Question

What is the vertex of `7y= 12(x-15)^2 +12`?

Answer 1

The vertex happens to be
`(x,y)=(15,12/7)`

Explanation:

The given equation is:
`7y=12(x-15)^2+12`
The curve is symmetrical about the x axis
Differentiating the equation wrt x
`7dy/dx=12(2)(x-15)+0`
The vertex coresponds to the point where the slope is zero.
Equating `dy/dx=0`
`7(0)=24(x-15)`
ie
`24(x-15)=0`
`x-15=0`
`x=15`
Substituting for x in the equation of the curve
`7y=12(15-15)+12`
`7y=12`
`y=12/7`
Thus, the vertex happens to be
`(x,y)=(15,12/7)`

Answer 2

`"vertex "=(15,12/7)`

Explanation:

`"divide both sides by 7"`

`rArry=12/7(x-15)^2+12/7`

`"the equation of a parabola in "color(blue)"vertex form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))`

`"where "(h,k)" are the coordinates of the vertex and a"`
`"is a multiplier"`

`y=12/7(x-15)^2+12/7" is in vertex form"`

`rArrcolor(magenta)"vertex "=(15,12/7)`