`y=3x^2+6x-1`, find Parabola Directrix
Solution:
`y=3x^2+6x-1`
1. Vertex :
`y=3x^2+6x-1`
Method-1: Find vertex using polynomial form
Comparing the equation 3x^2+6x-1 with `ax^2+bx+c`, we get
`a=3,b=6,c=-1`
`h=(-b)/(2a)=(-6)/(2 * 3)=-1`
Now, substitute value of h in f(x), to find value of k
`k=f(h)=f(-1)=3(-1)^2+6(-1)-1`
`:. k=3-6-1`
`:. k=-4`
Vertex `=(h,k)=(-1,-4)`
Method-2: Find vertex using vertex form `y=a(x-h)^2+k`
Completing the Square
`3x^2+6x-1`
`=3 (x^2+2x-1/3)`
The coefficient of the x is `2`, so now we divide this by 2 : `(2 -: 2 = 1)`
and square it `(1)^2=1`. So we add and subtract `1`
`=3 (x^2+2x + 1 - 1 - 1/3)`
`=3 [(x^2+2x+1) -4/3]`
`=3[( x + 1 )^2 -4/3 ]`
`:. 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. Directrix :
Directrix `y=k-p=-4=-49/12`
4. Graph :
some extra points to plot the graph
`y=f(x)=3x^2+6x-1`
`f(-5)=3(-5)^2+6(-5)-1=75-30-1=44`
`f(-4)=3(-4)^2+6(-4)-1=48-24-1=23`
`f(-3)=3(-3)^2+6(-3)-1=27-18-1=8`
`f(-2)=3(-2)^2+6(-2)-1=12-12-1=-1`
`f(-1)=3(-1)^2+6(-1)-1=3-6-1=-4`
`f(0)=3(0)^2+6(0)-1=0-1=-1`
`f(1)=3(1)^2+6(1)-1=3+6-1=8`
`f(2)=3(2)^2+6(2)-1=12+12-1=23`
`f(3)=3(3)^2+6(3)-1=27+18-1=44`
graph
This material is intended as a summary. Use your textbook for detail explanation.
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