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

We know that the tangent at point A is parallel to the line $2y - 7x + 1 = 0$. First, let's rearrange this line into slope-intercept form:

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

Thus, the slope of the line is $\frac{7}{2}$.

Since point $A$ has x-coordinate $-\frac{1}{2}$, we substitute it into the derivative to find the slope at A:

\frac{dy}{dx}\bigg|_{x = -\frac{1}{2}} = 3k\left(-\frac{1}{2}\right)^2 - 2\left(-\frac{1}{2}\right) + 1$$$$\frac{dy}{dx}\bigg|_{x = -\frac{1}{2}} = 3k \cdot \frac{1}{4} + 1 + 1 = \frac{3k}{4} + 2

Setting this equal to the slope of the line:

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

This gives:

$\frac{3k}{4} = \frac{7}{2} - 2$ $\frac{3k}{4} = \frac{3}{2}$ $3k = 6 \Rightarrow k = 2$

Now we substitute $k = 2$ back into the original equation to find the y-coordinate of point A:

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

Calculating each term:

1. $2 \cdot \left(-\frac{1}{2}\right)^3 = 2 \cdot -\frac{1}{8} = -\frac{1}{4}$
2. $-\left(-\frac{1}{2}\right)^2 = -\frac{1}{4}$
3. $-\frac{1}{2}$
4. $-5$

Putting it all together:

$y = -\frac{1}{4} -\frac{1}{4} - \frac{1}{2} - 5 = -\frac{1}{4} - \frac{1}{4} - \frac{2}{4} - \frac{20}{4} = -\frac{24}{4} = -6$

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