# What is the vertex form of y=x^2-12x+34?

Sep 19, 2016

$y = {\left(x - 6\right)}^{2} - 2$

The vertex is at $\left(6 , - 2\right)$

#### Explanation:

(I assumed the second term was -12x and not just -12 as given)

To find the vertex form, you apply the method of:
"completing the square".

This involves adding the correct value to the quadratic expression to create a perfect square.

Recall: ${\left(x - 5\right)}^{2} = {x}^{2} \textcolor{\to m a \to}{- 10} x \textcolor{\to m a \to}{+ 25} \text{ } \leftarrow \textcolor{\to m a \to}{{\left(\frac{- 10}{2}\right)}^{2} = 25}$

This relationship between $\textcolor{\to m a \to}{b \mathmr{and} c}$ will always exist.

If the value of $c$ is not the correct one, add on what you need. (Subtract it as well to keep the value of the expression the same)

$y = {x}^{2} \textcolor{\to m a \to}{- 12} x + 34 \text{ } \leftarrow {\left(\frac{- 12}{2}\right)}^{2} = 36 \ne 34$

Adding 2 will make the 36 that is needed.

$y = {x}^{2} \textcolor{\to m a \to}{- 12} x + 34 \textcolor{b l u e}{+ 2 - 2} \text{ } \leftarrow$ the value is the same

$y = {x}^{2} \textcolor{\to m a \to}{- 12} x + \textcolor{\to m a \to}{36} \textcolor{b l u e}{- 2}$

$y = {\left(x - 6\right)}^{2} - 2 \text{ } \leftarrow$ this is vertex form

The vertex is at $\left(6 , - 2\right) \text{ } \leftarrow$ note the signs

How do you get to it?

$y = \textcolor{\lim e}{{x}^{2}} \textcolor{\to m a \to}{- 12} x + 36 \textcolor{b l u e}{- 2}$

$y = {\left(\textcolor{\lim e}{x} \textcolor{\to m a \to}{- 6}\right)}^{2} \textcolor{b l u e}{- 2}$

$\textcolor{\lim e}{x = \sqrt{{x}^{2}}} \mathmr{and} \textcolor{\to m a \to}{\frac{- 12}{2} = - 6} \text{ check} \sqrt{36} = 6$