# How do you write the equation in vertex form y=x^2-6x+11?

Mar 2, 2018

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

#### Explanation:

A quadratic equation, for reference, is:

$a {x}^{2} + b x + c = 0$

$a {\left(x - h\right)}^{2} + k$

where the vertex is $\left(h , k\right)$

Here we can convert ${x}^{2} - 6 x + 11$ into vertex form by taking the perfect square of ${x}^{2} - 6 x$:

${x}^{2} - 6 x + 9 + 11 - 9$

(to find a perfect square, we halve the coefficient of $b x$, which is $b$, and then square the answer)

Here, we halved $6$, which equals $3$. Then, we squared $3$, which equals $9$.

When we add $9$, remember that we must also subtract $9$ at the end to keep the equation correct.

$\left({x}^{2} - 6 x + 9\right) + 2$

Here is a perfect square. All we need to do is halve the coefficient, add it with $x$, and square the whole thing.

$\left({x}^{2} - 6 x + 9\right) + 2 = {\left(x - 3\right)}^{2} + 2$

To finish it in the vertex form:

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

The vertex is $\left(h , k\right)$, so here the vertex will be:

$\left(3 , 2\right)$

Thus, solved.