What is the vertex form of y= 6x^2-9x+3 ?

Dec 14, 2017

$y = 6 {\left(x - \frac{3}{4}\right)}^{2} - \frac{3}{8}$

Explanation:

To complete the square of the equation, first take out the 6:

$y = 6 \left({x}^{2} - \frac{3}{2} x + \frac{1}{2}\right)$

Then do the bit in the brackets:

$y = 6 \left[{\left(x - \frac{3}{4}\right)}^{2} - \frac{9}{16} + \frac{1}{2}\right]$

$y = 6 \left[{\left(x - \frac{3}{4}\right)}^{2} - \frac{1}{16}\right]$

$y = 6 {\left(x - \frac{3}{4}\right)}^{2} - \frac{3}{8}$, as required.

Dec 14, 2017

$y = 6 {\left(x - \frac{3}{4}\right)}^{2} - \frac{3}{8}$

Explanation:

$\text{the equation of a parabola in "color(blue)"vertex form}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{where "(h,k)" are the coordinates of the vertex and a}$
$\text{is a multiplier}$

$\text{to obtain this form use the method of}$
$\textcolor{b l u e}{\text{completing the square}}$

• " the coefficient of the "x^2" term must be 1"

$\Rightarrow y = 6 \left({x}^{2} - \frac{3}{2} x\right) + 3$

• " add/subtract "(1/2"coefficient of x-term")^2" to"

${x}^{2} - \frac{3}{2} x$

$\Rightarrow y = 6 \left({x}^{2} + 2 \left(- \frac{3}{4}\right) x \textcolor{red}{+ \frac{9}{16}} \textcolor{red}{- \frac{9}{16}}\right) + 3$

$\Rightarrow y = 6 {\left(x - \frac{3}{4}\right)}^{2} - \frac{27}{8} + 3$

$\Rightarrow y = 6 {\left(x - \frac{3}{4}\right)}^{2} - \frac{3}{8} \leftarrow \textcolor{red}{\text{in vertex form}}$