Math II · F-IF.6

Calculating and Interpreting Average Rate of Change for Quadratic Functions

This objective teaches students how to measure how fast one quantity changes compared with another when the rate is not constant. It is the bridge from slope of a line to velocity, acceleration, trend analysis, cost change, profit change, and eventually calculus.

Concept Functions
Domain Interpreting Functions
Read time 9 minutes

What this learning objective is really asking you to learn

In Math I, students learn slope as the constant rate of change of a line. If a line has slope 3, then every time \(x\) increases by 1, \(y\) increases by 3. Linear functions have one rate that applies everywhere.

Quadratic functions are different. Their graphs curve, so their rate of change is not constant. A parabola may rise quickly, then slowly, then stop rising, then fall. Or it may fall, flatten, and rise. Because the rate changes, students need a way to describe change over an interval. That tool is average rate of change.

The average rate of change of a function \(f\) from \(x=a\) to \(x=b\) is

\[(f(b)-f(a))/(b-a)\].

This is the change in output divided by the change in input. Graphically, it is the slope of the secant line connecting the two points \((a,f(a))\) and \((b,f(b))\) on the graph. Numerically, it is the average change in \(y\) per one unit of \(x\) over that interval. Contextually, it answers, “Over this stretch, how much did the output change per input unit on average?”

For example, suppose \(h(t)=-16t^2+64t+5\) gives the height in feet of an object after \(t\) seconds. The average rate of change from \(t=0\) to \(t=2\) is

\[(h(2)-h(0))/(2-0)\].

Compute \(h(0)=5\) and \(h(2)=-16(4)+128+5=69\). The average rate is \((69-5)/2=32\) feet per second. That means that over the first two seconds, the object's height increased by an average of 32 feet each second. It does not mean the object was moving upward at exactly 32 feet per second at every moment. The instantaneous rate changes throughout the interval.

From \(t=2\) to \(t=4\), the same function gives \(h(2)=69\) and \(h(4)=-16(16)+256+5=5\). The average rate is \((5-69)/(4-2)=-32\) feet per second. The object decreased by an average of 32 feet per second over that interval. The sign tells direction. Positive means the output increased over the interval; negative means it decreased.

From \(t=1\) to \(t=3\), \(h(1)=53\) and \(h(3)=53\), so the average rate is 0. This does not mean the object was not moving. It means the object started and ended the interval at the same height. It rose during part of the interval and fell during part of the interval, with net change zero. This example is powerful because it shows why average rate must be interpreted carefully.

Why students should learn this math

Average rate of change is one of the most useful ideas in the entire high school function sequence. It turns “change” into a measurable quantity. Without it, students can say that something went up or down. With it, they can say how much it changed per unit, over which interval, and in what direction.

Real life rarely gives perfectly constant rates. A car speeds up and slows down. A business grows faster in some months than others. A ball rises and falls. A phone battery drains at different speeds depending on use. A population changes differently across decades. A student's test scores may improve quickly at first and then level off. Average rate of change gives a practical way to summarize these changing processes over a chosen interval.

In science, average rate of change is essential. Average speed is distance traveled divided by time elapsed. Average velocity is change in position divided by time elapsed. Average acceleration is change in velocity divided by time elapsed. A quadratic position function often appears when acceleration is constant, such as in ideal projectile motion. Students who calculate average rate of change for a quadratic are preparing for physics.

In business, average rate of change describes how revenue, cost, or profit changes as production or price changes. If a profit function is quadratic, the average rate of change over a price interval tells how profit changes per dollar of price increase across that range. A positive average rate suggests the price increase improved profit over that interval. A negative average rate suggests it hurt profit. Around the vertex, the rate changes sign, signaling the transition from improvement to decline.

In data interpretation, average rate of change prevents vague claims. Instead of saying “the graph is rising,” a student can say, “From 2010 to 2020, the quantity increased by an average of 4.2 units per year.” That is more precise and more useful. It also invites comparison: Was the growth faster from 2000 to 2010 or from 2010 to 2020?

