Find an equation of the curve that satisfies the given conditions: (d^2y/dx^2)=6x, the line y=5-3x is tangent to the curve at x=1

Answer 1

since y" = 6x
y' = 6x+a
y = 3x^2+ax+b

we know the slope is y'(1) = -3, so
6(1) + a = -3
a = -9
so, y = 3x^2-9x+b

since 5-3x=2 at x=1,
y(1) = 2, and we have

3-9+b = 2
b = 8

and y = 3x^2-9x+8

See the graphs at

http://www.wolframalpha.com/input/?i=plot+y%3D3x^2-9x%2B8%2C+y%3D5-3x+for+-1+%3C%3D+x+%3C%3D+3

Answer 2

But .... if the second derivative is linear, the original function must be a cubic.

y'' = 6x
y' = 3x^2 + a
at x = 1 we have y = 5 - 3x is a tangent, so y' = -3 when x=1
-3 = 3(1) + a
a = -6

y' = 3x^2 - 6
then y = x^3 - 6x + b

at (1,2)
2 = (1)^3 - 6(1) + b
b = 7

y = x^3 - 6x + 7

Wolfram confirmation:
http://www.wolframalpha.com/input/?i=plot+y%3Dx%5E3+-+6x+%2B+7%2C+y%3D5-3x+for+-3+%3C%3D+x+%3C%3D+4

Answer 3

all that anti-derivative stuff is too much for me that early. Thanks for the fix.

I'm sure Tim caught the error and fixed it though.
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Right, buddy?