# What is the vertex of y=-x^2 +12x - 4 ?

Nov 23, 2015

$x = 6$ I will let you solve for $y$ by substation.

$\textcolor{b r o w n}{\text{Look at the explanation. It shows you a short cut!}}$

#### Explanation:

Standard form: $y = a {x}^{2} + b {x}_{c} = 0 \textcolor{w h i t e}{\ldots .}$Where

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$a = - 1$
$b = 12$
$c = - 4$

$\textcolor{b l u e}{\approx \approx \approx \approx \approx \approx \text{ Short Cut } \approx \approx \approx \approx \approx \approx}$

$\textcolor{b r o w n}{\text{Change into format of " y= ax^2+bx+c " into:}}$
$\textcolor{b r o w n}{y = a \left({x}^{2} + \frac{b}{a} x + \frac{c}{a}\right) \textcolor{w h i t e}{\times x} \to \textcolor{w h i t e}{\ldots . .} \left(- 1\right) \left({x}^{2} - 12 x + 4\right)}$

$\textcolor{b l u e}{\text{THE TRICK!}}$ $\textcolor{w h i t e}{\ldots .} \textcolor{g r e e n}{{x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \left(\frac{b}{a}\right) = \left(- \frac{1}{2}\right) \left(- 12\right) = + 6}$

$\textcolor{b l u e}{\approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx}$

$\textcolor{red}{\text{ To demonstrate the point - 'The long way round !'}}$

The factors of 4 will not produce the sum of 12 so use the formula

The vertex $x$ will be the mean of the two $x ' s$ that are a solution the standard form

$a = - 1$
$b = 12$
$c = - 4$

Thus

$x = \frac{- \left(12\right) \pm \sqrt{{12}^{2} - \left(4\right) \left(- 1\right) \left(- 4\right)}}{2 \left(- 1\right)}$

$x = + 6 \pm \frac{\sqrt{144 - 16}}{- 2}$

$x = + 6 \pm \frac{\sqrt{128}}{- 2}$

$x = 6 \pm \frac{\sqrt{2 \times 64}}{- 2}$

$x = 6 \pm \frac{8 \sqrt{2}}{- 2}$

$x = 6 \pm \left(- 4 \sqrt{2}\right)$

The mean point is:

${x}_{\text{vertex}} = \frac{\left(6 - 4 \sqrt{2}\right) + \left(6 + 4 \sqrt{2}\right)}{2} = 6$

Substitute ${x}_{\text{vertex}} = 6$ into the original equation to find the value of ${y}_{\text{vertex}}$