The deeper reason to learn this objective is that it bridges Algebra and Calculus. The derivative, one of the central ideas of calculus, is an instantaneous rate of change. It is built by studying average rates of change over smaller and smaller intervals. A secant line becomes a tangent line in the limit. Students do not need calculus yet, but they are building the mental machinery for it. If they understand average rate now, derivatives later will feel like a natural refinement rather than a mysterious new topic.

The historical machinery: rates, secants, and the road to calculus

Average rate of change is one of the great bridge ideas in mathematics. Long before students meet formal calculus, they need language for how one quantity changes compared with another. Merchants cared about price per unit. Travelers cared about distance per time. Astronomers cared about changing positions of planets. Scientists cared about speed, acceleration, cooling, growth, and decay. In each case, a rate compares change in one quantity to change in another.

The graph version of average rate of change is the slope of a secant line: a line connecting two points on a curve. This idea became historically important because curved motion does not have one constant slope. A linear model has the same rate everywhere, but a quadratic model changes rate as the input changes. If a projectile's height is quadratic in time, its average velocity over one interval may be positive, over another interval zero-ish near the top, and over a later interval negative. The changing average rates tell the story of the motion.

This concept points directly toward calculus. Mathematicians such as Fermat, Newton, and Leibniz studied how to move from average rates over intervals to instantaneous rates at a point. That leap created derivatives, one of the central ideas of modern science and engineering. Objective 079 does not require students to take that leap yet. It prepares them for it. Students learn that the slope formula is not only for lines. It is a tool for measuring change over an interval on any function, especially a curve.

The technical machinery: slope, secant lines, and units

Average rate of change is slope applied to a function over an interval. For any two distinct input values \(a\) and \(b\), the formula \((f(b)-f(a))/(b-a)\) compares vertical change to horizontal change. The numerator is output change. The denominator is input change. The units are output units per input unit.

Units matter. If height is measured in feet and time is measured in seconds, the average rate is feet per second. If profit is measured in dollars and price is measured in dollars, the rate is dollars of profit per dollar of price increase. If area is measured in square meters and side length is measured in meters, the rate is square meters per meter. Units help students interpret the number rather than treating it as a naked calculation.

For a quadratic function, average rate of change depends on the interval. Consider \(f(x)=x^2\). From 0 to 2, the average rate is \((4-0)/(2-0)=2\). From 2 to 4, it is \((16-4)/(4-2)=6\). From -2 to 2, it is \((4-4)/(2-(-2))=0\). The function has no single slope. The interval determines the average rate.

Tables can also be used. If a table gives \(f(1)=7\) and \(f(5)=31\), then the average rate from 1 to 5 is \((31-7)/(5-1)=6\). Students do not need an equation if the needed values are in the table. If the exact inputs are not listed, students may estimate or interpolate, depending on instructions.

Graphs require estimation. To estimate average rate from a graph, identify the two points corresponding to the interval endpoints, estimate their coordinates, and compute the slope between them. The graph may not give exact values, so the answer should acknowledge approximation. A common mistake is estimating the slope of the curve at one endpoint instead of the slope of the secant line connecting both endpoints.

One elegant property of quadratics is that their average rate of change over symmetric intervals around the vertex may be zero if the outputs match. More generally, for a quadratic, the average rate of change over an interval equals the slope of the tangent line at the interval's midpoint. Students do not need to prove this in Integrated Math II, but the pattern can help them see why quadratic change is orderly. The rate is changing linearly.

How to interpret average rate in context

A correct interpretation should include four pieces: the interval, the direction of change, the amount per input unit, and the units.

