`y=3(x+1)^2-4`, find Parabola focus

Solution:
`y=3(x+1)^2-4`

1. Vertex :
`:. y=3(x-(-1))^2+(-4)`

Now compare with `y=a(x-h)^2+k`, we get

`a=3,h=-1,k=-4`

Vertex `=(h,k)=(-1,-4)`

If `a<0` then the vertex is a maximum value

If `a>0` then the vertex is a minimum value

Here `a=3>0`

So minimum Vertex = `(h,k)=(-1,-4)`

2. Focus :
Find `p`, distance from the vertex to a focus of the parabola

`p=1/(4a)=1/(4*3)=1/12`

Focus `=(h,k+p)=(-1,-4)=(-1,-47/12)`

3. Graph :
some extra points to plot the graph
`y=f(x)=3(x+1)^2-4`

`f(-5)=3(-5+1)^2-4=3(16)-4=44`

`f(-4)=3(-4+1)^2-4=3(9)-4=23`

`f(-3)=3(-3+1)^2-4=3(4)-4=8`

`f(-2)=3(-2+1)^2-4=3(1)-4=-1`

`f(-1)=3(-1+1)^2-4=3(0)-4=-4`

`f(0)=3(0+1)^2-4=3(1)-4=-1`

`f(1)=3(1+1)^2-4=3(4)-4=8`

`f(2)=3(2+1)^2-4=3(9)-4=23`

`f(3)=3(3+1)^2-4=3(16)-4=44`

graph

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