Final answer:
The equation of the tangent line at the point (-2, -1) for the given curve is y = -9x - 19, calculated by finding the derivative and using the point-slope form.
Step-by-step explanation:
To find the equation of the tangent line to the given curve at the point (-2, -1), we first need to calculate the derivative of the function to find the slope of the tangent line at that point.
Let's take the derivative of the implicit equation x³y - 1 = y² - 3x using implicit differentiation, which gives us:
3x²y + x³(dy/dx) = 2y(dy/dx) - 3
We then solve for dy/dx to find the slope at our point of interest:
x³(dy/dx) - 2y(dy/dx) = 3 - 3x²y
(dy/dx)(x³ - 2y) = 3 - 3x²y
dy/dx = (3 - 3x²y)/(x³ - 2y)
Substituting the coordinates of the point (-2, -1) into the derivative, we get:
dy/dx = (3 - 3(-2)²(-1))/((-2)³ - 2(-1))
dy/dx = -9
The slope of the tangent line at (-2, -1) is -9. Now, using the point-slope form of a line, y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point on the line, we get the equation of the tangent:
y - (-1) = -9(x - (-2))
y + 1 = -9(x + 2)
y = -9x - 18 - 1
y = -9x - 19
The equation of the tangent line at the point (-2, -1) is y = -9x - 19.