Algebra Help (multiple problems)

1. Find the vertex of the graph of the equation y = 3x^2 - 6x + 7.

The x coordinate of the vertex is  given by -b/(2a)  =  6 / (2*3)   =   6/6 =  1

The y coordinate of the vertex is given by substituting the x value of the vertex back into the function = 

3(1)^2 - 6(1) + 7  =    3 - 6 + 7 =    4

So.....the vertex lies at ( 1, 4)

Here's the graph :  https://www.desmos.com/calculator/q83yhluiyr

2. The graph of the equation y =ax^2 + bx + c, where a, b, and c are constants, is a parabola with axis of symmetry x = -3. Find b/a.

If the axis of symmetry is x = -3, then the x coordinate of the vertex lies at (-3, m)  and we know that the x coordinate of the vertex can be found  by :   -b/2a  =  -3

Multiplying by 2 on each side, we have that     -b/a  = -6   →    b / a = 6

3. The grid lines in the graph are one unit apart. The red parabola shown is the graph of the equation y=ax^2 + bx + c. Find abc.

It appears that the following 3 points are contained on the graph : (-3, 5), (-1, 3)  and (-5, 3)

So....using the fact that   -b/2a = the x coordinate of the vertex, we have that -b / 2a =  -3   →   -b = -6a  →  b = 6a

So we can set up the following equations to find abc

a(-3)^2 + 6a(-3) + c  =  5    →    9a - 18a + c  =  5  →   -9a + c  = 5    (1)

a(-1)^2 + 6a(-1) +  c = 3   →    -5a +  c  =    3        (2)

Multiply  (1) by -1 and add to (2)  and we have

4a = -2    →   a = -2/4   →   a = -1/2

So b = 6(-1/2)   = -3

And using (1) to find c, we have     -9(-1/2) + c  = 5    →   9/2 + c  = 5   →   c = 10/2  - 9/2 = 1/2

So  abc =  (-1/2)(-3)(1/2)    =  3/4

Here's the graph to show that our function is correct :  https://www.desmos.com/calculator/gtspbs8qjg

4. The graph of (x-3)^2 + (y-5)^2=16 is reflected over the line y=2. The new graph is the graph of the equation x^2 + Bx + y^2 + Dy + F = 0 for some constants B, D, and F. Find B + D + F.

The first graph  is a circle with its center at (3, 5)   and a radius of 4

If it is reflected across y = 2, the new center lies at (3,-1)   with the same radius

So....the new equation becomes

(x - 3)^2 + (y + 1)^2    = 4        expand this   and simplify

x^2 -6x + 9  + y^2 + 2y + 1   = 4

x^2 -6x + y^2 + 2y + 6   = 0

And if

x^2 + Bx + y^2 + Dy  + F  = 0

Then, equating coefficients.........B = -6 ,  D = 2   and F = 6   and B + D + F =  2

Here's the graph of the original circle and its reflection across y = 2 :   https://www.desmos.com/calculator/d6l7hx2ye5