C1 Differantiaton

I will rep you if im still online when you answer :redface:

The curve C has equation y = f(x). Given that

dy/dx = 3x^2 – 20x + 29

and that C passes through the point P(2, 6),

(a) find y in terms of x.

(b) Verify that C passes through the point (4, 0).

(c) Find an equation of the tangent to C at P.

The tangent to C at the point Q is parallel to the tangent at P.

(d) Calculate the exact x-coordinate of Q.

(n.b if y in terms x just differntiating then thats 6x^1 - 20 , i just cant memba if that same :P )

Reply 1

BCHL85

a) rewrite dy/dx = 3x^2 – 20x + 29
to dy = (3x^2 – 20x + 29)dx
Then integrate it
y = INT((3x^2 – 20x + 29)dx)
Remember there's a constant K in your function after integration
say y = g(x) + K
Find K by puting y = 6, x = 2 into y = g(x) + K.

(If you don't know how to integrate it, please open your notebook, it's quite simple)

b) Substitute x = 4 in y = g(x) + K, you'll get y = 0
So the curve passes (4, 0)

c)Tangent at P:
You know dy/dx = 3x^2 – 20x + 29
Plug x = 2 to get y'(2)
So y = y'(2)(x-2) + 6.

d) Tangent at Q is parallel to tangent at P
=> it has same gradient y'(2)
You'll find xQ by solving this equation
y'(2) = 3x^2 – 20x + 29
and eliminate x = 2 since xP = 2.
Then you'll get yQ = g(xQ) + K

Sorry I can't calculate everything for you, just show you the way to do it