Identify the Minimum value of the function y=x^2+6x+ 9

Explanation:

You can factor x^2+6x+9 into (x+3)^2, which turns into 1(x - (-3) )^2 + 0

So y = x^2+6x+9 becomes y = 1(x - (-3) )^2 + 0

That matches with vertex form y = a(x-h)^2 + k

The vertex is (-3,0). The y coordinate of the vertex directly tells us the minimum value. We know we have a minimum because a = 1 is positive. If a < 0, then we'd have a max y value instead.

Alternatively, compare y = x^2+6x+9 to the form y = ax^2+bx+c

We see that a = 1, b = 6, c = 9

The x coordinate of the vertex is

h = -b/(2a)

h = -6/(2*1)

h = -3

The x coordinate of the vertex is x = -3. Plug this into the original equation to find the y coordinate of the vertex

y = x^2 + 6x + 9

y = (-3)^2 + 6(-3) + 9

y = 9 - 18 + 9

y = -9 + 9

y = 0

The y coordinate of the vertex is y = 0. This is the smallest y output possible.

d? maybe

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