Write the equation in vertex form

Write the equation in vertex form.
y = x² + 6x + 9

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Vertex form of a quadratic equation is y = a(x - h) + k,
where h and k are the x and y coordinate of the vertex.

One Method
Find the x-coordinate of the vertex, which is x = -b/2a

In y = x² + 6x + 9,
a = 1  {the coefficient of x²}
b = 6 {the coefficient of x}
c = 9  {the constant term}

x = -b/2a {the x-coordinate of the vertex}
= -6/2(1)  {substituted 6 in for b and 1 in for a}
x = -3  {divided}
the x-coordinate of the vertex is -3

Then to find the y-coordinate, substitute -3 back in for x.

y = (-3)² + 6(-3) + 9  {substituted -3, in for x, into y = x² + 6x + 9}
y = 9 - 18 + 9  {evaluated exponent and multiplied}
y = 0  {subtracted and added}
the y-coordinate of the vertex is 0

Substitute the x and y coordinates back into vertex form, along with a,
a is the same in vertex form as it is in standard form {the original equation was in standard form}.

y = a(x - h)² + k  {vertex form of a quadratic equation}
y = [x - (-3)]² + 0  {substituted 1 in for a, -3 in for x, and 0 in for k, into vertex form}
y = (x + 3)²  {simplified}


Another Method
Complete the square within the equation:

y = x² + 6x + 9

Take 1/2 of the coefficient of x, square it and add it to the first two terms,
also subtract it from the 9 to keep the equation balanced.

y = x² + 6x + 9 + 9 - 9  {added 9 and subtracted 9 from the right side}
y = x² + 6x + 9  {simplified}
y = (x + 3)² {factored the binomial, which you could've done from the very beginning!}

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