Suppose \(P(p)=-2(p-10)^2+200\) models profit in dollars at price \(p\) dollars. From \(p=6\) to \(p=8\), \(P(6)=168\) and \(P(8)=192\), so the average rate is \((192-168)/(8-6)=12\). Interpretation: as the price increased from $6 to $8, profit increased by an average of $12 for each $1 increase in price. From \(p=12\) to \(p=14\), profit decreases from 192 to 168, so the average rate is -12. Interpretation: as price increased from $12 to $14, profit decreased by an average of $12 per $1 increase in price.

The sign is meaningful. Positive does not always mean “good,” and negative does not always mean “bad.” If the output is cost, a negative rate may mean cost is decreasing. If the output is height, a negative rate means the object is moving downward on average. If the output is error, a negative rate may be desirable. Context decides.

The interval is also meaningful. A company might have a positive average profit change over a large interval even though profit declined during part of that interval. A ball might have zero average rate over an interval while still moving. A data trend might look stable over ten years but volatile year to year. Average rate summarizes net change, not every moment inside the interval.

This is why students should avoid saying “the rate of change is” without naming the interval. For a nonlinear function, there are many average rates of change. The phrase must be completed: “from \(x=2\) to \(x=5\),” “over the first three seconds,” or “between prices of $8 and $12.”

Where this fits into the big map of math

Average rate of change connects several regions of the math map. It extends slope from lines to curves. It supports interpreting graphs and tables. It prepares students for derivatives. It links algebra to physics and economics. It also connects to statistics, where slopes of fitted lines summarize average trends in data. In Math III, students will compare average rates across polynomial, rational, radical, exponential, logarithmic, and trigonometric functions. The same formula will keep working, even as function types become more sophisticated.

For quadratics specifically, average rate of change helps students understand curvature. A line has constant first differences. A quadratic has changing first differences but constant second differences for equal input steps. That means the average rate changes in a predictable pattern. This is the beginning of understanding acceleration: not just position, not just velocity, but change in velocity.

Common student traps and how to avoid them

One trap is using \(f(b)-f(a)\) but forgetting to divide by \(b-a\). That gives total change, not rate of change.

A second trap is subtracting in inconsistent order. \((f(b)-f(a))/(b-a)\) and \((f(a)-f(b))/(a-b)\) both work, but mixing the order gives the wrong sign.

A third trap is ignoring units. The number alone is incomplete. An average rate of 32 could mean feet per second, dollars per item, meters squared per meter, or something else.

A fourth trap is treating average rate as instantaneous rate. The average rate over an interval describes net change across the interval, not the exact rate at every point.

A fifth trap is giving an answer without the interval. For a quadratic, the average rate depends on the interval, so the interval must be part of the statement.

Problem Library

Problems in the App From This Objective

150 problems across 12 archetypes in the app.

compute slope of secant line.
12 problems Warmup Practice Mixed Review Assessment
Problem 1

Calculate the average rate of change between points (1,3) and (4,15).

Problem 2

Calculate the average rate of change between points (-2,9) and (2,1).

Problem 3

Calculate the average rate of change between points (0,5) and (6,5).

Problem 4

Calculate the average rate of change between points (3,12) and (5,28).

Problem 5

Calculate the average rate of change between points (1,2) and (3,8).

Problem 6

Calculate the average rate of change between points (0,10) and (5,0).

Problem 7

Calculate the average rate of change between points (2,5) and (4,6).

Problem 8

Calculate the average rate of change between points (0,7) and (2,6).

Problem 9

Calculate the average rate of change between points (-3,-1) and (-1,3).

Problem 10

Calculate the average rate of change between points (10,50) and (20,100).

Problem 11

Calculate the average rate of change between points (-5,10) and (5,10).

Problem 12

Calculate the average rate of change between points (1,20) and (3,0).

Open in simulator
use endpoint values over an interval.
12 problems Warmup Practice Mixed Review Assessment
Problem 13

Calculate average rate of change from table x: 0,1,2,3; f(x): 1,4,9,16 over interval [1,3].

Problem 14

Calculate average rate of change from table x: -1,0,2; f(x): 8,5,1 over interval [-1,2].

