Algebra · Learn by Concept

Algebra Help: Average Rate Of Change

Review real example questions for Average Rate Of Change in Algebra.

Question 1 / 10

0 of 10 answered

The value of a used laptop (in dollars) is modeled by $V(t)=900-80t$ , where $t$ is the number of years since purchase. What is the average rate of change of $V(t)$ from $t=2$ to $t=5$ , and what does the sign mean?

All questions

Question 1

$80$ $/\text{year}$; the value is increasing.
$-80$ $/\text{year}$; the value is decreasing. (correct answer)
$-240$ dollars; the value is decreasing.
$\frac{-80}{3}$ $/\text{year}$; the value is decreasing.

Explanation: This question tests your understanding of average rate of change, which is a super important concept connecting slope, functions, and real-world rates like speed or growth. The average rate of change has units that come from dividing output units by input units: if value is in dollars and time is in years, the rate is in dollars per year ($/year). These units help us understand what the number means—it's not just abstract math! To find the average rate of change of V(t) = 900 - 80t from t = 2 to t = 5, we first evaluate at the endpoints: V(2) = 900 - 80(2) = 900 - 160 = 740 and V(5) = 900 - 80(5) = 900 - 400 = 500. Then we use the formula: average rate = [V(5) - V(2)]/(5 - 2) = (500 - 740)/(5 - 2) = -240/3 = -80. That's it! Choice B is correct because it properly calculates [V(5) - V(2)]/(5 - 2) = -240/3 = -80 $/year, getting both the arithmetic and the interpretation right! Choice C gives just the change in value (-240 dollars) instead of the rate. Remember: average rate of change = (change in y)/(change in x), not just one or the other. We're finding how much value changes per unit of time! The average rate of change of -80 $/year means that on average, the laptop value is decreasing by $80 for each year. In this context, that translates to the laptop depreciating at a constant rate of $80 per year—the negative sign tells us it's losing value, not gaining!

Question 2

A car's distance from home (in miles) after $t$ hours is $d(t)=50t+20$ . What is the average rate of change of $d(t)$ from $t=1$ to $t=4$ , and what are the units?

$70$ mi/hr
$50$ mi/hr (correct answer)
$200$ miles
$\frac{50}{3}$ mi/hr

Explanation: This question tests your understanding of average rate of change, which is a super important concept connecting slope, functions, and real-world rates like speed or growth. For a linear function, the average rate of change is the same as the slope and doesn't depend on which interval you choose—the function changes at a constant rate everywhere. But for nonlinear functions like quadratics, the average rate of change can be different over different intervals. To find the average rate of change of d(t) = 50t + 20 from t = 1 to t = 4, we first evaluate at the endpoints: d(1) = 50(1) + 20 = 50 + 20 = 70 and d(4) = 50(4) + 20 = 200 + 20 = 220. Then we use the formula: average rate = [d(4) - d(1)]/(4 - 1) = (220 - 70)/(4 - 1) = 150/3 = 50. That's it! Choice B is correct because it properly calculates [d(4) - d(1)]/(4 - 1) = 150/3 = 50 mi/hr, getting both the arithmetic and the units right! Choice C gives just 200 miles, which is d(4) - 20, not the rate of change. Remember: average rate of change = (change in y)/(change in x), not just one or the other. We're finding how much distance changes per unit of time! Units are your friend for understanding: if distance is in miles and time is in hours, then average rate of change is in miles per hour (mi/hr). The 'per' in the units reminds you that it's a ratio—change in output PER change in input. This helps you interpret what the number means!

Question 3

A runner's distance from the starting line (in miles) after $t$ hours is given by $d(t)=6t+2$ . What is the average rate of change of $d(t)$ from $t=1$ to $t=4$ ? Give your answer in mi/hr.

