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

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Answer 1 · 2018-05-30T20:03:52

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

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

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]}