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Home Q & A Board Science-Related Homework Help Mathematics
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rivera5454 rivera5454
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12 years ago
Can anybody tell me step by step how to do this calculus question? BTW ^ = to the power of.

The gradient function of a curve is f'(x)=3x^2-8x+1
The curve passes through the point (2,6)
Find the equation of the curve.
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wrote...
#1
12 years ago
if
f'(x)=3x^2-8x+1
then
f(x) = x^3 -4x^2 +x + A (some constant we don't know yet)
Since f(x) passes through (2,6)
6=2^3 -4(2^2) + 2 + A
6=8-16+2+A
6=-6+A
A=12

So the equation of f(x) is:
f(x) = x^3 -4x^2 +x + 12
wrote...
#2
12 years ago
f '(x) = 3x^2 - 8x + 1

integrating both sides

f(x) = x^3 - 4x^2 + x + C

since f(x) passes through (2.6),  f(2) = 6

8 - 16 + 2 + C = 6

C - 6 = 6

C = 12

so f(x) =  x^3 - 4x^2 + x + 12
wrote...
#3
12 years ago
f(x) = x^3 - 4x^2 + x + C

plug in (2, 6) into f(x) to solve for C

6 = 8 - 16 + 2 + C

C = 6 - 8 +16 - 2 = 12

Hence :  f(x) = x^3 - 4x^2 + x + 12
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