# Solve Tangent Lines Problems in Calculus

Tangent lines problems and their solutions, using first derivatives, are presented.

 Problem 1: Find all points on the graph of y = x 3 - 3x where the tangent line is parallel to the x axis (or horizontal tangent line). Solution to Problem 1: Lines that are parallel to the x axis have slope = 0. The slope of a tangent line to the graph of y = x 3 - 3x is given by the first derivative y '. y ' = 3x 2 - 3 We now find all values of x for which y ' = 0. 3x 2 - 3 = 0 Solve the above equation for x to obtain the solutions. x = -1 and x = 1 The above values of x are the x coordinates of the points where the tangent lines are parallel to the x axis. Find the y coordinates of these points using y = x 3 - 3x for x = -1 , y = 2 for x = 1 , y = -2 The points at which the tangent lines are parallel to the x axis are: (-1,2) and (1,-2). See the graph of y = x3 - 3x below with the tangent lines. Problem 2: Find a and b so that the line y = -3x + 4 is tangent to the graph of y = ax3 + bx at x = 1. Solution to Problem 2: We need to determine two algebraic equations in order to find a and b. Since the point of tangency is on the graph of y = ax3 + bx and y = -3x + 4, at x = 1 we have a(1)3 + b(1) = -3(1) + 4 Simplify to write an equation in a and b a + b = 1 The slope of the tangent line is -3 which is also equal to the first derivative y ' of y = ax3 + bx at x = 1 y ' = 3ax2 + x = -3 at x = 1. The above gives a second equation in a and b 3a + b = -3 Solve the system of equations a + b = 1 and 3a + b = -3 to find a and b a = -2 and b = 3. See graphs of y = ax3 + bx, with a = -2 and b = 3, and y = -3x + 4 below. Problem 3: Find conditions on a and b so that the graph of y = ae x + bx has NO tangent line parallel to the x axis (horizontal tangent). Solution to Problem 3: The slope of a tangent line is given by the first derivative y ' of y = ae x + bx. Find y ' y ' = ae x + b To find the x coordinate of a point at which the tangent line to the graph of y is horizontal, solve y ' = 0 for x (slope of a horizontal line = 0) ae x + b = 0 Rewrite the above equation as follows e x = -b/a The above equation has solutions for -a/b >0. Hence, the graph of y = ae x + bx has NO horizontal tangent line if -a/b <= 0 Exercises 1 - Find all points on the graph of y = x 3 - 3x where the tangent line is parallel to the line whose equation is given by y = 9x + 4. 2 - Find a and b so that the line y = -2 is tangent to the graph of y = ax2 + bx at x = 1. 3 - Find conditions on a, b and c so that the graph of y = ax 3 + bx 2 + cx has ONE tangent line parallel to the x axis (horizontal tangent). solutions to the above exercises 1 -    (2,2) and (-2,-2) 2 -    a = 2 and b = - 4 3 -    4b 2 - 12 ac = 0 More references on calculus problems

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Updated: February 2015