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4. Properties of a function example ( Enter your problem )
  1. `y=x^2+3x-4` Example-1
  2. `y=(x+2)^2-9` Example-2
  3. `y=3x^2+6x-1` Example-3
  4. `y=3(x+1)^2-4` Example-4
Other related methods
  1. Domain of a function
  2. Range of a function
  3. Inverse of a function
  4. Properties of a function
  5. Parabola Vertex of a function
  6. Parabola focus
  7. axis symmetry of a parabola
  8. Parabola Directrix
  9. Intercept of a function
  10. Parity of a function
  11. Asymptotes of a function

2. `y=(x+2)^2-9` Example-2
(Previous example)
4. `y=3(x+1)^2-4` Example-4
(Next example)

3. `y=3x^2+6x-1` Example-3





`y=3x^2+6x-1`, find Properties of a function

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. 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)=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

6. Intercepts :
Intercept :
To find the y-intercept put x=0 in `y=3x^2+6x-1`, we get

`y=3(0)^2+6(0)-1=-1`

`:.` y-intercept is `(0,-1)`


To find the x-intercept put y=0 in `y=3x^2+6x-1`, we get

`=>3x^2+6x-1=0`

factor is not possible for equation `3x^2+6x-1=0`

But we are trying find solution using the method of perfect square.

Comparing the given equation with the standard quadratic equation `ax^2+bx+c=0,`

we get, `a=3, b=6, c=-1.`

`:. Delta=b^2-4ac`

`=(6)^2-4 (3) (-1)`

`=36+12`

`=48`

`:. sqrt(Delta)=sqrt(48)=4sqrt(3)`



Now, `alpha=(-b+sqrt(Delta))/(2a)`

`=(-(6)+4sqrt(3))/(2*3)`

`=(-6+4sqrt(3))/6`

`=(-3+2sqrt(3))/3`



and, `beta=(-b-sqrt(Delta))/(2a)`

`=(-(6)-4sqrt(3))/(2*3)`

`=(-6-4sqrt(3))/6`

`=(-3-2sqrt(3))/3`


`=>x = (-3+2sqrt(3))/3" or "x = (-3-2sqrt(3))/3`

`:.` x-intercepts are `((-3+2sqrt(3))/3,0)` and `((-3-2sqrt(3))/3,0)`

`:.` x-intercepts are `(0.1547,0)` and `(-2.1547,0)`


This material is intended as a summary. Use your textbook for detail explanation.
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2. `y=(x+2)^2-9` Example-2
(Previous example)
4. `y=3(x+1)^2-4` Example-4
(Next example)





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