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

2. Factor out \(-\frac{1}{40}\) from the first two terms: \[ h(x) = -\frac{1}{40} \left(x^2 - 40x\right) + 6 \]

3. To complete the square, we need to find the value to add and subtract inside the parentheses. Take half of the coefficient of \(x\) (which is -40), square it, and add it inside the parentheses. \[ \left(\frac{-40}{2}\right)^2 = 20^2 = 400 \]

4. Add and subtract this value inside the parentheses: \[ h(x) = -\frac{1}{40} \left(x^2 - 40x + 400 - 400\right) + 6 \]

5. Rewrite it as: \[ h(x) = -\frac{1}{40} \left((x - 20)^2 - 400\right) + 6 \]

6. Distribute \(-\frac{1}{40}\): \[ h(x) = -\frac{1}{40}(x - 20)^2 + \frac{400}{40} + 6 \] \[ h(x) = -\frac{1}{40}(x - 20)^2 + 10 + 6 \] \[ h(x) = -\frac{1}{40}(x - 20)^2 + 16 \]

Now, we can see that the function in vertex form is: \[ h(x) = -\frac{1}{40}(x - 20)^2 + 16 \]

Therefore, substituting into the blanks provided in your question:

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

Final Answers:

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