1. `y=x^2+3x-4`, find axis symmetry of a parabola
Solution:
`y=x^2+3x-4`
1. Symmetry :
Axis of symmetry is the line that passes through the vertex and the focus
`x=h=-3/2`
2. Graph :
some extra points to plot the graph
`y=f(x)=x^2+3x-4`
`f(-5)=(-5)^2+3(-5)-4=25-15-4=6`
`f(-4)=(-4)^2+3(-4)-4=16-12-4=0`
`f(-3)=(-3)^2+3(-3)-4=9-9-4=-4`
`f(-2)=(-2)^2+3(-2)-4=4-6-4=-6`
`f(-1)=(-1)^2+3(-1)-4=1-3-4=-6`
`f(0)=(0)^2+3(0)-4=0-4=-4`
`f(1)=(1)^2+3(1)-4=1+3-4=0`
`f(2)=(2)^2+3(2)-4=4+6-4=6`
`f(3)=(3)^2+3(3)-4=9+9-4=14`
graph
This material is intended as a summary. Use your textbook for detail explanation.
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