$6$ mi/hr (correct answer)
$18$ mi/hr
$-6$ mi/hr
$2$ mi/hr

Explanation: This question tests your understanding of average rate of change, which is a super important concept connecting slope, functions, and real-world rates like speed or growth. The average rate of change has units that come from dividing output units by input units: if distance is in miles and time is in hours, the rate is in miles per hour (mi/hr). To find the average rate of change of d(t) = 6t + 2 from t = 1 to t = 4, we first evaluate at the endpoints: d(1) = 61 + 2 = 8 and d(4) = 64 + 2 = 26. Then we use the formula: average rate = [d(4) - d(1)]/(4 - 1) = (26 - 8)/3 = 18/3 = 6 mi/hr. Choice A is correct because it properly calculates [d(4) - d(1)]/(4 - 1) = 18/3 = 6 mi/hr, getting both the arithmetic and the units right! Choice C calculates the change in y correctly as 18, but forgets to divide by the change in t (which is 3). Units are your friend for understanding: if distance is in feet and time is in seconds, then average rate of change is in feet per second (ft/sec). The 'per' in the units reminds you that it's a ratio—change in output PER change in input. This helps you interpret what the number means!

Question 4

The value of a machine decreases over time according to $V(t)=200-15t$ , where $V$ is in dollars and $t$ is in years. What is the average rate of change of $V(t)$ from $t=2$ to $t=8$ ? Include units.

$-15$ dollars
$15$ $/\text{year}$
$-15$ $/\text{year}$ (correct answer)
$-90$ $/\text{year}$

Explanation: This question tests your understanding of average rate of change, which is a super important concept connecting slope, functions, and real-world rates like speed or growth. The average rate of change has units that come from dividing output units by input units: if value is in dollars and time is in years, the rate is in dollars per year ($/year). To find the average rate of change of V(t) = 200 - 15t from t = 2 to t = 8, we first evaluate at the endpoints: V(2) = 200 - 152 = 170 and V(8) = 200 - 158 = 80. Then we use the formula: average rate = [V(8) - V(2)]/(8 - 2) = (80 - 170)/6 = -90/6 = -15 $/year. Choice A is correct because it properly calculates [V(8) - V(2)]/(8 - 2) = -90/6 = -15 $/year, getting both the arithmetic and the units right! Choice B has the magnitude right but the wrong sign. When V(8) is less than V(2), the numerator is negative, giving a negative rate, not positive. Units are your friend for understanding: the 'per' in the units reminds you that it's a ratio—change in output PER change in input. This helps you interpret what the number means!

Question 5

A ball is thrown upward with initial velocity. The table shows its height $h$ (in feet) at various times $t$ (in seconds). During which interval does the ball have the most negative average rate of change?

From $t = 1$ to $t = 2$ seconds
From $t = 2$ to $t = 3$ seconds
From $t = 3$ to $t = 4$ seconds
From $t = 4$ to $t = 5$ seconds (correct answer)

Explanation: Calculate each average rate of change: From t=1 to t=2: $\frac{84-68}{1} = 16$ . From t=2 to t=3: $\frac{52-84}{1} = -32$ . From t=3 to t=4: $\frac{20-52}{1} = -32$ . From t=4 to t=5: $\frac{-28-20}{1} = -48$ . The most negative (steepest decline) is -48 from t=4 to t=5, indicating the ball is falling fastest during this interval.

Question 6

A water tank's volume (in gallons) after $t$ minutes is $V(t)=120-6t$ . What is the average rate of change of $V(t)$ from $t=5$ to $t=15$ ? (Compare $(5,V(5))$ and $(15,V(15))$ .)

$60$ gallons/min
$-60$ gallons/min
$-6$ gallons/min (correct answer)
$6$ gallons/min

Explanation: This question tests your understanding of average rate of change, which is a super important concept connecting slope, functions, and real-world rates like speed or growth. The average rate of change has units that come from dividing output units by input units: if volume is in gallons and time is in minutes, the rate is in gallons per minute (gallons/min). These units help us understand what the number means—it's not just abstract math! To find the average rate of change of V(t)=120-6t from t=5 to t=15, we first evaluate at the endpoints: V(5)=120-65=90 and V(15)=120-615=30. Then we use the formula: average rate = [30 - 90]/(15 - 5) = -60/10 = -6. That's it! Choice A is correct because it properly calculates [V(15) - V(5)]/(15 - 5) = -60/10 = -6, getting both the arithmetic and the sign/interpretation right! Choice B has the magnitude right but the wrong sign. When V(15) is less than V(5), the numerator is negative, giving a negative rate, not positive. Quick sanity check: if the function is increasing (going up) on your interval, the average rate should be positive. If decreasing (going down), it should be negative. If horizontal, it should be zero. Does your answer match what you see happening in the function? If not, recheck your work!

