The curve C has equation $y = kx^3 - x^2 + x - 5$, where $k$ is a constant - Edexcel - A-Level Maths Pure - Question 11 - 2008 - Paper 1

Given that the tangent at point A is parallel to the line $2y - 7x + 1 = 0$, we first rewrite the line in slope-intercept form:

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

Thus, the slope (gradient) of the line is $\frac{7}{2}$. Since the tangent line is parallel, we set:

$3k\left(-\frac{1}{2}\right)^2 - 2\left(-\frac{1}{2}\right) + 1 = \frac{7}{2}$

Calculating step-by-step:

1. Substitute $x = -\frac{1}{2}$ into $\frac{dy}{dx}$: $3k \cdot \frac{1}{4} + 1 + 1 = \frac{7}{2}$
2. This simplifies to: $\frac{3k}{4} + 2 = \frac{7}{2}$
3. Solving for $k$: $\frac{3k}{4} = \frac{7}{2} - 2$ $\frac{3k}{4} = \frac{3}{2}$ $3k = 6$ $k = 2$

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

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

Calculating:

1. First term: $2\left(-\frac{1}{2}\right)^3 = 2 \cdot -\frac{1}{8} = -\frac{1}{4}$
2. Second term: $-\left(-\frac{1}{2}\right)^2 = -\frac{1}{4}$
3. Third term: $-\frac{1}{2}$
4. Fourth term: $-5$

Summing these gives: $y = -\frac{1}{4} - \frac{1}{4} - \frac{1}{2} - 5$ Convert to a common denominator (4):

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

Therefore, the y-coordinate of A is $-6$.