differentiation (1 Viewer)

[bos1234]

A curve with gradient function f'(x)=cx+d has a turning pt at (2,0) and cross the y-axis when y=4. Find c and d, and hence find the eqn of the curve.

this what i done----------------------------

f'(x) = cx+d
0=2c+d
d=-2c

f'(x)=cx-2c
f(x) = 1/2cx^2 - 2cx + d
d=4
c=-2

y=-x^2+4x+4

answer:x^2-4x+4

 

[webby234]

I don't know where you get
f(x) = 1/2cx^2 - 2cx + d from

the constant is not d, but another number (call it k)
You then have f(0) = 4 = k

f(x) = 1/2cx² - 2cx + 4
f(2) = 0
2c - 4c + 4 = 0
c = 2

So you get f(x) = x² - 4x + 4

EDIT: Actually I think I might know where you were going wrong - d is the intercept of the gradient function, which is not necessarily the intercept of f(x)