# How do you write f(x) = x² - 8x +11 in vertex form?

Apr 12, 2017

$f \left(x\right) = {\left(x - 4\right)}^{2} - 5$

#### Explanation:

To go from Standard Form to Vertex Form, we need to complete the square. That means we need to find the constant that makes ${x}^{2} - 8 x$ a perfect square.

$y = {x}^{2} - 8 x + 11$

To find that constant, we use this equation: $c = {\left(\frac{1}{2} \cdot b\right)}^{2}$ or ${\left(\frac{1}{2} \cdot - 8\right)}^{2}$ or $16$.

We now have the constant that completes the square, but we can't just add a value to an equation. We can either add $16$ to both sides or add it and the immediately subtract it. Either way works

$y = {x}^{2} - 8 x \textcolor{b l u e}{+ 16 - 16} + 11$
$y = \left({x}^{2} - 8 x + 16\right) - 16 + 11$

${x}^{2} - 8 x + 16$ is a perfect square, so let's simplify:

$y = {\left(x - 4\right)}^{2} - 16 + 11$
$y = {\left(x - 4\right)}^{2} - 5$
$f \left(x\right) = {\left(x - 4\right)}^{2} - 5$

That's our vertex form, so now we're done. Good job!