Question

The points on the curve \( y = 2x^3 + 3x^2 - 8x \) where the tangents are parallel to the line \( y = 4x + 3 \) are

• A. \( (1, -3) \) and \( (0, 0) \)
• B. \( (0, 0) \) and \( (-2, 12) \)
• C. \( (1, -3) \) and \( (-2, 12) \)
• D. \( (0, 0) \) and \( (2, 12) \)

Hint:

Two lines are parallel if and only if their slopes are equal. Always equate the first derivative of the curve to the slope of the given line to find the points of tangency.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to find the specific points on the curve \( y = 2x^3 + 3x^2 - 8x \) where the tangent is parallel to the given line \( y = 4x + 3 \).
Lines that are parallel to each other have identical slopes.


Step 2: Key Formula or Approach:
The slope of the tangent to a curve \( y = f(x) \) at any point is given by its first derivative, \( \frac{dy}{dx} \).
Equate this derivative to the slope of the linear equation \( y = mx + c \), which is \( m \).


Step 3: Detailed Explanation:
The given line is \( y = 4x + 3 \), so its slope is \( m = 4 \).
Now, find the derivative of the curve's equation:
\[ \frac{dy}{dx} = \frac{d}{dx} (2x^3 + 3x^2 - 8x) = 6x^2 + 6x - 8 \] Set the derivative equal to the slope of the line:
\[ 6x^2 + 6x - 8 = 4 \] \[ 6x^2 + 6x - 12 = 0 \] Divide the entire equation by 6 to simplify:
\[ x^2 + x - 2 = 0 \] Factorize the quadratic equation:
\[ (x + 2)(x - 1) = 0 \] This gives us the x-coordinates: \( x = -2 \) and \( x = 1 \).
Now, find the corresponding \( y \)-coordinates by substituting \( x \) back into the curve's original equation.
For \( x = 1 \):
\[ y = 2(1)^3 + 3(1)^2 - 8(1) = 2 + 3 - 8 = -3 \] So, the first point is \( (1, -3) \).
For \( x = -2 \):
\[ y = 2(-2)^3 + 3(-2)^2 - 8(-2) = 2(-8) + 3(4) + 16 = -16 + 12 + 16 = 12 \] So, the second point is \( (-2, 12) \).


Step 4: Final Answer:
The required points are \( (1, -3) \) and \( (-2, 12) \).