To find the equation of the tangent line to the graph of the given equation at the point $(-1, 2)$, we need to follow these steps:

Step 1: Rewrite the equation for clarity

The equation given is: $-81 = -9y^3 - 2y^2 + xy^2 - 9x + 5x^3 - x^2$

Step 2: Differentiate implicitly

To find the slope of the tangent line, we differentiate both sides of the equation implicitly with respect to $x$.

• Differentiate the left-hand side: $\frac{d}{dx}(-81) = 0$

• Differentiate the right-hand side: $\frac{d}{dx}\left( -9y^3 - 2y^2 + xy^2 - 9x + 5x^3 - x^2 \right)$

Using the chain rule and the product rule where necessary, we get:

$0 = -27y^2 \frac{dy}{dx} - 4y \frac{dy}{dx} + y^2 + 2xy \frac{dy}{dx} - 9 + 15x^2 - 2x$

Step 3: Plug in the point $(-1, 2)$

Now substitute $x = -1$ and $y = 2$ into the equation to find $\frac{dy}{dx}$, the slope of the tangent line.

Substitute $x = -1$ and $y = 2$ into the differentiated equation: $0 = -27(2)^2 \frac{dy}{dx} - 4(2) \frac{dy}{dx} + (2)^2 + 2(-1)(2)\frac{dy}{dx} - 9 + 15(-1)^2 - 2(-1)$

Simplify: $0 = -108 \frac{dy}{dx} - 8 \frac{dy}{dx} + 4 - 4 \frac{dy}{dx} - 9 + 15 + 2$ $0 = -120 \frac{dy}{dx} + 4 + 8$ $0 = -120 \frac{dy}{dx} + 12$

Solving for $\frac{dy}{dx}$: $120 \frac{dy}{dx} = 12$ $\frac{dy}{dx} = \frac{12}{120} = \frac{1}{10}$

So the slope of the tangent line at the point $(-1, 2)$ is $\frac{1}{10}$.

Step 4: Write the equation of the tangent line

Now, using the point-slope form of the equation of a line: $y - y_1 = m(x - x_1)$ where $m$ is the slope and $(x_1, y_1)$ is the point $(-1, 2)$, we plug in the values: $y - 2 = \frac{1}{10}(x - (-1))$ $y - 2 = \frac{1}{10}(x + 1)$

This is the equation of the tangent line in point-slope form.

Final Answer:

$y - 2 = \frac{1}{10}(x + 1)$

Would you like to go over any specific step or need more details?

Relative questions:

1. How do we differentiate implicitly for more complex equations involving both $x$ and $y$?
2. Why is the point-slope form useful for tangent line problems?
3. Can you explain the product rule and how it applied to the differentiation step?
4. How does the chain rule function in implicit differentiation?
5. How can we verify if the calculated slope is correct?

Tip:

Always verify the differentiation carefully, especially when using implicit differentiation, since small errors can lead to incorrect slopes for tangent lines.