How do you convert the following equation from standard to vertex form by completing the square: #y=3x^2+12x+5#?

1 Answer
Douglas K.
Nov 15, 2016

Please see the explanation.

Explanation:

The given equation is the equation of a parabola that opens upward (or downward). The vertex form of the equation of a parabola that opens upward (or downward) is:

#y = a(x - h)^2 + k#

where #(h, k)# is the vertex and "a" is the coefficient of the #x^2# term.

Given: #y = 3x^2 + 12x + 5#

#a = 3#, therefore, add 0 in the form #3h^2 - 3h^2# to the equation:

#y = 3x^2 + 12x + 3h^2 - 3h^2 + 5#

Factor out 3 from the first 3 terms:

#y = 3(x^2 + 4x + h^2) - 3h^2 + 5#

Set the middle term in right side of the pattern, #(x - h)^2 = x^2 - 2hx + h^2#, equal to the middle term in the equation:

#-2hx = 4x#

Solve for h:

#h = -2#

Substitute the left side of the pattern into the equation:

#y = 3(x - h)^2 - 3h^2 + 5#

Substitute -2 for h:

#y = 3(x - -2)^2 - 3(-2)^2 + 5#

Combine the constant terms:

#y = 3(x - -2)^2 - 7#

The above is the vertex form.

The vertex can be read directly from the equation; it is at:

#(-2, -7)#

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