We are given the quadratic equation $y = 2 + 8x - x^2$, and we need to address each part of the question.

Step-by-step Solution:

(a) Find the vertex of the graph.

A quadratic equation in the form $y = ax^2 + bx + c$ has its vertex at $x = -\frac{b}{2a}$. Here, we can rewrite the given equation as:

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

Comparing with $y = ax^2 + bx + c$, we have:

• $a = -1$
• $b = 8$
• $c = 2$

Now, find the $x$-coordinate of the vertex:

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

To find the $y$-coordinate, substitute $x = 4$ into the original equation:

$y = 2 + 8(4) - (4)^2 = 2 + 32 - 16 = 18$

So, the vertex is $(x, y) = (4, 18)$.

(b) Determine whether the vertex is a maximum or minimum point.

Since the coefficient of $x^2$ (i.e., $a = -1$) is negative, the parabola opens downwards. This means that the vertex is a maximum point.

(c) Determine what value of $x$ gives the optimal value of the function.

The optimal value occurs at the vertex, so $x = 4$.

(d) Determine the optimal (maximum or minimum) value of the function.

The optimal value of the function is the $y$-coordinate of the vertex, which is $y = 18$.

Final Answers:

(a) The vertex is ( (x, y) =