# How do you find the axis of symmetry, graph and find the maximum or minimum value of the function f(x)=-x^2+6x+6?

May 19, 2018

$\frac{- b}{2 a}$ gives the x coordinate of the max/min point.

$\frac{- 6}{- 2} = 3$

$x = 3$ is the line of symmetry.

When $x = 3 , y = - {3}^{2} + 6 \times 3 + 6$

$y = - 9 + 18 + 6 = 15$

(3,15) is the vertex, as the ${x}^{2}$ is negative the parabola will be $\cap$ shaped.

May 19, 2018

See explanation

#### Explanation:

As this is in calculus we do the following:

$\textcolor{b l u e}{\text{Determine the general shape and vertex}}$

Shortcut approach:

Given: $f \left(x\right) = - {x}^{2} + 6 x + 6 \textcolor{w h i t e}{\text{d}} \to f ' \left(x\right) = - 2 x + 6 = 0$ at the turning point.

Set $f ' \left(x\right) = 0 = - 2 x + 6 \implies {x}_{\text{vertex}} = \frac{6}{2} = 3$

By substitution set $y = - {\left(3\right)}^{2} + 6 \left(3\right) + 6 = - 9 + 18 + 6 = + 15$

Vertex $\to \left(x , y\right) = \left(3 , 15\right)$

$f ' ' \left(x\right) = - 2$ and as this is negative the vertex is a maximum.

So the graph is of form $\cap$ which is compatible with the ${x}^{2}$ term being negative. This also indicates the general form of $\cap$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{\text{The y-intercept}}$

This is at $x = 0$ so set $y = - {x}^{2} + 6 x + 6 = - {\left(0\right)}^{2} + 6 \left(0\right) + 6 = 6$

${y}_{\text{intercept}} = 6$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{\text{The x-intercept}}$

As the vertex is $\left(x , y\right) = \left(3 , 15\right)$ and the general form is $\cap$ then the plot crosses the x-axis so ${x}_{\text{intercept}}$ exists.

Using Completing the square:

Set $y = 0 = - {x}^{2} + 6 x + 6$

$0 = - 1 {\left(x - 3\right)}^{2} + k + 6$

Set $- 1 {\left(- 3\right)}^{2} + k = 0 \implies k = + 9$ giving:

$0 = - 1 {\left(x - 3\right)}^{2} + 15$

$x - 3 = \pm \sqrt{15}$

$x = 3 \pm \sqrt{15}$

$x \approx + 6.87298 \ldots .$
$x \approx - 0.87298 \ldots .$

May 19, 2018

Just for reference: The first calculus part only using first principles.

#### Explanation:

Given: $f \left(x\right) = - {x}^{2} + 6 x + 6$

Set: $y = - {x}^{2} + 6 x + 6 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots . . E q u a t i o n \left(1\right)$

Increment $x$ by the very small amount of $\delta x$

Consequently the value of $y$ would have also changed by the small amount of $\delta y$

By substitution $E q n \left(1\right)$ becomes:

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

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

$y + \delta y = - {x}^{2} - 2 x \delta x + {\left(\delta x\right)}^{2} + 6 x + 6 \delta x + 6 \text{ } . . . E q n \left({1}_{a}\right)$

$E q n \left({1}_{a}\right) - E q n \left(1\right)$

$y + \delta y = - {x}^{2} - 2 x \delta x + {\left(\delta x\right)}^{2} + 6 x + 6 \delta x + 6$
ul(ycolor(white)("dddd")=-x^2color(white)("ddddddddddd.d")+6xcolor(white)("ddddd")+6)larr Subtract"
$\textcolor{w h i t e}{\text{ddd")deltay= color(white)("dd")0color(white)("d")-2xdeltax +(deltax)^2+0color(white)("d}} + 6 \delta x + 0$

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

Divide both sides bu $\delta x$

$\textcolor{w h i t e}{\text{ddddd")(deltay)/(deltax) =-2x+color(white)("dd")deltaxcolor(white)("ddd}} + 6$

${\lim}_{\delta x \to 0} \left[\frac{\delta y}{\delta x}\right] = - 2 x + \underbrace{{\lim}_{\delta x \to 0} \left[\delta x\right] \textcolor{w h i t e}{.}} \textcolor{w h i t e}{\text{d}} + 6$

$\textcolor{w h i t e}{\mathrm{dd} \mathrm{dd} \text{d")dy/dx= color(white)("dd")-2xcolor(white)("dd.d")+0color(white)("ddddd}} + 6$