The curve y = ax 2 + bx + c passes through the point (2, 28) and is tangent to the line at the origin. Find a, b, and c.?

Do you mean "and its tangent at that point passes through the origin", and "Calculus"?

Tangent to what line at the origin?

28 = 4a + 2b + c

It passes through the origin, too

0 = 0a + 0b + c

0 = c

28 = 4a + 2b

14 = 2a + b

b = 14 - 2a

y = ax^2 + bx + c

y = ax^2 + (14 - 2a) * x + 0

y = ax^2 + (14 - 2a) * x

Assuming it's tangent to y = 0 at (0 , 0)

y' = 2ax + (14 - 2a)

y' = 0 when x = 0

0 = 2a * 0 + 14 - 2a

0 = 14 - 2a

0 = 7 - a

a = 7

y = ax^2 + (14 - 2a) * x

y = 7x^2 + (14 - 14) * x

y = 7x^2 + 0x

y = 7x^2

The question mentions a line that is tangent to the curve at the origin, but gives none of the necessary details about that line, like its slope.