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 \( (-2, 12) \).

Answer 2

To find the points where the tangent to \( y = 2x^3 + 3x^2 - 8x \) is parallel to \( y = 4x + 3 \), we can check each candidate point directly: a valid point must both lie on the curve and give a tangent slope of \( 4 \) there. The slope of the curve at any \( x \) is \( \frac{dy}{dx} = 6x^2 + 6x - 8 \).

1. \( (1, -3) \) and \( (0, 0) \): Check \( (0,0) \): substituting \( x = 0 \) gives \( y = 0 \), so this point does lie on the curve. But the slope there is \( 6(0)^2 + 6(0) - 8 = -8 \), not \( 4 \), so the tangent at \( (0,0) \) is not parallel to the given line. This pair fails.
2. \( (0, 0) \) and \( (-2, 12) \): This pair also includes \( (0,0) \), whose slope is \( -8 \), not \( 4 \), as shown above. This pair fails for the same reason.
3. \( (1, -3) \) and \( (-2, 12) \): Check \( (1,-3) \): substituting \( x = 1 \) gives \( y = 2(1) + 3(1) - 8(1) = -3 \), so the point lies on the curve, and the slope there is \( 6(1)^2 + 6(1) - 8 = 4 \), matching the required slope. Check \( (-2, 12) \): substituting \( x = -2 \) gives \( y = 2(-8) + 3(4) - 8(-2) = -16 + 12 + 16 = 12 \), so this point also lies on the curve, and the slope there is \( 6(4) + 6(-2) - 8 = 24 - 12 - 8 = 4 \), also matching. Both points check out.
4. \( (0, 0) \) and \( (2, 12) \): \( (0,0) \) again fails the slope test as shown above, and even though \( (2,12) \) lies on the curve (since \( 2(8) + 3(4) - 16 = 16 + 12 - 16 = 12 \)), its slope is \( 6(4) + 6(2) - 8 = 24 + 12 - 8 = 28 \), also not \( 4 \). This pair fails on both counts.

Only the pair \( (1, -3) \) and \( (-2, 12) \) satisfies both conditions, lying on the curve and having a tangent slope of \( 4 \).

Therefore, the correct answer is \( (1, -3) \) and \( (-2, 12) \).