2. `y=(x+2)^2-9`, find Properties 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. 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. Symmetry :
Axis of symmetry is the line that passes through the vertex and the focus
`x=h=-2`
4. Directrix :
Directrix `y=k-p=-9-1/4=-37/4`
5. 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
6. Intercepts :
Intercept :
To find the y-intercept put x=0 in `y=(x+2)^2-9`, we get
`y=(0+2)^2-9=4-9=-5`
`:.` y-intercept is `(0,-5)`
To find the x-intercept put y=0 in `y=(x+2)^2-9`, we get
`=>(x+2)^2-9=0`
`=>(x+2)^2=0+9`
`=>(x+2)^2=9`
`=>x+2=+- 3`
Now, `x+2=3`
`=>x=3-2`
`=>x=1`
Now, `x+2=-3`
`=>x=-3-2`
`=>x=-5`
`:.` x-intercepts are `(1,0)` and `(-5,0)`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
