Question

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

Answer 1

Answer given in detail so you can see where everything comes from.

`y_("intercept")=c=+7" "x_("intercept")->x=+1" and "x=-7`

The coefficient of `x^2` is negative so the graph is of form `nn` thus the vertex is a maximum.

`"vertex "->(x,y)=(-3,16)`

Explanation:

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
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`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`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`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)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determining the y-intercept")`

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

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