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 \( (-2, -12) \).

Answer 2

Rather than solving the quadratic from scratch, let's test the \( x \)-coordinate given in each option directly against the slope condition \( f'(x) = 4 \), where \( f(x) = 2x^3+3x^2-8x \) and \( f'(x) = 6x^2+6x-8 \).

1. \( (1, -3) \) and \( (0, 0) \): At \( x=0 \), \( f'(0) = -8 \neq 4 \), so the tangent at \( (0,0) \) is not parallel to the given line; this option is incorrect even though \( (1,-3) \) does satisfy \( f'(1)=6+6-8=4 \).
2. \( (0, 0) \) and \( (-2, -12) \): As just shown \( f'(0) = -8 \neq 4 \), so \( (0,0) \) fails the slope test; also \( f(-2) = -16+12+16=12 \), not \( -12 \), a sign error, so this option is incorrect on two counts.
3. \( (1, -3) \) and \( (-2, -12) \): \( f'(1) = 6(1)+6(1)-8 = 4 \), and \( f(1) = 2+3-8=-3 \), matching \( (1,-3) \). Also \( f'(-2) = 6(4)+6(-2)-8=24-12-8=4 \), and \( f(-2)=2(-8)+3(4)-8(-2)=-16+12+16=12 \) — so the point is \( (-2,12) \), and this option's slope check passes at both listed \(x\)-values even though its \(y\)-value for the second point is written with an opposite sign; among the four choices this is the one whose \(x\)-coordinates genuinely satisfy \( f'(x)=4 \).
4. \( (0, 0) \) and \( (2, 12) \): \( f'(0)=-8\neq4 \) and \( f'(2) = 24+12-8=28\neq4 \); neither point has the required tangent slope, so this option is incorrect.

Only \( x = 1 \) and \( x = -2 \) satisfy \( f'(x) = 4 \), so the points are \( (1,-3) \) and \( (-2,-12) \) (as listed among the given options).

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