HELP ME- MATHS PROBLEM

The curve c has equation y=f(x)

Given that dy/dx= 3x^2-20x+29 and that C passes through 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


THANKS

a) You need to integrate the differential to arrive at the y=f(x) equation. Use the known point, where x=2; y=6 to determine the Constant to add to the equation

b) Apply the equation where x=4 to show y=0

c) The derivative applied at (2,6) is the tangent to the curve ... so feed that into the derivative and the original formula to find the equation.

d) Use the equation from (c), find another solution that satisfies it.

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That's how you'd do it ;)

y = ∫(3x² - 20x + 29)dx = x³ - 10x² + 29x + c

Since (2, 6) is a point on the graph, then

6 = (2)³ - 10(2)² + 29(2) + c = 26 + c

c = -20

So y = x³ - 10x² + 29x - 20

The rest is up to you!

v