Question

What is the vertex of `y=-x^2 - 2x - 3(x/3-2/3)^2`?

Answer 1

Hence, the vertex is
I have approached by the method of calculus (maxima and minima)
`V-=(x,y)=V-=(-1/4,-34/16)`

Explanation:

I have approached by the method of calculus (maxima and minima)
The curve is symmetrical about an axis parallel to y axis.
The vertex is the point where `dy/dx=0`

Given:

`y=-x^2-2x-3(x/3-2/3)^2`

Differentiating wrt x

`dy/dx=-2x-2-3xx2(x/3-2/3)xx1/3`

`dy/dx=0`

`-2x-2-3xx2(x/3-2/3)xx1/3=0`

`-2x-2-2/3x+4/3=0`

`-2x-2/3x=2-4/3`
`-6/3x-2/3x=6/3-4/3`

`-6x-2x=6-4`

`-8x=2`

`8/8x=-2/8`

`x=-1/4`

`y=-x^2-2x-3(x/3-2/3)^2`

`y=-(-1/4)^2-2(-1/4)-3((-1/4)/3-2/3)^2`

`=-1/16+1/2-3(-1/12-2/3)^2`

`=-1/16+8/16-3(-1/12-8/12)^2`

`=(-1+8)/16-3((-1-8)/12)^2`

`=7/16-3(-9/12)^2`

`=7/16-3(-3/4)^2`

`=7/16-3xx9/16`

`7/16-27/16`

`y=-34/16`

Hence, the vertex is

`V-=(x,y)=V-=(-1/4,-34/16)`