Alan

If $y=ax^2+bx+c$ then usingthe vertex values we have 

$4=4a+2b+c$        (1)

Using the point (7,7) we have

$7=49a+7b+c$      (2)

We need a thid equation.  Because of the symmetry about the vertex, the distance on the x-axis of point (7,7) from the vertex is matched by a point the same distance from th vertex in the opposite x direction.  Since point(7,7) is a distance of +5 in the x-direction from (2,4), we must have a point with y-value 7 at x-value 2 - 5 = -3. ie. the point (-3, 7) is also on the curve.  Hence our third equation becomes:

$7=9a-3b+c$        (3)

There are now three equations in three unknowns.  Can you take it from here?

This should help:

This can be factorized as:  $2(t-6)(t+1)$

The minimum occurs half way between the two roots.

Like so:

This should help:

Find x from these two equations.  Then you should be able to find the required area by subtracting the area of the two identical triangles from the area of the rectangle.

Like so:

Can you finish?

Set -4.9t^2 + 14t - 0.4 = 6, solve the quadratic for two values of t and take the difference between them.

Here's one way:

See https://web2.0calc.com/questions/pre-calc-problem#r5

Numerators that are multiples of three are able to express the fraction in lowest form, but aren't actually so expressed here.  For example, 3/2019 isn't in lowest form.

And what about numerators that are multiples of 673?