The curve C has equation $y = 2x^3 + kx^2 + 5x + 6$, where $k$ is a constant - Edexcel - A-Level Maths Pure - Question 1 - 2015 - Paper 1

To find the value of $k$, we calculate the slope of the tangent to the line $2y - 17x - 1 = 0$. Rewriting it in slope-intercept form gives:

$y = \frac{17}{2}x + \frac{1}{2}$

Thus, the gradient is ( \frac{17}{2} ). Now substitute ( x = -2 ) into the derivative:

$\frac{dy}{dx} = 6(-2)^2 + 2k(-2) + 5$

This simplifies to:

$\frac{dy}{dx} = 24 - 4k + 5 = 29 - 4k$

Setting this equal to the gradient of the line gives:

$29 - 4k = \frac{17}{2}$

Solving for $k$:

$4k = 29 - \frac{17}{2}$ $4k = \frac{58}{2} - \frac{17}{2} = \frac{41}{2}$ $k = \frac{41}{8}$

To find the y-coordinate of point $P$, substitute ( x = -2 ) back into the original equation:

$y = 2(-2)^3 + k(-2)^2 + 5(-2) + 6$

Substituting $k = \frac{41}{8}$ results in:

$y = 2(-8) + \frac{41}{8}(4) - 10 + 6$

Calculating step by step:

1. Calculate $2(-8) = -16$.
2. $\frac{41}{8}(4) = \frac{164}{8} = 20.5$.
3. Therefore:

$y = -16 + 20.5 - 10 + 6 = 0.5$

The equation of the tangent line at point $P$ can be expressed using the point-slope form:

$y - y_1 = m(x - x_1)$

Where:

• m is the slope, (m = \frac{17}{2})
• (x_1 = -2)
• (y_1 = 0.5)

Substituting these values into the equation gives:

$y - 0.5 = \frac{17}{2}(x + 2)$

Rearranging gives:

$y = \frac{17}{2}x + 17 + 0.5$

Combining the terms:

$y = \frac{17}{2}x + \frac{35}{2}$

Multiplying through by 2 to eliminate the fraction:

$2y = 17x + 35$

Thus, rearranging gives:

$17x - 2y + 35 = 0$