The equation of the curve is $y = 7x - x^2 - 5$.

To find the turning point, we first differentiate the equation to obtain the gradient function and then solve for $x$ when the gradient is zero (since the turning point occurs when the gradient is zero).

1. Differentiate $y = 7x - x^2 - 5$ with respect to $x$: $\frac{dy}{dx} = 7 - 2x$

2. Set the derivative equal to zero to find the $x$-coordinate of the turning point: $7 - 2x = 0$ $2x = 7$ $x = \frac{7}{2}$

3. Substitute $x = \frac{7}{2}$ back into the original equation to find the corresponding $y$-coordinate: $y = 7\left(\frac{7}{2}\right) - \left(\frac{7}{2}\right)^2 - 5$ $y = \frac{49}{2} - \frac{49}{4} - 5$ To simplify, express all terms with a common denominator of 4: $y = \frac{98}{4} - \frac{49}{4} - \frac{20}{4}$ $y = \frac{98 - 49 - 20}{4}$ $y = \frac{29}{4}$

Thus, the coordinates of the turning point are: $\left( \frac{7}{2}, \frac{29}{4} \right)$

Would you like further details or have any questions?

Here are 5 related questions for further exploration:

1. How do you determine whether the turning point is a maximum or minimum?
2. Can you sketch the graph of this quadratic equation?
3. What are the roots of the equation $7x - x^2 - 5 = 0$?
4. How does completing the square help in finding the turning point of a quadratic function?
5. How would the turning point change if the equation were $y = 7x - 2x^2 - 5$?

Tip: The second derivative of a function helps determine whether a turning point is a maximum or a minimum.