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:

To find where the tangents are parallel to a line, set the derivative equal to the slope of the line and solve for the values of \( x \).

Answer 1

The Correct Option is C

Step 1: Slope of the tangent line.
The equation of the line is \( y = 4x + 3 \). The slope of the line is 4. For the tangents on the curve to be parallel to this line, their slopes must also be 4. The derivative of the given function \( y = 2x^3 + 3x^2 - 8x \) gives the slope of the tangent at any point \( x \): \[ y' = \frac{d}{dx}(2x^3 + 3x^2 - 8x) = 6x^2 + 6x - 8 \]
Step 2: Set the derivative equal to the slope of the line.
We set the derivative equal to 4 to find the points where the tangent is parallel to the line: \[ 6x^2 + 6x - 8 = 4 \] \[ 6x^2 + 6x - 12 = 0 \] \[ x^2 + x - 2 = 0 \] Factoring the quadratic: \[ (x - 1)(x + 2) = 0 \] Thus, \( x = 1 \) and \( x = -2 \).
Step 3: Find the corresponding \( y \)-coordinates.
For \( x = 1 \), substitute into the original equation: \[ y = 2(1)^3 + 3(1)^2 - 8(1) = 2 + 3 - 8 = -3 \] So the point is \( (1, -3) \). For \( x = -2 \), substitute into the original equation: \[ y = 2(-2)^3 + 3(-2)^2 - 8(-2) = -16 + 12 + 16 = 12 \] So the point is \( (-2, 12) \).
Step 4: Conclusion.
The points on the curve where the tangents are parallel to the line are \( (1, -3) \) and \( (-2, -12) \). Thus, the correct answer is (3) \( (1, -3) \) and \( (-2, -12) \).