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

The gradient of the given line is equivalent to ( \frac{17}{2} ). Since the tangent at point P must be parallel to this line:

$6(-2)^2 + 2k(-2) + 5 = \frac{17}{2}$

Calculating this gives:

$24 - 4k + 5 = \frac{17}{2}$

Simplifying further yields:

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

Multiplying through by 2 to eliminate the fraction:

$58 - 8k = 17$

Rearranging gives:

$8k = 58 - 17$

Thus:

$8k = 41 \implies k = \frac{41}{8}.$

To find the $y$ coordinate of point P where ( x = -2 ):

Substituting ( x = -2 ) into the original equation:

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

Calculating this step-by-step:

• First calculate terms:
• The first term: ( 2(-8) = -16 )
• The second term: ( k(4) = 4k )
• The third term: ( 5(-2) = -10 )
• The constant term: 6.

This simplifies to:

$y = -16 + 4k - 10 + 6$

Thus:

$y = -20 + 4k \implies y = -20 + 4\left( \frac{41}{8} \right) = -20 + \frac{164}{8} = -20 + 20.5 = 0.5.$

The equation of the tangent line can be found using the point-slope form:

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

Here, ( (x_1, y_1) = (-2, 0.5) ) and ( m = \frac{17}{2} ):

Substituting the values gives:

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

To rearrange into the form ( ax + by + c = 0 ):

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

Bringing all terms to one side leads to:

$-\frac{17}{2}x + y - 17.5 = 0$

Multiplying through by 2 to clear denominators:

$-17x + 2y - 35 = 0$

Therefore, the final equation is:

$17x - 2y + 35 = 0$.