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.

I've answered the first one elsewhere today........

Geometrically speaking, a parabola is defined as the set of points that are the same distance from a given point and a given line. The point is called the focus of the parabola and the line is called the directrix of the parabola. Suppose P is a parabola with focus (4,3) and directrix y=1 . The point (8,6) is on P because (8,6)  is 5 units away from both the focus and the directrix. If we write the equation whose graph is P in the form y=ax^2 + bx + c, then what is a*b*c ?

The vertex  will be found at (4,2)   and we can write that :

(y - 2) = (a)(x - 4)^2       and since (8,6)   is on the curve, we can solve for a

(6 -2)  = (a) (8 - 4)^2    simplify

4 = (a)(4)^2

4 = (a)16   →    a = 4/16  = 1/4

Since the x coordinate of the vertex is given by -b/ (2a)  we have that   -b/[2 (1/4)]  = 4   → -b / (1/2) = 4 → -b = 2 → b = -2

And using the fact that  (4,2)  is on the graph, we can find c, thusly :

y = ax^2 + bx + c  ......so......

2 = (1/4)(4)^2 -2(4) + c

2 = 4 - 8 + c

2  = -4 + c

c = 6

Then a*b*c =  (1/4) (-2) (6)    =  (1/4)(-12)    = -3

Here's the graph of the function :  https://www.desmos.com/calculator/1hi34pnvzw

Find the area of the region enclosed by the graph of the equation x^2 + y^2 = 4x + 6y+13.

This will be a circle....we need to find the radius to find the area enclosed by the circle

Let's write this as    x^2 -4x + y^2 -6y - 13  = 0        complete the square on x and y 

x^2 - 4x + 4  + y^2 - 6y + 9  - 13  =  4 + 9

(x -2)^2 + (y -3)^2  - 13  = 13       add 13 to both sides

(x -2) ^2  + (y - 3)^2   = 26        26 = (radius)^2   so

The area of a circle =  pi * (radius) ^2

So....the area =  pi * 26 =   26* pi  ≈  81.68 units^2