Thread:

1. 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.

2. Re: Finding the values of a, b and c.

Hello, grooverandshaker!

You were off to a good start . . .

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

$\text{The slope of the line }y\,=\,4x\text{ is }4.$

The slope of a tangent to the curve is: . $y' \:=\:2ax + b$
When $x = 2$, the slope of the tangent is $4.$
. . $\text{We have: }\:4a + b \:=\:4$

There is a stationary point at $(1,-3).$
That is, when $x = 1$, the derivative equals zero.
. . We have: . $2a + b \:=\:0$

The point $(1,-3)$ is on the curve.
That is, when $x = 1,\:y = -3$
. . We have: . $a + b + c \:=\:-3$

We have a system of equations: . $\begin{Bmatrix}4a + b\qquad &=& 4 & [1] \\ 2a + b \qquad &=& 0 & [2] \\ a + b + c &=& \text{-}3 & [3] \end{Bmatrix}$

Subtract [1] - [2]: . $2a \,=\,4 \quad\Rightarrow\quad \boxed{a \,=\,2}$

Substitute into [2]: . $4 + b \:=\:0 \quad\Rightarrow\quad \boxed{b \:=\:-4}$

Substitute into [3]: . $2 - 4 + c \:=\:-3 \quad\Rightarrow\quad \boxed{c \:=\:-1}$

Therefore, the function is: . $y \:=\:2x^2 - 4x - 1$

3. Re: Finding the values of a, b and c.

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