Problem 15

Calculate average rate of change from table x: 2,4,6; f(x): 10,18,34 over interval [2,6].

Problem 16

Calculate average rate of change from table x: 0,5,10; f(x): 100,75,0 over interval [0,10].

Problem 17

Calculate average rate of change from table x: 1,3,5,7; f(x): 2,8,14,20 over interval [1,5].

Open in simulator
Problem 18

Calculate average rate of change from table x: -2,-1,0,1,2; f(x): 0,3,4,3,0 over interval [-1,1].

Problem 19

Calculate average rate of change from table x: 0,1,2,3; f(x): 1,2,4,8 over interval [0,2].

Problem 20

Calculate average rate of change from table x: -3,-1,0,2; f(x): 9,1,-2,-8 over interval [-3,2].

Problem 21

Calculate average rate of change from table x: 10,20,30; f(x): 5,5,5 over interval [10,30].

Problem 22

Calculate average rate of change from table x: 5,10,15,20; f(x): 25,100,225,400 over interval [5,15].

Problem 23

Calculate average rate of change from table x: -5,-3,-1,0; f(x): 20,10,2,1 over interval [-5,-1].

Problem 24

Calculate average rate of change from table x: 0.5,1.0,1.5,2.0; f(x): 1,3,5,7 over interval [0.5,1.5].

evaluate endpoints and divide differences.
12 problems Warmup Practice Mixed Review Assessment
Problem 25

Calculate average rate of change of f(x)=x^2 over interval [1,4].

Problem 26

Calculate average rate of change of g(x)=-(x-2)^2+9 over interval [0,2].

Problem 27

Calculate average rate of change of h(x)=2x^2-3 over interval [2,5].

Problem 28

Calculate average rate of change of p(x)=x^2-6x+1 over interval [1,7].

Problem 29

Calculate average rate of change of f(x)=x^2+3x over interval [0,3].

Problem 30

Calculate average rate of change of f(x)=-x^2+5 over interval [1,3].

Problem 31

Calculate average rate of change of f(x)=3x^2-x+2 over interval [0,2].

Open in simulator
Problem 32

Calculate average rate of change of f(x)=(x-1)^2 over interval [0,4].

Problem 33

Calculate average rate of change of f(x)=-2x^2+x over interval [-1,1].

Problem 34

Calculate average rate of change of f(x)=x^3 over interval [0,2].

Problem 35

Calculate average rate of change of f(x)=x^2+2x-5 over interval [-2,0].

Problem 36

Calculate average rate of change of f(x)=4x-7 over interval [1,5].

read endpoint coordinates and calculate.
12 problems Warmup Practice Mixed Review Assessment
Problem 37

Estimate average rate of change from graph description The graph passes near (1,2) and (5,18).

Problem 38

Estimate average rate of change from graph description The graph passes near (0,10) and (4,2).

Problem 39

Estimate average rate of change from graph description The graph shows points (2,6) and (6,6).

Open in simulator
Problem 40

Estimate average rate of change from graph description The endpoint readings are approximately (-1,9) and (3,1).

Problem 41

Estimate average rate of change from graph description The graph passes near (3,5) and (7,17).

Problem 42

Estimate average rate of change from graph description The graph passes near (-2,10) and (2,0).

Problem 43

Estimate average rate of change from graph description The graph shows points (0,0) and (10,5).

Problem 44

Estimate average rate of change from graph description The endpoint readings are approximately (-5,-2) and (1,4).

Problem 45

Estimate average rate of change from graph description The graph passes near (10,20) and (15,5).

Problem 46

Estimate average rate of change from graph description The graph passes near (-3,7) and (0,7).

Problem 47

Estimate average rate of change from graph description The graph shows points (1,1) and (2,10).

