y = 3x^{2} + 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(x^{2}+6x)

Now let's "complete the square" by adding (6/2)^{2} to the x^{2} + 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(x^{2}+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 x^{2} 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)