PLEASE help with this Calculus problem?

Question

Consider the curve defined by -8x^2+5xy+y^3=(-)149
a) Find dy/dx
b) Write an equation for the line tangent to the curve at (4,-1)
c) There is a number k so that the point (4.2, k) is on the curve. Using the tangent line found in part (b), approximate the value of k.
d) Write an equation that can be solved to find the actual value for k so that the point (4.2, k) is on the curve.
e) Solve the equation found in part (d) for the value of k.

I'm pretty sure i got parts a and b..... but need major help w/ the rest.....

Answer 1

- 8x^2 + 5xy + y^3 = - 149
a)
- 16xdx + 5ydx + 5xdy + 3y^2dy = 0
(5x + 3y^2)dy = (16x - 5y)dx
dy/dx = (16x - 5y) / (5x + 3y^2)
b)
m = (16*4 + 5) / (5*4 + 3) = 69/15 = 23/5
y = (23/5)(x - 4) - 1
y = (23/5)x - (92 + 5)/5
y = 4.6x - 19.4
c)
k ≈ 4.6*4.2 - 19.4
k ≈ - 0.08
d)
- 8(4.2)^2 + 5(4.2)k + k^3 = - 149
k^3 + 21k + 7.88 = 0
e)
k ≈ - 0.37277143918

Answer 2

I don't think you should have people do you homework for you.