Problem 48

Estimate average rate of change from graph description The endpoint readings are approximately (5,-1) and (7,-7).

describe average velocity over interval.
12 problems Warmup Practice Mixed Review Assessment
Problem 49

Interpret average rate of change 32 ft/s for projectile context height changes from 0 ft to 64 ft over 2 seconds.

Problem 50

Interpret average rate of change -32 ft/s for projectile context height changes from 64 ft to 0 ft from t=2 to t=4.

Problem 51

Interpret average rate of change 0 ft/s for projectile context height is 0 ft at t=0 and 0 ft at t=4.

Problem 52

Interpret average rate of change 16 ft/s for projectile context height changes from 48 ft to 64 ft over 1 second.

Problem 53

Interpret average rate of change 10 m/s for projectile context height increases from 10 meters to 50 meters in 4 seconds.

Problem 54

Interpret average rate of change -10 m/s for projectile context height decreases from 80 meters to 30 meters over 5 seconds.

Problem 55

Interpret average rate of change 0 ft/s for projectile context height is 20 feet at t=1 second and 20 feet at t=3 seconds.

Problem 56

Interpret average rate of change 40 ft/s for projectile context height goes from 100 ft to 120 ft in 0.5 seconds.

Problem 57

Interpret average rate of change -80 ft/s for projectile context height drops from 150 ft to 130 ft in 0.25 seconds.

Open in simulator
Problem 58

Interpret average rate of change 50 ft/s for projectile context height changes from 1000 ft to 1500 ft over 10 seconds.

Problem 59

Interpret average rate of change -50 m/s for projectile context height changes from 2000 m to 1000 m over 20 seconds.

Problem 60

Interpret average rate of change 20 ft/s for projectile context height is 5 ft at t=0.5s and 25 ft at t=1.5s.

describe output change per input unit.
12 problems Warmup Practice Mixed Review Assessment
Problem 61

Interpret average rate of change 20 dollars per dollar in context Revenue changes from 200 dollars to 260 dollars as price goes from 5 to 8 dollars.

Problem 62

Interpret average rate of change 10 square feet per foot in context Area changes from 24 to 54 square feet as side length goes from 2 to 5 feet.

Problem 63

Interpret average rate of change -6 dollars per item in context Profit changes from 100 to 40 as quantity goes from 10 to 20.

Open in simulator
Problem 64

Interpret average rate of change 5 dollars per unit in context Cost changes from 300 to 450 as production goes from 20 to 50 units.

Problem 65

Interpret average rate of change 60 miles per hour in context Distance changes from 100 miles to 340 miles as time goes from 1 hour to 5 hours.

Problem 66

Interpret average rate of change 2000 users per month in context The number of active users on a platform changes from 10,000 to 16,000 over 3 months.

Problem 67

Interpret average rate of change -600 people per year in context A town's population changes from 15,000 people to 12,000 people over 5 years.

Problem 68

Interpret average rate of change 25 miles per gallon in context The distance traveled by a car changes from 50 miles to 250 miles as fuel consumed goes from 2 gallons to 10 gallons.

Problem 69

Interpret average rate of change -2000 dollars per year in context The value of a machine changes from 10,000 dollars to 4,000 dollars over 3 years.

Problem 70

Interpret average rate of change 4 cm per week in context A plant's height changes from 10 cm to 34 cm over 6 weeks.

Problem 71

Interpret average rate of change -200 liters per minute in context The volume of water in a pool changes from 10,000 liters to 7,000 liters over 15 minutes.

Problem 72

Interpret average rate of change 4 dollars per item in context The total cost of production changes from 500 dollars to 900 dollars as the number of items produced goes from 100 to 200 items.

show rate changes across intervals.
15 problems Warmup Practice Mixed Review Assessment
Problem 73

Compare average rates of change for f(x)=x^2 over intervals [0,2] and [2,4].

Problem 74

Compare average rates of change for g(x)=-(x-3)^2+9 over intervals [0,3] and [3,6].

Problem 75

Compare average rates of change for h(x)=x^2-4x over intervals [0,2] and [2,4].

