The curve C has equation $y = kx^3 - x^2 + x - 5$, where $k$ is a constant - Edexcel - A-Level Maths Pure - Question 10 - 2008 - Paper 1

Given that the tangent to C at point A is parallel to the line $2y - 7x + 1 = 0$, we first determine the slope of this line. Rearranging it gives:

$y = \frac{7}{2}x - \frac{1}{2}$

Thus, the gradient of the line is $m = \frac{7}{2}$.

Now, using the point $A$ with the x-coordinate $x = -\frac{1}{2}$:

Substituting into the derivative:

$\frac{dy}{dx} = 3k(-\frac{1}{2})^2 - 2(-\frac{1}{2}) + 1$

This simplifies to:

$\frac{dy}{dx} = 3k(\frac{1}{4}) + 1 + 1 = \frac{3k}{4} + 2$

Now, we set this equal to the slope $\frac{7}{2}$:

$\frac{3k}{4} + 2 = \frac{7}{2}$

To solve for $k$, we rearrange:

$\frac{3k}{4} = \frac{7}{2} - 2$

Converting $2$ into quarters gives:

$\frac{7}{2} - \frac{4}{2} = \frac{3}{2}$

Thus:

$\frac{3k}{4} = \frac{3}{2}$

Multiplying both sides by $4$ results in:

$3k = 6\Rightarrow k = 2$

To find the y-coordinate at point A, we substitute $x = -\frac{1}{2}$ and $k = 2$ back into the original curve equation:

$y = kx^3 - x^2 + x - 5$

Substituting the known values gives:

$y = 2(-\frac{1}{2})^3 - (-\frac{1}{2})^2 + (-\frac{1}{2}) - 5$

This simplifies to:

$y = 2(-\frac{1}{8}) - \frac{1}{4} - \frac{1}{2} - 5$

Carrying out the calculations:

$y = -\frac{1}{4} - \frac{1}{4} - \frac{2}{4} - 5$

Combining like terms results in:

$y = -\frac{4}{4} - 5 = -1 - 5 = -6$

Thus, the y-coordinate of point A is $-6$.