Consider the following graph of a quadratic function write the equation for the quadratic function in standard form.

The equation is y=x²+6x+7.

We start out writing this in vertex form:

y = a(x-h)² + k, where (h, k) are the coordinates of the vertex.  Our vertex is at (-3, -2):

y = a(x--3)²+-2
y = a(x+3)²-2
y = a(x+3)(x+3)-2
y = a(x*x + 3*x + 3*x + 3*3) - 2
y = a(x²+3x+3x+9)-2
y = a(x²+6x+9)-2

We know that the y-intercept is at (0, 7); substituting these values into x and y in our equation we can find the value of a:
7 = a(0²+6(0) + 9) - 2
7 = a(9) - 2

Adding 2 to both sides, we have
9 = 9a

Divide both sides by 9:
9/9 = 9a/9 
1 = a

This means our equation is 
y = 1(x²+6x+9)-2
y = x²+6x+9-2
y = x²+6x+7

I'm sure the answer is a