How to write f ( x ) = x ² - 8 x + 11 in vertex form?

Question

How to write             f              (        x        )              =    x    ²    -    8    x    +    11   in vertex form?

Lindsay Banks · Accepted Answer

We must finish the square in order to transition from Standard Form to Vertex Form.
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 = {(\frac{1}{2} \cdot b)}^{2}

or

{(\frac{1}{2} \cdot - 8)}^{2}

or

16

Now that the square is complete, we have the necessary constant, but we cannot just add a number 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 + 16 - 16 + 11

y = (x^{2} - 8 x + 16) - 16 + 11

x^{2} - 8 x + 16

is a perfect square, so let's simplify:

y = {(x - 4)}^{2} - 16 + 11

y = {(x - 4)}^{2} - 5

f (x) = {(x - 4)}^{2} - 5

That's our final vertex form

How to write f(x) = x^2- 8x +11 in vertex form?