Final answer:
To find the tangent line to the curve x² + y³ = 9 at the point (1, 2), we first differentiate the equation and find dy/dx. Then, substituting the given values, we find the slope of the tangent line to be -1/6. Finally, using point-slope form, we find the equation of the tangent line to be y = -(1/6)x + 5/6.
Step-by-step explanation:
To find the tangent line to the curve x² + y³ = 9 at the point (1, 2), we need to find the derivative of the curve and evaluate it at the given point. Let's first differentiate the equation with respect to x:
2x + 3y² * dy/dx = 0
Now we can solve for dy/dx:
dy/dx = -2x / (3y²)
Substituting the values x = 1 and y = 2, the slope of the tangent line at the point (1, 2) is:
dy/dx = -2 * 1 / (3 * 2²) = -1/6
Therefore, the tangent line has a slope of -1/6. We can use the point-slope form of a line to find the equation of the tangent line:
y - 2 = (-1/6)(x - 1)
Simplifying the equation gives:
y = -(1/6)x + 5/6
So, the equation of the tangent line to the curve x² + y³ = 9 at the point (1, 2) is y = -(1/6)x + 5/6.