Question

How do you find the vertex and intercepts for `(x − 2)^2 = -12(y − 2)`?

Answer 1

So the curve has:

Vertex: maximum at `(2,2)`

Y-intercept: `y=13/6`

X-intercepts: `x=2+sqrt6` and `x=2-sqrt6`

Explanation:

`(x-2)^2=-12(y-2)`

Rearrange for y.

`-1/12(x-2)^2=y-2`

`y=-1/12(x-2)^2+2`

From the completed square form, the maximum turning point is at `(2, 2)`

For the y-intercept, let `x=0`

`y=-1/12(-2)^2+2`
`y=13/6`

For the x-intercepts, let `y=0`

`0=-1/12(x-2)^2+2`
`0=(x-2)^2-6`
`6=(x-2)^2`

`x-2=pmsqrt6`

`x=2pmsqrt6`

So the curve has:

Vertex: maximum at `(2,2)`

Y-intercept: `y=13/6`

X-intercepts: `x=2+sqrt6` and `x=2-sqrt6`

graph{(x-2)^2=-12(y-2) [-10, 10, -5, 5]}

Answer 2

`"see explanation"`

Explanation:

`"the quadratic is in standard translated form"`

`•color(white)(x)(x-h)^2=4p(y-k)`

`"where "(h,k)" are the coordinates of the vertex and p"`

`"is the distance from the vertex to the focus/directrix"`

`rArrcolor(magenta)"vertex "=(2,2)`

`"to find intercepts"`

`• " let x = 0, in the equation for y-intercept"`

`• " let y = 0, in the equation for x-intercepts"`

`x=0to4=-12y+24rArry=5/3larrcolor(red)"y-intercept"`

`y=0to(x-2)^2=24`

`color(blue)"take square root of both sides"`

`x-2=+-sqrt24larrcolor(blue)"note plus or minus"`

`rArrx=2+-2sqrt6larrcolor(red)"x-intercepts"`