A curve has equation y = 3x^3 - 7x + 10. Point A(-1, 14) lies on this curve. Find the equation of the tangent to the curve at the point A.
Firstly, we define a tangent as a straight line touching a curve at 1 point only.We can describe a straight line using this linear relation:y - y1 = m.(x - x1)We know a point on this line A(-1, 14) = (x1, y1). Hence, all we need is the gradient m, to find it's equation.The tangent and curve share the same gradient at point A . We have the curve's equation and can use it to find this gradient.We need an equation that describes the curve's gradient, dy/dx, in terms x. Hence we must differentiate our equation y = 3x3 -7x +10 as follows:dy/dx = 9x2 - 7To find the gradient at A(-1, 14), we sub in x = -1, hence,dy/dx = 9(-1)2 - 7 = 2Now substituting the gradient m = 2 and coordinates of point A into the original linear relation, we can find our equation in the form y = mx + c,y - 14 = 2(x + 1)Answer: y = 2x +16
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