Question

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`?

Answer 1

`(-b)/(2a)` gives the x coordinate of the max/min point.

`(-6)/(-2)=3`

`x=3` is the line of symmetry.

When `x=3, y=-3^2+6xx3+6`

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

(3,15) is the vertex, as the `x^2` is negative the parabola will be `nn` shaped.

Answer 2

See explanation

Explanation:

As this is in calculus we do the following:

`color(blue)("Determine the general shape and vertex")`

Shortcut approach:

Given: `f(x)=-x^2+6x+6 color(white)("d")->f'(x)=-2x+6=0` at the turning point.

Set `f'(x)=0=-2x+6 => x_("vertex")=6/2=3`

By substitution set `y=-(3)^2+6(3)+6 = -9+18+6 = +15`

Vertex `->(x,y)=(3,15)`

`f''(x)=-2` and as this is negative the vertex is a maximum.

So the graph is of form `nn` which is compatible with the `x^2` term being negative. This also indicates the general form of `nn`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)("The y-intercept")`

This is at `x=0` so set `y=-x^2+6x+6 =-(0)^2+6(0)+6=6`

`y_("intercept")=6`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)("The x-intercept")`

As the vertex is `(x,y)=(3,15)` and the general form is `nn` then the plot crosses the x-axis so `x_("intercept")` exists.

Using Completing the square:

Set `y=0=-x^2+6x+6`

`0=-1(x-3)^2+k+6`

Set `-1(-3)^2+k=0 => k= +9` giving:

`0=-1(x-3)^2+15`

`x-3=+-sqrt(15)`

`x=3+-sqrt(15)`

`x~~ +6.87298....`
`x~~-0.87298....`

Answer 3

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

Explanation:

Given: `f(x)=-x^2+6x+6`

Set: `y=-x^2+6x+6" "....................Equation(1)`

Increment `x` by the very small amount of `delta x`

Consequently the value of `y` would have also changed by the small amount of `deltay`

By substitution `Eqn(1)` becomes:

`y+deltay=-(x+deltax)^2+6(x+deltax)+6`

`y+deltay=-(x^2+2xdeltax+(deltax)^2)+(6x+6deltax)+6`

`y+deltay=-x^2-2xdeltax + (deltax)^2+6x+6deltax+6" ". ..Eqn(1_a)`

`Eqn(1_a)-Eqn(1)`

`y+deltay=-x^2-2xdeltax + (deltax)^2+6x+6deltax+6`
`ul(ycolor(white)("dddd")=-x^2color(white)("ddddddddddd.d")+6xcolor(white)("ddddd")+6)larr Subtract"`
`color(white)("ddd")deltay= color(white)("dd")0color(white)("d")-2xdeltax +(deltax)^2+0color(white)("d")+6deltax+0`

`deltay=-2xdeltax+(deltax)^2+6deltax`

Divide both sides bu `deltax`

`color(white)("ddddd")(deltay)/(deltax) =-2x+color(white)("dd")deltaxcolor(white)("ddd")+6`

`lim_(deltax->0)[(deltay)/(deltax)]=-2x+ubrace(lim_(deltax->0)[ deltax ]color(white)(.))color(white)("d")+6`

`color(white)(dddd"d")dy/dx= color(white)("dd")-2xcolor(white)("dd.d")+0color(white)("ddddd")+6`