Question 237055
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ x^2\ +\ x\ -\ 12]


Complete the square.  Step 1, divide the coefficient on the 1st degree term by 2 and square the result: *[tex \LARGE \left(\frac{1}{2}\right)^2\ =\ \frac{1}{4}]


Add and subtract this value in the RHS


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ x^2\ +\ x\ +\ \frac{1}{4}\ -\ \frac{1}{4}\ -\ 12]


The first three terms in the RHS are now a perfect square:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \left(x\ +\ \frac{1}{2}\right)^2\ -\ \frac{49}{4}]


Vertex at *[tex \LARGE \left(-\frac{1}{2}\,,\,-\frac{49}{4}\right)]



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