Question 7

For the function $p(x)=-3x+4$ , calculate the average rate of change of $p(x)$ over the interval $[-2,2]$ .

$-12$
$-3$ (correct answer)
$3$
$0$

Explanation: This question tests your understanding of average rate of change, which is a super important concept connecting slope, functions, and real-world rates like speed or growth. For a linear function, the average rate of change is the same as the slope and doesn't depend on which interval you choose—the function changes at a constant rate everywhere. To find the average rate of change of p(x) = -3x + 4 from x = -2 to x = 2, we first evaluate at the endpoints: p(-2) = -3*(-2) + 4 = 10 and p(2) = -3*2 + 4 = -2. Then we use the formula: average rate = [p(2) - p(-2)]/(2 - (-2)) = (-2 - 10)/4 = -12/4 = -3. Choice A is correct because it properly calculates [p(2) - p(-2)]/(2 - (-2)) = -12/4 = -3, getting both the arithmetic and the sign right! Choice B has the magnitude right but the wrong sign. When p(2) is less than p(-2), the numerator is negative, giving a negative rate, not positive. The key formula for average rate of change is (y₂ - y₁)/(x₂ - x₁), which you might recognize as the slope formula! To use it: (1) identify your two points or endpoints, (2) subtract the y-values (later minus earlier), (3) subtract the x-values (later minus earlier), (4) divide. Keep the order consistent and you'll get the right answer every time!

Question 8

The cost function for producing $x$ widgets is $C(x) = 0.1x^2 + 5x + 100$ . A manufacturer increases production from 20 widgets to 40 widgets. What is the average rate of change in cost per widget over this interval?

$11$ dollars per widget (correct answer)
$9$ dollars per widget
$13$ dollars per widget
$15$ dollars per widget

Explanation: Calculate $C(20) = 0.1(400) + 5(20) + 100 = 40 + 100 + 100 = 240$ and $C(40) = 0.1(1600) + 5(40) + 100 = 160 + 200 + 100 = 460$ . The average rate is $\frac{460 - 240}{40 - 20} = \frac{220}{20} = 11$ dollars per widget. Choice B results from computational errors in the quadratic term. Choice C comes from using wrong interval endpoints. Choice D results from errors in the linear coefficient calculation.

Question 9

For the quadratic function $h(x)=x^2$ , which interval has the larger average rate of change: $[0,2]$ or $[2,4]$ ?

The interval $[0,2]$ has the larger average rate of change.
The intervals have the same average rate of change.
The interval $[2,4]$ has the larger average rate of change. (correct answer)
Neither interval has an average rate of change because $h(x)$ is not linear.

Explanation: This question tests your understanding of average rate of change, which is a super important concept connecting slope, functions, and real-world rates like speed or growth. But for nonlinear functions like quadratics, the average rate of change can be different over different intervals. Over interval [0,2], the rate is [h(2) - h(0)]/(2 - 0) = (4 - 0)/2 = 2. Over interval [2,4], the rate is [h(4) - h(2)]/(4 - 2) = (16 - 4)/2 = 6. Comparing these: 6 > 2, so interval [2,4] has the larger average rate of change. This makes sense because the function is steeper there. Choice C is correct because it properly calculates and compares the rates to identify [2,4] as larger, getting the interpretation right! Choice A states the interpretation backwards: [0,2] has the smaller rate, not larger. Quick sanity check: if the function is increasing (going up) on your interval, the average rate should be positive. If decreasing (going down), it should be negative. If horizontal, it should be zero. Does your answer match what you see happening in the function? If not, recheck your work!

Question 10

Based on the table shown, what is the average rate of change of $f(x)$ from $x = -1$ to $x = 5$ ?

$\frac{7}{6}$ (correct answer)
$\frac{6}{7}$
$\frac{7}{3}$
$\frac{3}{7}$

Explanation: The average rate of change is $\frac{f(5) - f(-1)}{5 - (-1)} = \frac{9 - 2}{6} = \frac{7}{6}$ . Choice B incorrectly inverts the fraction. Choice C uses the wrong denominator (3 instead of 6). Choice D combines both errors from choices B and C.