Question

[VVVCCCEEE]

Help?

[S200]

Hey there!
You have two points, so you can use rise over run for the gradient.
Equate this to the \(\frac{dy}{dx}\) and you should be able to work out k.

Once you have k, you can integrate to get the curve.

So...

So k must be zero?

The integrated curve would then be \(f(x)=x^2+c\), and c must logically be \((-3)\)

EDIT: Wrong working, see below.

[S_R_K]

Hey there!
You have two points, so you can use rise over run for the gradient.
Equate this to the \(\frac{dy}{dx}\) and you should be able to work out k.

Once you have k, you can integrate to get the curve.

So...

So k must be zero?

The integrated curve would then be \(f(x)=x^2+c\), and c must logically be \((-3)\)

Spoiler

k = 4, since the gradient of the tangent at x = 3 is equal to 2. Hence 2(3) + k = 2, implying that k = 4.

[fun_jirachi]

Help?

Sorry to rain on your parade SRK, you've forgotten the negative sign!! 
And for part b) integrate and substitute back (3,6) to find the constant of integration.
Answers in spoiler

Spoiler

a) k = -4
b) y = x²-4x+9

[S200]

Hmmm. Lucky we've got you guys around...

Listen to what they said VVVCCCEEE...