A point on -8x^2+5xy+y^3=-149

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Question from Vivian, a student:

Consider the curve defined by -8x2+5xy+y3=-149
a) find dy/dx
b) Write an equation for the line tangent to the curve at the point (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 of 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

Hi Vivian,

find dy/dx

Differentiate both sides of -8x²+5xy+y³ = -149 using implicit differentiation. Solve the resulting expression for dy/dx.
Write an equation for the line tangent to the curve at the point (4,-1)

Use (y - y₁) = m (x - x₁) where m is dy/dx evaluated at x = 4, y = -1.
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.

k is approximately the y-value of the point on the tangent line with x = 4.2.
write an equation that can be solved to find the actual value of k so that the point (4.2,k) is on the curve

Substitute (4.2,k) into -8x²+5xy+y³ = -149.
Solve the equation found in part d) for the value of k

Solve for k.

Harley

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