Problem 76

Compare average rates of change for p(x)=2x^2 over intervals [1,2] and [3,4].

Problem 77

Compare average rates of change for f(x)=x^2+3 over intervals [1,3] and [3,5].

Problem 78

Compare average rates of change for f(x)=-x^2 over intervals [-2,0] and [0,2].

Problem 79

Compare average rates of change for f(x)=(x-1)^2 over intervals [0,1] and [1,2].

Problem 80

Compare average rates of change for f(x)=3x^2-1 over intervals [-1,0] and [0,1].

Open in simulator
Problem 81

Compare average rates of change for f(x)=-2x^2+5 over intervals [0,1] and [1,2].

Problem 82

Compare average rates of change for f(x)=x^2+x over intervals [-2,-1] and [0,1].

Problem 83

Compare average rates of change for f(x)=(x+2)^2-1 over intervals [-3,-2] and [-2,-1].

Problem 84

Compare average rates of change for f(x)=-x^2+4x over intervals [0,1] and [1,2].

Problem 85

Compare average rates of change for f(x)=0.5x^2 over intervals [0,2] and [2,4].

Problem 86

Compare average rates of change for f(x)=-(x-1)^2+10 over intervals [0,1] and [1,2].

Problem 87

Compare average rates of change for f(x)=x^2-6x+5 over intervals [0,3] and [3,6].

use graph or table endpoints.
12 problems Warmup Practice Mixed Review Assessment
Problem 88

Classify the average rate of change of f(x)=x^2 over [-2,2] as positive, negative, or zero.

Problem 89

Classify the average rate of change of g(x)=-(x-1)^2+4 over [-1,1] as positive, negative, or zero.

Problem 90

Classify the average rate of change of h(x)=(x-3)^2 over [3,6] as positive, negative, or zero.

Problem 91

Classify the average rate of change of p(x)=-(x+2)^2 over [-2,1] as positive, negative, or zero.

Problem 92

Classify the average rate of change of f(x)=3x+1 over [0,2] as positive, negative, or zero.

Problem 93

Classify the average rate of change of f(x)=-2x+5 over [1,3] as positive, negative, or zero.

Open in simulator
Problem 94

Classify the average rate of change of f(x)=x^2-4x over [1,3] as positive, negative, or zero.

Problem 95

Classify the average rate of change of f(x)=(x-1)^2 over [1,4] as positive, negative, or zero.

Problem 96

Classify the average rate of change of f(x)=-(x+1)^2+3 over [0,2] as positive, negative, or zero.

Problem 97

Classify the average rate of change of f(x)=x^3 over [-1,2] as positive, negative, or zero.

Problem 98

Classify the average rate of change of f(x)=-x^3 over [0,2] as positive, negative, or zero.

Problem 99

Classify the average rate of change of f(x)=5 over [-3,3] as positive, negative, or zero.

connect graph geometry to calculation.
12 problems Warmup Practice Mixed Review Assessment
Problem 100

Relate average rate of change between (1,2) and (5,18) to secant line slope.

Open in simulator
Problem 101

Relate average rate of change between (-2,6) and (2,-2) to secant line slope.

Problem 102

Relate average rate of change between (0,3) and (4,3) to secant line slope.

Problem 103

Relate average rate of change between (3,1) and (6,19) to secant line slope.

Problem 104

Relate average rate of change between (0,0) and (3,9) to secant line slope.

Problem 105

Relate average rate of change between (-1,-1) and (1,1) to secant line slope.

Problem 106

Relate average rate of change between (2,5) and (4,1) to secant line slope.

Problem 107

Relate average rate of change between (-3,7) and (0,1) to secant line slope.

Problem 108

Relate average rate of change between (10,20) and (12,24) to secant line slope.

Problem 109

Relate average rate of change between (-5,10) and (-1,2) to secant line slope.

Problem 110

Relate average rate of change between (1,10) and (7,-2) to secant line slope.

