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#?

3 Answers
May 19, 2018

#(-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.

May 19, 2018

Answer:

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....#

Tony B

May 19, 2018

Answer:

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#