`y=3(x+1)^2-4`, find Properties of a function
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. Focus :
Find `p`, distance from the vertex to a focus of the parabola
`p=1/(4a)=1/(4*3)=1/12`
Focus `=(h,k+p)=(-1,-4)=(-1,-47/12)`
3. Symmetry :
Axis of symmetry is the line that passes through the vertex and the focus
`x=h=-1`
4. Directrix :
Directrix `y=k-p=-4=-49/12`
5. 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
6. Intercepts :
Intercept :
To find the y-intercept put x=0 in `y=3(x+1)^2-4`, we get
`y=3(0+1)^2-4=3(1)-4=-1`
`:.` y-intercept is `(0,-1)`
To find the x-intercept put y=0 in `y=3(x+1)^2-4`, we get
`=>3*(x+1)^2-4=0`
`=>3*(x+1)^2=4`
`=>(x+1)^2=4/3`
`=>x+1=+- 2/(sqrt(3))`
Now, `x+1=2/(sqrt(3))`
`=>x=2/(sqrt(3))-1`
`=>x=0.1547`
Now, `x+1=-2/(sqrt(3))`
`=>x=-2/(sqrt(3))-1`
`=>x=-2.1547`
`:.` x-intercepts are `(0.1547,0)` and `(-2.1547,0)`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
