Question

the curve (x^{2}+y^{2}+7x - 7y = 0) is shown in the graph below in blue. use the graph below to draw line tangent to the curve at the point ((-1,-1)). submit your answer by dragging the black point such that the drawn tangent line (shown in red) matches the equation for it. provide your answer below:

Explanation:

Step1: Differentiate the curve implicitly

Differentiate $x^{2}+y^{2}+7x - 7y=0$ with respect to $x$.
Using the power - rule and chain - rule, we have:
$2x + 2y\frac{dy}{dx}+7-7\frac{dy}{dx}=0$.

Step2: Solve for $\frac{dy}{dx}$

Rearrange the terms to isolate $\frac{dy}{dx}$:
$2y\frac{dy}{dx}-7\frac{dy}{dx}=-2x - 7$.
Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(2y - 7)=-2x - 7$.
So, $\frac{dy}{dx}=\frac{-2x - 7}{2y - 7}$.

Step3: Evaluate $\frac{dy}{dx}$ at the point $(-1,-1)$

Substitute $x=-1$ and $y = - 1$ into $\frac{dy}{dx}$:
$\frac{dy}{dx}\big|_{x=-1,y=-1}=\frac{-2(-1)-7}{2(-1)-7}=\frac{2 - 7}{-2 - 7}=\frac{-5}{-9}=\frac{5}{9}$.
The equation of the tangent line at the point $(x_0,y_0)=(-1,-1)$ using the point - slope form $y - y_0=m(x - x_0)$ (where $m=\frac{5}{9}$) is:
$y+1=\frac{5}{9}(x + 1)$.
$y+1=\frac{5}{9}x+\frac{5}{9}$.
$y=\frac{5}{9}x+\frac{5}{9}-1$.
$y=\frac{5}{9}x-\frac{4}{9}$.

Answer:

$y=\frac{5}{9}x-\frac{4}{9}$