The equation of the tangent to the curve x³ + y² + 3y + x = 0 and passing through the point (2, -1).

Concept:

Steps to find the equation of the tangent to the curve:

Find the first derivative of f(x).

Use the point-slope formula to find the equation for the tangent line.

Point-slope is the general form: y - y₁=m(x - x₁), Where m = slope of tangent = \(\rm \frac {dy}{dx}\) 

Calculation:

Given curve x3 + y2 + 3y + x = 0

Differentiating w.r.t x

3x² + 2y\(\rm dy\over dx\) + 3\(\rm dy\over dx\) + 1 = 0

(2y + 3)\(\rm dy\over dx\) = -3x² - 1

\(\rm dy\over dx\) = \(\rm -{3x^2+1\over2y+3}\)

At (2, -1)

\(\rm dy\over dx\) = \(\rm -{3(2)^2+1\over2(-1)+3}\)

\(\rm dy\over dx\) = \(-{12+1\over 1}\) = -13

The equation of the tangent is

(y - (-1)) = -13(x - 2)

y + 1 = -13x + 26

y + 13x - 25 = 0