The problem you uploaded involves a quadratic equation modeling the path of a toy airplane. Let's break down each part:

a) Maximum height of the toy airplane

The equation modeling the path is:

$y = -x^2 + 8x + 1$

This is a parabola, and since the coefficient of $x^2$ is negative, it opens downwards. The vertex of this parabola will give us the maximum height of the toy airplane.

For any quadratic equation of the form $y = ax^2 + bx + c$, the x-coordinate of the vertex can be found using the formula:

$x = \frac{-b}{2a}$

In your equation, $a = -1$ and $b = 8$. Let's substitute these values into the formula:

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

So, the x-coordinate of the vertex is $x = 4$. To find the maximum height, substitute $x = 4$ back into the equation for $y$:

$y = -(4)^2 + 8(4) + 1 = -16 + 32 + 1 = 17$

Thus, the maximum height of the toy airplane is 17 meters.

b) Horizontal distance when height is 8 meters

We are given that the height, $y$, is 8 meters. To find the corresponding horizontal distances, we substitute $y = 8$ into the equation:

$8 = -x^2 + 8x + 1$

Rearranging:

$0 = -x^2 + 8x + 1 - 8$ $0 = -x^2 + 8x - 7$

Now, we solve this quadratic equation using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = -1$, $b = 8$, and $c = -7$. Substituting the values:

$x = \frac{-8 \pm \sqrt{8^2 - 4(-1)(-7)}}{2(-1)}$ $x = \frac{-8 \pm \sqrt{64 - 28}}{-2}$ $x = \frac{-8 \pm \sqrt{36}}{-2}$ $x = \frac{-8 \pm 6}{-2}$

This gives two solutions:

1. $x = \frac{-8 + 6}{-2} = \frac{-2}{-2} = 1$
2. $x = \frac{-8 - 6}{-2} = \frac{-14}{-2} = 7$

So, the two possible horizontal distances are $x = 1$ meter and $x = 7$ meters.

Would you like further clarification or details on any of these steps?

Here are some follow-up questions for you:

1. What does the negative coefficient of $x^2$ imply about the shape of the parabola?
2. How would the path change if the coefficient of $x^2$ was positive instead of negative?
3. What happens to the maximum height if the coefficient $b$ were increased?
4. How can we graph this quadratic equation to visually identify the vertex?
5. If the height were to reach 20 meters, what would the horizontal distance be?

Tip: When working with quadratics, remember the quadratic formula is a powerful tool for finding roots, but you can also factor when the equation is simple enough!