y=3(x+1)^2-4, find axis symmetry of a parabola
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. Symmetry :
Axis of symmetry is the line that passes through the vertex and the focus
x=h=-1
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
This material is intended as a summary. Use your textbook for detail explanation.
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