Originally Posted by
masters I assume you want to put this equation of a parabola in vertex form by completing the square.
$\displaystyle y=2x^2-3x-1$
First, factor out the 2 from the x terms:
$\displaystyle y=2\left(x^2-\frac{3}{2}x\right)-1$
Next, take half the coefficient of x, $\displaystyle \left(-\frac{3}{4}\right)$, square it, $\displaystyle \left(\frac{9}{16}\right)$, and add it to complete the perfect square trinomial in parentheses. But, you really added twice that amount since we have a factor of $\displaystyle 2$ outside the parentheses. So, to keep things balanced in the equation, we must subtract twice that amount, $\displaystyle \left(\frac{9}{8}\right)$, outside the parentheses. 'What one addeth to just one side of an equation, one must taketh away'.
$\displaystyle y=2\left(x^2-\frac{3}{2}x+\frac{9}{16}\right)-1-\frac{9}{8}$
$\displaystyle y=2\left(x-\frac{3}{4}\right)^2-\frac{17}{8}$