2. `y=(x+2)^2-9`, find Parabola focus

Solution:
`y=(x+2)^2-9`

1. Vertex :
`:. y=1(x-(-2))^2+(-9)`

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

`a=1,h=-2,k=-9`

Vertex `=(h,k)=(-2,-9)`

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

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

Here `a=1>0`

So minimum Vertex = `(h,k)=(-2,-9)`

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

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

Focus `=(h,k+p)=(-2,-9+1/4)=(-2,-35/4)`

3. Graph :
some extra points to plot the graph
`y=f(x)=(x+2)^2-9`

`f(-6)=(-6+2)^2-9=16-9=7`

`f(-5)=(-5+2)^2-9=9-9=0`

`f(-4)=(-4+2)^2-9=4-9=-5`

`f(-3)=(-3+2)^2-9=1-9=-8`

`f(-2)=(-2+2)^2-9=0-9=-9`

`f(-1)=(-1+2)^2-9=1-9=-8`

`f(0)=(0+2)^2-9=4-9=-5`

`f(1)=(1+2)^2-9=9-9=0`

`f(2)=(2+2)^2-9=16-9=7`

graph

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