What is the equation of the tangent drawn to the curve y = x^3 - 2x + 1 at x = 2?
The equation of a two-dimensional non-vertical line can easily be determined if its gradient 'm' and a point on the line (x0, y0) is known, using the formula m = (y - y0) / (x - x0). The gradient of the tangent can be easily determined as, by definition, it is value of the derivative of the curve at x = 2. Differentiating with respect to x yields that y' = 3x2 - 2. Since this y'(x) function gives for any x the gradient of the tangent at x, calculating y'(2) should give the gradient at x = 2. y'(2) = 3 * 22 - 2 = 10. Now a point needs to be found on the tangent. A trivial choice is to determine the intersection of the curve and the tangent. It is known that the x-coordinate of this point is 2, and the y-coordinate can be calculated using the curve's equation (as this point of intersection should satisfy y = x3 - 2x + 1 too). Substituting x = 2 results in y = 23 - 2*2 + 1 = 5. Therefore, a point on the line is (2, 5) and the gradient is 10. Using the formula, the equation of the tangent is 10 = (y - 5) / (x - 2)
