Find the curve, y that satisfies the conditions:
2
2
d y = 6x
dx
; the line
y = 5- 3x
is
tangent to the curve at the point where
x = 1.
maybe you mean
d^2 y/dx^2 = 6 x
dy/dx = 3 x^2 + a
y = x^3 + a x + b
at x = 1, y = 2 and dy/dx = 5
your turn
You have said y" = 6x
and y'(1) = -3
y(1) = 2
so now we can proceed.
y' = 3x^2 + C
and since y'(1) = -3, C = -6
y' = 3x^2 - 6
y = x^3 - 6x + C
since y(1) = 2
y = x^3 - 6x + 7
see the graphs at
www.wolframalpha.com/input?i=plot+y+%3D+x%5E3+-+6x+%2B+7%2C+y%3D5-3x