Question

1. the function ( m(x) ) represents a storage account balance ( m ) (measured in dollars) and ( x ) represents the number of years since the account was opened with an initial sum. describe in words what ( m(20)-m(10)=100 ) represents.
2. alex is driving from city a to city b. the distance between the cities is 800 miles.

the cars fuel efficiency is 20 miles per gallon. alexs car has a full tank at the start of the journey using both a graph and all equations.
alexs distance (in miles) after driving for ( x ) hours is modeled by the equation.
( d(x)=80 x-2 x^{2} ) for values of ( x ) at ( 0 leq x leq 20 )
the graph and table of ( d(x) ) is shown below. use the equation, graph, and table to answer the following questions:

time (hours)

0

2

4

6

10

:---:

:---:

:---:

:---:

:---:

:---:

distance (miles)

0

112

208

288

400

a) between which two hours is alex driving the fastest? how do you know?
b) what does ( d(15)=450 ) represent in this situation?
c) if alex wants to stop at city b which is exactly 300 miles from city a, how long will he need to drive? explain using the graph, table, or equation how you figured this out.

Explanation:

Step1: Calculate the average speed for each time interval

The average speed \(v=\frac{\Delta d}{\Delta t}\).
For \(t = 0\) to \(t=2\): \(\Delta d=117 - 0=117\), \(\Delta t = 2-0 = 2\), \(v_1=\frac{117}{2}=58.5\)
For \(t = 2\) to \(t = 4\): \(\Delta d=208 - 117 = 91\), \(\Delta t=4 - 2=2\), \(v_2=\frac{91}{2}=45.5\)
For \(t = 4\) to \(t = 6\): \(\Delta d=288 - 208=80\), \(\Delta t=6 - 4 = 2\), \(v_3=\frac{80}{2}=40\)
For \(t = 6\) to \(t = 10\): \(\Delta d=400 - 288 = 112\), \(\Delta t=10 - 6=4\), \(v_4=\frac{112}{4}=28\)

Step2: Analyze the meaning of \(d(15) = 450\)

The function \(d(t)\) gives the distance Alex has driven after \(t\) hours. So \(d(15)=450\) means that after driving for 15 hours, Alex has covered a distance of 450 miles.

Step3: Solve for \(t\) when \(d(t)=300\)

Set \(d(t)=300\), so \(60t-2t^{2}=300\)
Rearrange to \(2t^{2}-60t + 300=0\), divide by 2: \(t^{2}-30t + 150=0\)
Using the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a = 1\), \(b=-30\), \(c = 150\)
\(t=\frac{30\pm\sqrt{900 - 600}}{2}=\frac{30\pm\sqrt{300}}{2}=\frac{30\pm10\sqrt{3}}{2}=15\pm5\sqrt{3}\)
\(t_1=15 - 5\sqrt{3}\approx15-8.66 = 6.34\), \(t_2=15 + 5\sqrt{3}\approx15 + 8.66=23.66\) (but \(t\leq30\) and we consider the non - negative value within the domain of the function. Also, from the table and the shape of the parabola \(y = 60x-2x^{2}\), we can check values.

Answer:

a) Alex is driving the fastest between \(t = 0\) and \(t=2\) hours. Because the average speed (change in distance over change in time) is the highest in this interval.
b) It means that after driving for 15 hours, Alex has traveled 450 miles.
c) \(t = 15\pm5\sqrt{3}\approx6.34\) hours or \(23.66\) hours. We use the equation \(d(t)=60t - 2t^{2}\), set \(d(t) = 300\) and solve the quadratic equation \(60t-2t^{2}=300\) (or we can check the values around the table values and use the symmetry of the parabola \(y=-2x^{2}+60x\) (axis of symmetry \(x =-\frac{b}{2a}=-\frac{60}{2\times(- 2)}=15\))).