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## Q(x)=3x^2+18x+26

The vertex is

y = 3x2 + 18x +26

Let's put this quadratic into the "vertex form":

(y-k) = a(x-h)2

Where (h,k) is the location of the vertex.  To do this, first subtract 26 from both sides of the equation then factor the 3 out of the RHS:

y-26 = 3(x2+6x)

Now let's "complete the square" by adding (6/2)2 to the x2 + 6x on the RHS.  We'll have to add 3*(6/2)2 to the LHS to keep things equal:

y-26+3(6/2)2 = 3(x2+6x+(6/2)2)

y -26+27 = 3(x+3)2

(y+1) = 3(x+3)2

Hence (h,k) = (-3,-1)

RHS = Right Hand Side of the equation
LHS =  Left Hand Side of the equation

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Now here's the easier way.  The x-coordinate of the vertex is always equal to -b/2a, where a is the coefficient of the x2 term (3 in this case) and b is the coefficient of the x term (18 in this case):

x = -b/2a = -18/2*3 = -18/6 = -3

To get the y-coordinate, plug x = -3 into the original quadratic:

y = 3*(-3)2 + 18(-3) + 26 = 27 -54 + 26 = -1

Vertex is located at (-3,-1)
Philip's answer is great, and really the way to go with this.
I'll present an advanced alternative once you learn calculus:

Take the derivative of the equation, you get 6x+18.  When you have this remember that the derivative tells you the slope at any point on the curve.  When you are at the vertex the slope should be 0, so you can set 6x+18=0 and solve for x getting x = -18/6 = -3

Plugging this back into your equation you can get the value for y: 3(9)+18(-3) + 26 = 27 - 54+26 = 53-54 = -1
So you quickly get that the vertex is at (-3,-1).

Hope this is useful in the future.