# How do you write  y =5x^2 - 3x + 2 in vertex form?

Nov 5, 2017

$y = 5 {\left(x - \frac{3}{10}\right)}^{2} + \frac{31}{20}$

#### Explanation:

Vertex form: $y = a {\left(x - h\right)}^{2} + k$

$y = 5 {x}^{2} - 3 x + 2$

$y = 5 \left({x}^{2} - \frac{3}{5} x\right) + 2$

$y = 5 \left({x}^{2} - 2 \cdot \frac{3}{10} x + \frac{9}{100}\right) + 2 - 5 \cdot \frac{9}{100}$

$y = 5 {\left(x - \frac{3}{10}\right)}^{2} + 2 - \frac{9}{20}$

$y = 5 {\left(x - \frac{3}{10}\right)}^{2} + \frac{31}{20}$

Nov 9, 2017

$y = 5 {\left(x - \frac{3}{10}\right)}^{2} + \frac{31}{20}$

#### Explanation:

Vertex form is a useful way of writing the equation of a parabola because it tells you the co-ordinates of the turning point, (the vertex).

It is written as $y = p {\left(x + q\right)}^{2} + r$
The vertex is at $\left(- q , r\right)$

$y = 5 {x}^{2} - 3 x + 2 \text{ } \leftarrow$ in the form $a {x}^{2} + b x + c$

y = 5[x^2-3/5x" "+2/5]" "larr you must have $1 {x}^{2}$

Now you need to add in a term to create the square of the binomial.
The missing term comes from ${\left(\frac{b}{2}\right)}^{2}$ where $b = \left(- \frac{3}{5}\right)$

$\frac{- 3}{5} \div 2 = \frac{- 3}{10}$

If you add an extra term, you must subtract it as well to keep the equation the same.

y = 5[x^2-3/5x + color(green)(((-3)/10)^2)+2/5 -color(green)(((-3)/10)^2]

$y = 5 \left[{x}^{2} - \frac{3}{5} x + \frac{9}{100} + \frac{2}{5} - \frac{9}{100}\right]$

$y = 5 \left[\left(\textcolor{b l u e}{{x}^{2} - \frac{3}{5} x + \frac{9}{100}}\right) + \left(\textcolor{red}{\frac{2}{5} - \frac{9}{100}}\right)\right]$

$y = 5 \left[{\textcolor{b l u e}{\left(x - \frac{3}{10}\right)}}^{2} + \left(\textcolor{red}{\frac{40}{100} - \frac{9}{100}}\right)\right]$

$y = 5 \left[{\textcolor{b l u e}{\left(x - \frac{3}{10}\right)}}^{2} + \left(\textcolor{red}{\frac{31}{100}}\right)\right] \text{ } \leftarrow$ distribute the $5$

$y = 5 {\textcolor{b l u e}{\left(x - \frac{3}{10}\right)}}^{2} + 5 \left(\textcolor{red}{\frac{31}{100}}\right)$

$y = 5 {\left(x - \frac{3}{10}\right)}^{2} + \frac{31}{20}$

The vertex is $\left(\frac{3}{10} , \frac{31}{20}\right)$