# 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 | 1 | 2 | 3 | 4 | 5 | More on calculus questions with answers, tutorials and problems .