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 - 3 x 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 - 3 x is given by the first derivative y '.
    y ' = 3 x 2 - 3
  • We now find all values of x for which y ' = 0.
    3 x 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 - 3 x
    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 - 3 x below with the tangent lines.

    tangent lines to the graph of y = x<sup>3</sup> - 3x



Problem 2: Find a and b so that the line y = - 3 x + 4 is tangent to the graph of y = a x3 + b x 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 = a x3 + b x and y = - 3 x + 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 = a x3 + b x at x = 1
    y ' = 3 a x2 + x = - 3 at x = 1.
  • The above gives a second equation in a and b
    3 a + b = -3
  • Solve the system of equations a + b = 1 and 3 a + b = - 3 to find a and b
    a = - 2 and b = 3.
  • See graphs of y = a x3 + b x, with a = - 2 and b = 3, and y = - 3 x + 4 below.

    tangent line y = -3 x + 4 to the graph of y = a x<sup>3</sup> + b x


Problem 3: Find conditions on a and b so that the graph of y = a e 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 = a e x + bx. Find y '
    y ' = a e 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)
    a e 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 = a e x + bx has NO horizontal tangent line if -a/b <= 0
Exercises
1) Find all points on the graph of y = x
3 - 3 x where the tangent line is parallel to the line whose equation is given by y = 9 x + 4.
2) Find a and b so that the line y = - 2 is tangent to the graph of y = a x
2 + b x at x = 1.
3) Find conditions on a, b and c so that the graph of y = a x
3 + b x 2 + c x 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) 4 b
2 - 12 a c = 0
More references on
calculus problems