Question 1.P.3: Find the equation of the tangent line to the cubic polynomia......

Let Q = ( x, x³ ) be another point on the graph of the cubic polynomial y = x³ . The slope of the line joining (-1,-1) and (x,x³) is $m=\frac{x^{3}-(-1)}{x-(-1)}=\frac{x^{3}+1}{x+1}=\frac{\left(x+1\right)\left(x^{2}-x+1\right)}{x+1}=x^{2}-x+1,x\neq-1\;.$

Notice how we factored the numerator, which is a sum of 2 cubics. As Q approaches (-1, -1), x approaches -1, and the slope approaches (-1)² – (-1) + 1 = 3. Hence, the slope of the tangent line is 3, and the equation becomes y – (-1) = 3( x-(-1)), or y = 3x + 2.