How do you find the axis of symmetry, graph and find the maximum or minimum value of the function #y = 7 – 6x – x^2#?

Conventional format:#-> y=ax^2+bx+c #

So we have: #" " y=-x^2-6x+7................Equation(1)#

#color(blue)("Determining the x-intercepts")#

This factorises making the calculations more straight forward.

To make #-x^2# we need: #(-x)xx(+x)#. Also, where the graph crosses the x-axis, we have the value #y=0#. So we write:

#(-x+?)(+x+?)=0#

I spot that #1xx7=7" and that "7-1=6# but we need the bigger value to be negative as #-7+1=-6" to give us "-6x#

If we place #+7# at #(-x+?)(x+7)# we end up with #-7x#

Try out: #color(blue)((-x+1))color(brown)((x+7))=0 .........Equation(2)#

CHECK:
Multiply the right hand brackets by everything in the left hand brackets giving:

#" "color(brown)( color(blue)(-x)(x+7)color(blue)(" "+" "1)(x+7))#
#" "-x^2-7x" "+" "x+7#

#=-x^2-6x+7# as required so this is the correct factorisation

Using Equation(2):

#" "(-x+1)(x+7)=0 => x=+1" and "x=-7#

Are solutions for this condition
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#color(blue)("Determining the axis of symmetry")#

This will be in the middle of the x-intercepts.

#x_("symmetry")=(1-7)/2=(-6)/2=-3#
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#color(blue)("Determining the vertex")#

#x_("vertex")=x_("symmetry") = -3#

Substitute #x=3# into Equation(1)

#y_("vertex")=-(-3)^2-6(-3)+7" "=" "+16#

#"vertex "->(x,y)=(-3,16)#
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#color(blue)("Determining the y-intercept")#

Consider: #y=ax^2+bx+c" "->" "y=-x^2-6x+7#

#y_("intercept")=c=+7#