Question

How do you sketch the parabola `(y-2)^2=-12(x+3)` and find the vertex, focus, and directrix?

Answer 1

`"see explanation"`

Explanation:

`"the equation of the parabola is of the form"`

`•color(white)(x)(y-k)^2=4p(x-h)`

`"this parabola opens horizontally"`

`• " if "4p>0" opens to the right"`

`• " if "4p<0" opens to the left"`

`"vertex "=(h,k)" and "4p=-12rArrp=-3`

`"here "(h,k)=(-3,2)larrcolor(red)" vertex"`

`"since "4p<0" then opens to the left"`

`"the vertex is midway between the focus and directrix"`

`p=-3" is the distance from the vertex to the focus"`

`rArr"focus "=(-3-3,2)=(-6,2)larrcolor(red)" focus"`

`"the directrix is also 3 units from the vertex in the"`
`"opposite side from the focus"`

`rArrx=0larrcolor(red)" equation of directrix"`
graph{(y-2)^2=-12(x+3) [-10, 10, -5, 5]}