# If the standard form of an equation is y=1/2x^2-2x+5, what is the vertex form?

Nov 17, 2016

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

#### Explanation:

To convert from the standard form to the vertex form, you must use completing the square, and in order to use it, the value of $a$ must be "$1$". In this case

$a = \frac{1}{2} ,$ $b = - 2 ,$ $c = 5$

The value of $a$ here is $\frac{1}{2}$, so we must factor out $\frac{1}{2}$ from the expression as follows

$y = \frac{1}{2} {x}^{2} - 2 x + 5 \to y = \frac{1}{2} \left({x}^{2} - 4 x + 10\right)$

The values of $a , b , c$ in the expression ${x}^{2} - 4 x + 10$ are

$a = 1 ,$ $b = - 4 ,$ $c = 10$

The expression ${x}^{2} - 4 x + 10$ is not a perfect square trinomial. We can make it a perfect square by finding the value of $c$ which is equal to ${\left(\frac{b}{2}\right)}^{2}$

$\textcolor{red}{c} = {\left(\frac{b}{2}\right)}^{2} = {\left(\frac{- 4}{2}\right)}^{2} = {\left(- 2\right)}^{2} = 4$

Next, we add $\textcolor{red}{c}$ to the expression ${x}^{2} - 4 x + 10$, but to keep the equation balanced, we must subtract $\textcolor{red}{c}$ also, but without cancelling

${x}^{2} - 4 x + 10 = {x}^{2} - 4 x \textcolor{red}{+ 4} + 10 \textcolor{b l u e}{- 4} = {x}^{2} - 4 x + 4 + 6$

So

$y = \frac{1}{2} \left({x}^{2} - 4 x + 4 + 6\right)$

Bring $6$ outside from the brackets, and in order to do that you must multiply it by $\frac{1}{2}$. When you do that you will get

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

Finally, now you can factor ${x}^{2} - 4 x + 4$ into ${\left(x - 2\right)}^{2}$ and you get

$y = \frac{1}{2} {\left(x - 2\right)}^{2} + 3$ Which is the vertex form.