How do you write f(x) = -x^2 + 6x + 8 in vertex form?

Jun 20, 2016

$\text{ } y = - {\left(x - 3\right)}^{2} + 17$
Or if you prefer: $y = {\left(- x + 3\right)}^{2} + 17$

Explanation:

This process introduces an error that has to be compensated for by the inclusion of the correction factor of $k$. This has to be so to maintain the truth of the expression being equal that of y.

In that $f \left(x\right) = - {x}^{2} + 6 x + 8 + k$ at this stage $k = 0$

Write as:$y = \left(- 1 {x}^{2} + 6 x\right) + 8 + k$

Factor out the $- 1$ (Sometimes this value is not 1)

$y = - 1 \left({x}^{2} - 6 x\right) + 8 + k$

take the exponent (power of 2) outside the bracket

$y = - 1 {\left(x - 6 x\right)}^{2} + 8 + k \text{ "larr" now "k" comes into effect.}$

Note that each step the value of $k$ changes until we finish modifying at which point it becomes a constant.

Halve the coefficient of $- 6 x$

$y = - 1 {\left(x - 3 x\right)}^{2} + 8 + k$

Discard the $x$ from $- 3 x$

$y = - 1 {\left(x - 3\right)}^{2} + 8 + k \text{ "larr" nearly there. "k" is now fixed}$
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If you were to expand the bracket and compere to the original equation you would observe that there is an extra value. This is from squaring the $- 3$ from inside the bracket. $k$ cancels that out.

Let ${\left(- 3\right)}^{2} + k = 0 \text{ "=>" } k = - 9$

$\textcolor{b r o w n}{\text{However we have a -1 outside the bracket so this becomes.}}$

Let $\left(- 1\right) {\left(- 3\right)}^{2} + k = 0 \text{ "=>" } k = + 9$

color(brown)(y=-1(x-3)^2+8+k)color(blue)(->y=-1(x-3)^2+17

Write as:$\text{ } y = - {\left(x - 3\right)}^{2} + 17$