Calculus Tangent Line Problems – Worked Solutions (Part 5)

Calculus problems focusing on tangent lines, presented with full explanations and step-by-step solutions.

Question 1

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

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 \( p \): \[ p = \frac{9}{4}. \]

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.

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: \[ x = -\frac{4}{6} = -\frac{2}{3}. \]

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

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