Thread: Finding the values of a, b and c.

The tangent to the curve of y=ax^2 + bx +c at the point where x=2 is parallel to the line y=4x. There is a stationary point at (1, -3). Find the value of a, b and c.

I was really unsure of how to work out this question, so first I differentiated y.
dy/dx=2ax+b
The point (1, -3) is on the curve and therefore:
-3=a(1)^2+b(1)+c
-3=a^2+b+c

But I don't know how to find another equation, and I honestly don't even know if what I've done so far is on the right track.

Hello, grooverandshaker!

You were off to a good start . . .

. .

The slope of a tangent to the curve is: .
When , the slope of the tangent is
. .

There is a stationary point at
That is, when , the derivative equals zero.
. . We have: .

The point is on the curve.
That is, when
. . We have: .


We have a system of equations: .

Subtract [1] - [2]: .

Substitute into [2]: .

Substitute into [3]: .


Therefore, the function is: .

Thank you so much mate! That makes perfect sense. =)