Question

Find an equation of the tangent line to the curve 9x³ – y³ = 1 at the point (0, -1).

Answer 1

Let us find an equation of the tangent line to the curve $9x^3 – y^3 = 1$ at the point $(0, -1).$ It follows that $27x^2 – 3y^2y' = 0,$ and hence $y'(0)=\frac{27\cdot 0^2}{3(-1)^2}=0.$ Using the formula $y=y_0+y'(x_0)(x-x_0)$ for $x_0=0$ and $y_0=-1,$ we conclude that $y=-1+0(x-0)=-1$, that is $y=-1$ is an equation of the tangent line to the curve $9x^3 – y^3 = 1$ at the point $(0, -1).$