Calculus Tangent Line Problems – Worked Solutions (Part 5)

Calculus problems focusing on tangent lines, presented with full explanations. Each solution is hidden in a collapsible dropdown so you can attempt the problems independently before reviewing the steps[cite: 1].

Questions and Solutions

Question 1

Find the parameter \( p \) such that the line \[ y = 3x \] is tangent to the curve \[ y = x^2 + p. \]

View Solution
  • The slope of the tangent line is \( 3 \). The derivative of the curve is \[ y' = 2x. \]
  • At the point of tangency: \[ 2x = 3 \quad \Rightarrow \quad x = \frac{3}{2}. \]
  • The corresponding \( y \)-value on the line: \[ y = 3\left(\frac{3}{2}\right) = \frac{9}{2}. \]
  • Since the point lies on the curve: \[ \frac{9}{2} = \left(\frac{3}{2}\right)^2 + p. \]
  • Solving for \( p \): \[ p = \frac{9}{4}. \]

Question 2

a) Find \( p \) so that the curve \[ y = x^3 + 2x^2 + px + 3 \] has exactly one horizontal tangent line.

b) Find the value of \( x \) where this tangent occurs.

View Solution
  • A horizontal tangent occurs when \[ y' = 0. \]
  • Compute the derivative: \[ y' = 3x^2 + 4x + p. \]
  • For exactly one solution, the discriminant must be zero: \[ D = 4^2 - 4(3)(p) = 16 - 12p = 0. \]
  • Solving for \( p \): \[ p = \frac{4}{3}. \]
  • With \( D = 0 \), the solution for \( x \) is: \[ x = -\frac{4}{6} = -\frac{2}{3}. \]

Question 3

Find \( p \) and \( q \) such that the line \[ y = 2x \] is tangent to the curve \[ y = px^2 + qx + 2 \] at \( x = 3 \).

View Solution
  • The point of tangency lies on the line: \[ (3,\, 2 \cdot 3) = (3,6). \]
  • Since it lies on the curve: \[ 6 = 9p + 3q + 2. \]
  • The derivative of the curve is: \[ y' = 2px + q. \]
  • At \( x = 3 \), the slope equals the slope of the line: \[ 2 = 6p + q. \]
  • Solve the system: \[ \begin{cases} 9p + 3q = 4 \\ 6p + q = 2 \end{cases} \]
  • Solution: \[ p = \frac{2}{9}, \quad q = \frac{2}{3}. \]

Question 4

Find \( a \) and \( b \) such that the line \[ y = ax + b \] is tangent to the curve \[ y = x^2 + 3x + 2 \] at \( x = 3 \).

View Solution
  • Derivative of the curve: \[ y' = 2x + 3. \]
  • Slope at \( x = 3 \): \[ a = 2(3) + 3 = 9. \]
  • Point of tangency: \[ y = 3^2 + 3(3) + 2 = 20. \]
  • Substitute into the equation of the line: \[ 20 = 9(3) + b. \]
  • Solve: \[ b = -7. \]

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