Calculus Questions with Answers (5)

Calculus questions, on tangent lines, are presented along with detailed solutions and explanations.

Question 1:
Find parameter p so that y = 3x is a tangent line to the curve
y = x 2 + p

Solution to Question 1:

  • The slope of the tangent is equal to 3. At the point of tangency y ' = 3.
    y ' = 2 x = 3
  • Solve the equation 2 x = 3 to find the x coordinate of the point of tangency.
    x = 3 / 2
  • The point of tangency is on the line y = 3 x, therefore the y coordinate of the point of tangency is equal to
    y = 3 (3/2) = 9 / 2
  • The point of tangency (3 / 2 , 9 / 2) is on the curve
    y = x 2 + p and is therefore used to find p as follows

    9 / 2 = (3 / 2) 2 + p
  • Solve the above for p to obtain
    p = 9 / 4


Question 2:
a) Find p so that the curve y = x 3 + 2 x 2 + p x + 3 has one horizontal tangent line only.
b) Find the value of x where this tangent line occurs.

Solution to Question 2:

  • a) The horizontal tangent line has a slope equal to zero. We first need to find the first derivative of y and equate to 0.
    y ' = 3 x 2 + 4 x + p = 0
  • For the above quadratic function to have one solution only, it needs to have its discriminant D equal to zero.
    D = 16 - 12 p = 0
  • Solve the above for p to find
    p = 4 / 3
  • b) To find the value of x where the horizontal tangent line occurs, we need to solve the above quadratic equation 3 x 2 + 4 x + p = 0 for x. It had the discriminant equal to zero, hence
    x = - 4 / 6 = - 2 / 3



Question 3:
Find p and q so that y = 2 x is a tangent, line at x = 3, to the curve
y = p x 2 + q x + 2

Solution to Question 3:

  • The point of tangency is on the tangent line y = 2x, hence its y coordinate can also be calculated and its coordinates given by
    (3 , 2(3)) = (3 , 6)
  • The above point of tangency is also on the curve
    y = p x 2 + q x + 2, hence

    6 = p(3) 2 + q (3) + 2
  • Let us now find y '
    y ' = 2 p x + q
  • At x = 3 y ' is equal to the slope of the tangent line y = 2 x. Hence
    2 = 2 p (3) + q
  • We now solve the two linear equations
    9 p + 3 q = 4 and 6 p + q = 2
  • to find 9 p + 3 q = 4 and 6 p + q = 2
    p = 2 / 9 and q = 2 / 3


Question 4:
Find a and b so that y = a x + b is a tangent line to the curve
y = x 2 + 3 x + 2 at x = 3.

Solution to Question 4:

  • Find y ' of the curve
    y ' = 2 x + 3
  • The slope of the tangent line is equal to a and also to y ' at x = 3, hence
    a = 2(3) + 3 = 9
  • The point of tangency has x = 3 and is on the curve, hence the y coordinate of the point of tangency is given by
    y = (3) 2 + 3(3) + 2 = 20
  • The point of tangency is also on the tangent line, hence
    20 = 3 a + b
  • a has already been calculated, b is equal to
    b = - 7


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