Help with quadratic equations?

Question

I do not understand these problems in my math packet:

Find an equation of the quadratic function described:

1. The minimum value of f is f(3) = -5 and f(1)=2

2. Its graph is a parabola that is tangent to the x-axis at (4,0) and has y-intercept 6.

Answer 1

1.

The general equation of a parabola with a vertical axis is:

(x - h)^2 = 4a(y - k)

with vertex (h, k).

As (x - h)^2 is a square, its minimum value is 0.

This is when x = h and y = k.

Therefore h = 3, k = - 5, and a must be positive.

The equation is:

(x - 3)^2 = 4a(y + 5).

Putting x = 1 and y = 2:

(- 2)^2 = 4a(7)

4a = 4 / 7.

The equation is:

(x - 3)^2 = (4 / 7)(y + 5).

2.

(x - h)^2 = 4a(y - k) as before.

h = 4, k = 0.

(x - 4)^2 = 4ay

When x = 0, y = 6:

(- 4)^2 = 24a

4a = 8 / 3.

The equation is:

(x - 4)^2 = (8 / 3)y.