Let's address the given problem step by step. The function provided is $y = -x^2 + 8x$. Let's solve both parts:

(i) Find the value of $y$ when $x = 5$:

Substitute $x = 5$ into the function $y = -x^2 + 8x$: $y = -(5)^2 + 8(5)$ $y = -25 + 40$ $y = 15$

Answer: $y = 15$ when $x = 5$.

(ii) Find the values of $x$ when $y = 5$:

We solve the equation $5 = -x^2 + 8x$. Rearrange it into standard quadratic form: $-x^2 + 8x - 5 = 0$ Multiply through by $-1$ to simplify: $x^2 - 8x + 5 = 0$

This is a quadratic equation. Solve using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = -8$, and $c = 5$. Substitute into the formula: $x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(5)}}{2(1)}$ $x = \frac{8 \pm \sqrt{64 - 20}}{2}$ $x = \frac{8 \pm \sqrt{44}}{2}$ $x = \frac{8 \pm 2\sqrt{11}}{2}$ $x = 4 \pm \sqrt{11}$

Thus, the two solutions are: $x = 4 + \sqrt{11} \quad \text{and} \quad x = 4 - \sqrt{11}.$

Final Answer:

1. $y = 15$ when $x = 5$.
2. $x = 4 + \sqrt{11}$ and $x = 4 - \sqrt{11}$ when $y = 5$.

Would you like me to plot this or provide a further breakdown? Here are 5 related questions you might explore:

1. What is the vertex of the parabola $y = -x^2 + 8x$?
2. At what points does the parabola intersect the x-axis (find the roots)?
3. What is the maximum height the graph reaches, and at what $x$ value?
4. How does changing the coefficient of $x^2$ affect the graph's shape?
5. Can we express $x = 4 \pm \sqrt{11}$ as decimal approximations?

Tip: Always double-check calculations when working with the quadratic formula for accuracy!