Question

What is the vertex form of `y=6x^2+14x-2`?

Answer 1

`y=6(x+7/6)^2 - 61/6`

So your vertex = `(-7/6, -61/6)`

Explanation:

Vertex form is:

`y=a(x+h)^2 + k` and the vertex is: #(-h,k)

To put the function in vertex for we have to complete the square with the x values:

`y=6x^2+14x-2`

first isolate the term with x:

`y+2=6x^2+14x`

to complete the square the following must be done:

`ax^2 + bx + c`

`a=1`

`c=(b/2)^2`

the Square is: `(x + b/2)^2`

In your function `a=6` so we need to factor that out:

`y+2 =6(x^2+14/6x)`

`y+2 =6(x^2+7/3x)`

now add the c in to both sides of the equation, remember on the left we must add in 6c since the c on the right in inside the factored portion:

`y+2+6c =6(x^2+7/3x + c)`

now solve for c:

`c=(b/2)^2 = ((7/3)/2)^2= (7/6)^2=49/36`

`y+2+6(49/36) =6(x^2+7/3x + 49/36)`

`y+2+49/6 =6(x+7/6)^2`

`y+61/6 =6(x+7/6)^2`

Finally we have vertex form:

`y=6(x+7/6)^2 - 61/6`

So your vertex = `(-7/6, -61/6)`

graph{6x^2+14x-2 [-19.5, 20.5, -15.12, 4.88]}