first time poster so sorry if I'm doing something wrong.
"Consider the closed curve in the xy-plane given by $2x^2 - xy + y^3 + x = 9$. Find equation(s) of all tangent lines to the curve at $y = 1$."
So after I took the derivative I got:
$$\frac{y - 4x - 1}{3y^2 - x}$$
After that I have no clue what I'm supposed to do next. Sorry about the formatting and thanks for the help.
On
First, you want to find the points on the curve where $y = 1$. In this case, they are $x = 2$ and $x = -2$.
The equation of a line passing through $(x_0, y_0)$ is $y - y_0 = m (x - x_0)$. The slope, $m$, is just the derivative (which you've already found) evaluated at $(x_0, y_0)$. So now you can find the line for each point.
EDIT:
You have $2x^2−xy+y^3+x=9$, and you want to find points on it where $y = 1$. Plug in $1$ for $y$ and get $2x^2 - x + 1 + x = 9$, which simplifies to $x^2 = 4$.
Plug in $y = 1$ to your original equation: $2x^2 - xy + y^3 + x = 9$ and obtain a quadratic equation in $x$, which can be solved. At $y = 1$, we have $$2x^2 - x(1) + (1)^3 + x = 9\iff 2x^2 - 8 = 0 \iff x^2 - 4 = (x - 2)(x + 2) = 0$$ So we have two points on the curve where $y = 1$: one point corresponding to each solution: $x_1 = 2$, $x_2 = -2$. That is, you need to find the equations of the tangent line at the points $(2, 1)$ and the tangent line to the curve at $(-2,1)$.
Use $\dfrac{dy}{dx}= \dfrac{y - 4x - 1}{3y^2 - x}$, which you've already found, to compute the slope of tangent line at each point; i.e., evaluate $\frac {dy}{dx}$ at the point $(2, 1)$ to find $m_1$, and again at the point $(-2, 1)$ to find $m_2$.
$$\text{At}\;\;(2, 1),\quad\; m_1= -8$$
$$\text{At}\;\;(-2, 1),\quad m_2 = \dfrac {8}{5}$$
Then you'll have, for each point you've found $(x_i, 1)$, and the corresponding slope $m_i$ of the tangent line, a point on the line, and its slope. So you can write the corresponding equation of the tangent line using point-slope form of the equation of a line:
$$y - 1 = m_i(x - x_i)$$