Calculus Questions with Answers (5)

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


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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