The function U(t)=5t^2 -20t+20759
 models the annual consumption of beef, in tons, in the U.S from 2000 to 2017. where t
 represents the number of years since 2000.

Note: For the problems below, set the window on your calculator to x: [0..18] and y: [20000..22000]
a) Use a graphing method to determine the years in which the consumtion of beef was greater that 21305 tons per year.

Write your answer in interval notation rounding to one decimal place. 


Complete the following sentence using the information from your graph.
The annual consumption of beef was above 21305 tons per year starting in year 
.
b) Use a graphing method to determine the years in which the annual consumption of beef was less than 21135 tons per year.

Write your answer in interval notation rounding to one decimal place. 


Complete the following sentence using the information from your graph.
The annual consumption of beef was below 21135 tons per year until the year 
.

To solve these problems, we need to analyze the function:

\[
U(t) = 5t^2 - 20t + 20759
\]

where \( t \) represents the number of years since 2000, and \( U(t) \) represents the annual consumption of beef in tons. We will graph this function and use the graphing method to find the relevant time intervals for parts (a) and (b).

### Part (a): Consumption above 21305 tons per year
We are tasked with determining when \( U(t) > 21305 \). To do this, we solve the inequality:

\[
5t^2 - 20t + 20759 > 21305
\]

Simplify the inequality:

\[
5t^2 - 20t + 20759 - 21305 > 0
\]

\[
5t^2 - 20t - 546 > 0
\]

We now solve this inequality. First, solve the equation \( 5t^2 - 20t - 546 = 0 \) to find the points of intersection, which will help define the intervals where the consumption is above 21305.

Using the quadratic formula:

\[
t = \frac{-(-20) \pm \sqrt{(-20)^2 - 4(5)(-546)}}{2(5)}
\]

Simplifying further:

\[
t = \frac{20 \pm \sqrt{400 + 10920}}{10}
\]

\[
t = \frac{20 \pm \sqrt{11320}}{10}
\]

\[
t = \frac{20 \pm 106.4}{10}
\]

This gives:

\[
t_1 = \frac{20 - 106.4}{10} = -8.64 \quad \text{(not relevant since t must be non-negative)}
\]
\[
t_2 = \frac{20 + 106.4}{10} = 12.64
\]

The beef consumption exceeds 21305 tons between \( t = 12.6 \) (i.e., the year 2012.6 or mid-2012) and beyond.

Thus, the interval where beef consumption exceeds 21305 tons is:

\[
[12.6, 18] \quad \text{(in interval notation, since t ranges up to 18 representing the year 2017)}.
\]

So, the annual consumption of beef was above 21305 tons per year starting in year **2012.6**.

### Part (b): Consumption below 21135 tons per year
We are tasked with determining when \( U(t) < 21135 \). To do this, we solve the inequality:

\[
5t^2 - 20t + 20759 < 21135
\]

Simplify the inequality:

\[
5t^2 - 20t + 20759 - 21135 < 0
\]

\[
5t^2 - 20t - 376 < 0
\]

We now solve this inequality. First, solve the equation \( 5t^2 - 20t - 376 = 0 \) to find the points of intersection.

Using the quadratic formula:

\[
t = \frac{-(-20) \pm \sqrt{(-20)^2 - 4(5)(-376)}}{2(5)}
\]

Simplifying further:

\[
t = \frac{20 \pm \sqrt{400 + 7520}}{10}
\]

\[
t = \frac{20 \pm \sqrt{7920}}{10}
\]

\[
t = \frac{20 \pm 89}{10}
\]

This gives:

\[
t_1 = \frac{20 - 89}{10} = -6.9 \quad \text{(not relevant since t must be non-negative)}
\]
\[
t_2 = \frac{20 + 89}{10} = 10.9
\]

Thus, the beef consumption was below 21135 tons before \( t = 10.9 \), or mid-2010.

So, the interval where beef consumption is below 21135 tons is:

\[
[0, 10.9] \quad \text{(in interval notation)}.
\]

Thus, the annual consumption of beef was below 21135 tons per year until the year **2010.9**.

---

### Summary:
- **Part (a)**: The annual consumption of beef was above 21305 tons per year starting in year **2012.6**. The interval is \([12.6, 18]\).
- **Part (b)**: The annual consumption of beef was below 21135 tons per year until year **2010.9**. The interval is \([0, 10.9]\).

Let me know if you want more details or have any questions!

### Related Questions:
1. How does changing the constant term in the quadratic function affect the graph of the beef consumption model?
2. What happens to the intervals if the consumption threshold is lowered to 21000 tons?
3. How would the graph change if the quadratic term in the function was negative?
4. Can you find the maximum or minimum point of the function \( U(t) \), and what does it represent in this context?
5. How would you interpret the model if the function were linear instead of quadratic?

### Tip:
When analyzing quadratic models like this, always check if the parabola opens upwards or downwards by examining the coefficient of the \( t^2 \) term. This helps you predict the behavior of the function.

The function U(t) = 5t^2 - 20t + 20759 models the annual consumption of beef, in tons, in the U.S from 2000 to 2017. where t represents the number of years since 2000. a) Use a graphing method to determine the years in which the consumption of beef was greater than 21305 tons per year. Write your answer in interval notation rounding to one decimal place. Complete the following sentence using the information from your graph. The annual consumption of beef was above 21305 tons per year starting in year __. b) Use a graphing method to determine the years in which the annual consumption of beef was less than 21135 tons per year. Write your answer in interval notation rounding to one decimal place. Complete the following sentence using the information from your graph. The annual consumption of beef was below 21135 tons per year until the year __.

Analyze the quadratic function U(t) = 5t^2 - 20t + 20759, which models U.S. beef consumption from 2000 to 2017. Solve inequalities to determine when consumption exceeded 21305 tons and fell below 21135 tons. This problem involves graphing and applying the quadratic formula, suitable for Grades 10-12.

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