# What is the vertex form of y= 4t^2-12t+8 ?

Aug 18, 2017

$y = 4 {\left(t - \frac{3}{2}\right)}^{2} - 1$

#### Explanation:

Vertex form is given as $y = a {\left(x + b\right)}^{2} + c$,

where the vertex is at $\left(- b , c\right)$

Use the process of completing the square.

$y = 4 {t}^{2} - 12 t + 8$

$y = 4 \left({t}^{2} - \textcolor{b l u e}{3} t + 2\right) \text{ } \leftarrow$ take out the factor of $4$

$y = 4 \left({t}^{2} - 3 t \textcolor{b l u e}{+ {\left(\frac{3}{2}\right)}^{2} - {\left(\frac{3}{2}\right)}^{2}} + 2\right)$

$\left[\textcolor{b l u e}{+ {\left(\frac{3}{2}\right)}^{2} - {\left(\frac{3}{2}\right)}^{2} = 0}\right] \text{ } \leftarrow + {\left(\frac{b}{2}\right)}^{2} - {\left(\frac{b}{2}\right)}^{2}$

$y = 4 \left(\textcolor{red}{{t}^{2} - 3 t + {\left(\frac{3}{2}\right)}^{2}} \textcolor{f \mathmr{and} e s t g r e e n}{- {\left(\frac{3}{2}\right)}^{2} + 2}\right)$

$y = 4 \left(\textcolor{red}{{\left(t - \frac{3}{2}\right)}^{2}} \textcolor{f \mathmr{and} e s t g r e e n}{- \frac{9}{4} + 2}\right)$

$y = 4 \left(\textcolor{red}{{\left(t - \frac{3}{2}\right)}^{2}} \textcolor{f \mathmr{and} e s t g r e e n}{- \frac{1}{4}}\right)$

Now distribute the $4$ into the bracket.

$y = \textcolor{red}{4 {\left(t - \frac{3}{2}\right)}^{2}} + \textcolor{f \mathmr{and} e s t g r e e n}{4 \left(- \frac{1}{4}\right)}$

$y = 4 {\left(t - \frac{3}{2}\right)}^{2} - 1$