Question

The table below represents the distance of an airplane from its destination as a function of time:

Time (hours) x Distance (miles) y
0 2700
1 2160
2 1620
3 1080
4 540

Part A: What is the y-intercept of the function, and what does this tell you about the airplane? (4 points)

Part B: Calculate the average rate of change of the function represented by the table between x = 1 to x = 4 hours, and tell what the the average rate represents. (4 points)

Part C: What would be the domain of the function if the airplane continued to travel until it reached its destination?(2 points)

Answer 1

decreasing is the answer i aced the quiz

Explanation:

Answer 2

`\bf \begin{array}{ccllll} hours&miles\\ \text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\ 0&2700\\ 1&2160\\ 2&1620\\ 3&1080\\ 4&540 \end{array}`

a)

at 0hours, the plane is 2700miles away from its destination, so the plane is on a city which is 2700miles from its destination and the pilots are sipping coffee

b)


`\bf slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ f(x_2)}}-{{ f(x_1)}}}{{{ x_2}}-{{ x_1}}}\impliedby \begin{array}{llll} average\ rate\\ of\ change \end{array}\\\\ -------------------------------\\\\`


`\bf \begin{array}{ccllll} hours[x]&miles[f(x)]\\ \text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\ 0&2700\\ \boxed{1}&\boxed{2160}\\ 2&1620\\ 3&1080\\ \boxed{4}&\boxed{540} \end{array} \qquad \begin{array}{llll} \begin{cases} x_1=1\\ x_2=4 \end{cases}\implies \cfrac{f(4)-f(1)}{4-1}\\\\\\ \cfrac{2160-540}{4-1} \end{array}`

c)

well, notice from b), the distance is dropping a constant value, 1 more hour, and the plane is at its destination, so the domain will be from [0, 5]