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:
The topic of this problem is the Application of Derivatives, specifically finding points of tangency. We are looking for points on a given cubic curve where the slope of the tangent line is identical to the slope of a specified linear equation. This requires understanding that the geometric interpretation of the first derivative of a function at any point is the slope of the tangent line at that point. Because the tangent must be parallel to a given line, their slopes must be mathematically equal.
Step 2: Key Formulas and approach:
The primary formula used here is the power rule for differentiation: $\frac{d}{dx}(x^n) = nx^{n-1}$.
The slope of a line in the form $y = mx + c$ is simply $m$.
The approach follows these logical steps:
1. Differentiate the curve equation to find the general expression for the slope of the tangent.
2. Determine the slope of the given line.
3. Set the derivative equal to that slope and solve the resulting quadratic equation for $x$.
4. Substitute the $x$ values back into the original curve equation to find the corresponding $y$ coordinates.
Step 3: Detailed Explanation:

We start with the curve equation $y = 2x^3 + 3x^2 - 8x$. Calculating the derivative gives $\frac{dy}{dx} = 6x^2 + 6x - 8$.

The target line is $y = 4x + 3$. By comparing this to $y = mx + c$, we find the required slope is $m = 4$.

Now, we equate the slope of the tangent to the slope of the line: $6x^2 + 6x - 8 = 4$.

Rearrange the equation to form a standard quadratic equation: $6x^2 + 6x - 12 = 0$.

Simplify the equation by dividing all terms by 6: $x^2 + x - 2 = 0$.

Factoring the quadratic expression, we get $(x + 2)(x - 1) = 0$, which yields two roots: $x = 1$ and $x = -2$.

For $x = 1$, we find $y$ by substituting into the curve: $y = 2(1)^3 + 3(1)^2 - 8(1) = 2 + 3 - 8 = -3$. This gives the point $(1, -3)$.

For $x = -2$, we find $y$ by substituting into the curve: $y = 2(-2)^3 + 3(-2)^2 - 8(-2) = 2(-8) + 3(4) + 16 = -16 + 12 + 16 = 12$. This gives the point $(-2, 12)$.

Step 4: Final Answer:
The two points on the curve where the tangents are parallel to the given line are $(1, -3)$ and $(-2, 12)$.

Answer 2

Instead of factoring the quadratic obtained from equating slopes, we can use the relationship between its coefficients and its roots to quickly identify which pair of points is consistent.

Setting the derivative of \( y = 2x^3 + 3x^2 - 8x \) equal to the slope of \( y = 4x + 3 \) gives \( 6x^2 + 6x - 8 = 4 \), which simplifies to \( 6x^2 + 6x - 12 = 0 \), or \( x^2 + x - 2 = 0 \) after dividing through by \( 6 \).

For a quadratic \( x^2 + bx + c = 0 \), the sum of its roots is \( -b \) and the product of its roots is \( c \). Here, \( b = 1 \) and \( c = -2 \), so the two valid \( x \)-values must satisfy:

\[ x_1 + x_2 = -1, \qquad x_1 \times x_2 = -2 \]

Now check the \( x \)-coordinates implied by each option against these two conditions:

• \( x = 1 \) and \( x = 0 \): sum \( = 1 \), which does not match \( -1 \).
• \( x = 0 \) and \( x = -2 \): sum \( = -2 \), which does not match \( -1 \).
• \( x = 1 \) and \( x = -2 \): sum \( = 1 + (-2) = -1 \) and product \( = 1 \times (-2) = -2 \). Both conditions match exactly.
• \( x = 0 \) and \( x = 2 \): sum \( = 2 \), which does not match \( -1 \).

Only \( x = 1 \) and \( x = -2 \) satisfy both the sum and product relations, so these must be the roots of the quadratic. Substituting back into the curve's equation confirms \( y = -3 \) at \( x = 1 \) and \( y = 12 \) at \( x = -2 \).

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