`y=3(x+1)^2-4`, find Parabola Vertex form
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. 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
This material is intended as a summary. Use your textbook for detail explanation.
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