \Large 5x\ -\ 1] is really *[tex \Large 5x\ +\ (-1)].


If you remember it this way, you will seldom, if ever, get confused about signs.


The *[tex \Large y]-intercept is at the point *[tex \Large \left(0,\,b\right)], so if you are given the point *[tex \Large \left(0,\,\frac{3}{4}\right)] then *[tex \Large b\ =\ \frac{3}{4}].  You are also given that *[tex \Large m\ =\ -2], so just replace the values.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -2x\ +\ \frac{3}{4}]


The standard form is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ Ax\ +\ By\ =\ C]


So, multiply both sides of your equation by 4 (remembering to multiply BOTH terms of the RHS) and then add *[tex \Large 2x] to both sides.  Done.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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