Emit threw a football during practice. The function h(x)= -1/40x2 + x +6

To rewrite the function \( h(x) = -\frac{1}{40}x^2 + x + 6 \) in vertex form, we need to complete the square.

1. Start with the function: \[ h(x) = -\frac{1}{40}x^2 + x + 6 \]

2. Factor out \(-\frac{1}{40}\) from the quadratic and linear terms: \[ h(x) = -\frac{1}{40}(x^2 - 40x) + 6 \]

3. Now, complete the square inside the parentheses. To do that, we take half of the coefficient of \(x\) (which is \(-40\)), square it, and then add and subtract that value:

   • Half of \(-40\) is \(-20\), and squaring it gives \(400\).

   So we rewrite: \[ h(x) = -\frac{1}{40}(x^2 - 40x + 400 - 400) + 6 \]

4. This simplifies to: \[ h(x) = -\frac{1}{40}((x - 20)^2 - 400) + 6 \]

5. Distributing inside the parentheses gives: \[ h(x) = -\frac{1}{40}(x - 20)^2 + 10 + 6 \]

6. Combine like terms: \[ h(x) = -\frac{1}{40}(x - 20)^2 + 16 \]

Now, we have the vertex form of the equation, which is: \[ h(x) = -\frac{1}{40}(x - 20)^2 + 16 \]

Thus, the function can be rewritten as: \[ h(x) = -\frac{1}{40}(x - 20)^2 + 16 \]

The maximum height the football reached was: \[ 16 \text{ feet}. \]

So, filling in the blanks:

The function can be rewritten as \( h(x) = -\frac{1}{40}(x - 20)^2 + 16 \) to reveal its vertex. The maximum height the football reached was \( 16 \) feet.