`y=(x+2)^2-9`, find Vertex of a function
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. 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
This material is intended as a summary. Use your textbook for detail explanation.
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