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
 1
 2
 3
 4
 5

More on calculus
questions with answers, tutorials and problems .
