Question

The graphs $y = 2x^3 - 4x + 2$ and $y = x^3 + 2x - 1$ intersect in exactly 3 distinct points. The slope of the line passing through two of these points is

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

Answer: (D)

8

Answer 2

Let the two curves be $C_1: y = 2x^3 - 4x + 2$ and $C_2: y = x^3 + 2x - 1$. The points of intersection are found by setting the y-values equal: $2x^3 - 4x + 2 = x^3 + 2x - 1$ $x^3 - 6x + 3 = 0$

Let the roots of this cubic equation be $x_1, x_2, x_3$. Since the problem states there are exactly 3 distinct intersection points, these roots are distinct real numbers. Let the intersection points be $P_1(x_1, y_1)$, $P_2(x_2, y_2)$, and $P_3(x_3, y_3)$. The y-coordinates can be found using either equation. Let's use $C_2$: $y_i = x_i^3 + 2x_i - 1$.

The slope of the line passing through two points $P_i(x_i, y_i)$ and $P_j(x_j, y_j)$ is given by $m = \frac{y_j - y_i}{x_j - x_i}$. Let's calculate the slope for $P_1$ and $P_2$: $m_{12} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{(x_2^3 + 2x_2 - 1) - (x_1^3 + 2x_1 - 1)}{x_2 - x_1}$ $m_{12} = \frac{x_2^3 - x_1^3 + 2x_2 - 2x_1}{x_2 - x_1} = \frac{(x_2^3 - x_1^3) + 2(x_2 - x_1)}{x_2 - x_1}$ Using the difference of cubes formula $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$: $m_{12} = \frac{(x_2 - x_1)(x_2^2 + x_1x_2 + x_1^2) + 2(x_2 - x_1)}{x_2 - x_1}$ Since $x_1 \neq x_2$, we can cancel the term $(x_2 - x_1)$: $m_{12} = x_1^2 + x_1x_2 + x_2^2 + 2$.

Similarly, the slope for $P_2$ and $P_3$ is $m_{23} = x_2^2 + x_2x_3 + x_3^2 + 2$, and for $P_1$ and $P_3$ is $m_{13} = x_1^2 + x_1x_3 + x_3^2 + 2$. The problem asks for "The slope", implying a unique value. This suggests that the three intersection points are collinear. If they are collinear, the slope between any two points must be the same, i.e., $m_{12} = m_{23} = m_{13}$.

Let's check the condition for collinearity using the slope formula $m_{12} = m_{23}$: $x_1^2 + x_1x_2 + x_2^2 + 2 = x_2^2 + x_2x_3 + x_3^2 + 2$ $x_1^2 + x_1x_2 = x_2x_3 + x_3^2$ $x_1^2 - x_3^2 + x_1x_2 - x_2x_3 = 0$ $(x_1 - x_3)(x_1 + x_3) + x_2(x_1 - x_3) = 0$ $(x_1 - x_3)(x_1 + x_3 + x_2) = 0$. Since the roots $x_1, x_2, x_3$ are distinct, $x_1 \neq x_3$. Therefore, we must have $x_1 + x_2 + x_3 = 0$.

Let's check Vieta's formulas for the cubic equation $x^3 - 6x + 3 = 0$. The sum of the roots is $x_1 + x_2 + x_3 = -\frac{\text{coefficient of } x^2}{\text{coefficient of } x^3} = -\frac{0}{1} = 0$. The sum of the products of roots taken two at a time is $x_1x_2 + x_2x_3 + x_3x_1 = \frac{\text{coefficient of } x}{\text{coefficient of } x^3} = \frac{-6}{1} = -6$. The product of the roots is $x_1x_2x_3 = -\frac{\text{constant term}}{\text{coefficient of } x^3} = -\frac{3}{1} = -3$.

Since $x_1 + x_2 + x_3 = 0$, the condition for collinearity derived from the slopes is satisfied. Thus, the three intersection points $P_1, P_2, P_3$ are collinear. The line passing through any two of these points is the same line containing all three points. The slope of this line is unique.

We can also find the equation of the line containing these points directly. Suppose the line is $y = mx + c$. If the three intersection points lie on this line, then their coordinates $(x_i, y_i)$ must satisfy $y_i = mx_i + c$. Since $y_i = x_i^3 + 2x_i - 1$, we have $x_i^3 + 2x_i - 1 = mx_i + c$ for $i=1, 2, 3$. This means $x_i$ are the roots of the equation $x^3 + 2x - 1 - mx - c = 0$, or $x^3 + (2-m)x - (1+c) = 0$. We know that $x_i$ are the roots of $x^3 - 6x + 3 = 0$. For two cubic equations to have the same roots, their coefficients must be proportional. Since the coefficient of $x^3$ is 1 in both equations, the coefficients must be equal. Comparing the coefficients of $x$: $2 - m = -6 \implies m = 2 + 6 = 8$. Comparing the constant terms: $-(1+c) = 3 \implies 1+c = -3 \implies c = -4$. So, the line containing the three intersection points is $y = 8x - 4$. The slope of this line is $m=8$.

Alternatively, we can use the slope formula $m = x_i^2 + x_ix_j + x_j^2 + 2$. Using Vieta's formulas, $x_i + x_j + x_k = 0$ (where $i, j, k$ are distinct). So $x_i + x_j = -x_k$. $x_i^2 + x_j^2 = (x_i + x_j)^2 - 2x_ix_j = (-x_k)^2 - 2x_ix_j = x_k^2 - 2x_ix_j$. The slope is $m_{ij} = (x_i^2 + x_j^2) + x_ix_j + 2 = (x_k^2 - 2x_ix_j) + x_ix_j + 2 = x_k^2 - x_ix_j + 2$. This expression still involves specific roots.

Let's go back to $m_{ij} = x_i^2 + x_ix_j + x_j^2 + 2$. We know $x_i+x_j+x_k=0$ and $x_ix_j+x_jx_k+x_kx_i=-6$. Consider $m_{12} = x_1^2 + x_1x_2 + x_2^2 + 2$. We need to show this value is constant for any pair. $x_1^2 + x_2^2 = (x_1+x_2)^2 - 2x_1x_2 = (-x_3)^2 - 2x_1x_2 = x_3^2 - 2x_1x_2$. $m_{12} = x_3^2 - 2x_1x_2 + x_1x_2 + 2 = x_3^2 - x_1x_2 + 2$. This does not immediately show the value is constant.

Let's reconsider the equation $x^3 - 6x + 3 = 0$. For any root $x_i$, we have $x_i^3 - 6x_i + 3 = 0$, so $x_i^3 = 6x_i - 3$. The y-coordinate is $y_i = x_i^3 + 2x_i - 1$. Substitute $x_i^3$: $y_i = (6x_i - 3) + 2x_i - 1 = 8x_i - 4$. This equation $y_i = 8x_i - 4$ holds for all three roots $x_1, x_2, x_3$. This means the three points $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ all lie on the line $y = 8x - 4$. The slope of this line is 8. The line passing through any two of these collinear points is this same line $y = 8x - 4$, and its slope is 8.

The final answer is $\boxed{8}$.