Find and equation to the curve y=3x^3+2x^2-1 at x = -1

First substitute x = -1 into the equation for the line to find the coordinates at the point where x = -1y = -3+2-1 = -2Next differentiate to find dy/dxdy/dx = 9x²+4xSubstitute x = -1 to find the gradient of the curve at (-1,-2)dy/dx = 9-4 = 5Finally, use the equation y-y₁ = m(x-x₁) to find an equation of the tangenty+2 = 5(x+1)