Problem 111

Relate average rate of change between (0,-5) and (5,0) to secant line slope.

solve using rate equation.
12 problems Warmup Practice Mixed Review Assessment
Problem 112

Find the missing endpoint value given average rate information: Average rate from x=2 to x=6 is 5 and f(2)=3.

Problem 113

Find the missing endpoint value given average rate information: Average rate from x=1 to x=4 is -2 and f(1)=10.

Problem 114

Find the missing endpoint value given average rate information: Average rate from x=0 to x=8 is 0 and f(0)=12.

Problem 115

Find the missing endpoint value given average rate information: Average rate from x=-1 to x=3 is 6 and f(3)=20.

Problem 116

Find the missing endpoint value given average rate information: Average rate from x=1 to x=5 is 3 and f(1)=7.

Problem 117

Find the missing endpoint value given average rate information: Average rate from x=0 to x=10 is -1.5 and f(0)=20.

Open in simulator
Problem 118

Find the missing endpoint value given average rate information: Average rate from x=-2 to x=4 is 2 and f(4)=15.

Problem 119

Find the missing endpoint value given average rate information: Average rate from x=3 to x=7 is -3 and f(7)=1.

Problem 120

Find the missing endpoint value given average rate information: Average rate from x=-5 to x=5 is 0 and f(5)=-8.

Problem 121

Find the missing endpoint value given average rate information: Average rate from x=0 to x=3 is 0.5 and f(0)=1.

Problem 122

Find the missing endpoint value given average rate information: Average rate from x=2 to x=7 is -0.4 and f(7)=10.

Problem 123

Find the missing endpoint value given average rate information: Average rate from x=-10 to x=10 is 0.7 and f(-10)=-5.

connect changing slope to nonlinear behavior.
15 problems Warmup Practice Mixed Review Assessment
Problem 124

Explain why average rate of change varies for quadratic f(x)=x^2.

Problem 125

Explain why average rate of change varies for quadratic g(x)=-(x-2)^2+5.

Problem 126

Explain why average rate of change varies for quadratic h(x)=2x^2.

Problem 127

Explain why average rate of change varies for quadratic p(x)=x^2-6x.

Problem 128

Explain why average rate of change varies for quadratic f(x)=x^2+3x+2.

Problem 129

Explain why average rate of change varies for quadratic f(x)=-x^2.

Problem 130

Explain why average rate of change varies for quadratic f(x)=(x+1)^2.

Problem 131

Explain why average rate of change varies for quadratic f(x)=-3x^2+1.

Problem 132

Explain why average rate of change varies for quadratic f(x)=x^2-4.

Problem 133

Explain why average rate of change varies for quadratic f(x)=0.5x^2.

Problem 134

Explain why average rate of change varies for quadratic f(x)=x(x-3).

Problem 135

Explain why average rate of change varies for quadratic f(x)=-(x+2)(x-1).

Problem 136

Explain why average rate of change varies for quadratic f(x)=x^2+5.

Open in simulator
Problem 137

Explain why average rate of change varies for quadratic f(x)=(x-4)^2-2.

Problem 138

Explain why average rate of change varies for quadratic f(x)=-x^2+2x-1.

catch endpoint, subtraction, interval, and unit mistakes.
12 problems Warmup Practice Mixed Review Assessment
Problem 139

Correct the average-rate error in From (1,3) to (5,19), average rate is (5-1)/(19-3)=1/4.

Open in simulator
Problem 140

Correct the average-rate error in For f(0)=2 and f(4)=10, average rate is 10-2=8.

Problem 141

Correct the average-rate error in Average rate from x=2 to x=6 uses f(2)+f(6).

Problem 142

Correct the average-rate error in A negative average rate means the function is always decreasing.

Problem 143

Correct the average-rate error in From f(1)=3 to f(5)=11, the average rate is (11-3)/5 = 8/5.

Problem 144

Correct the average-rate error in Given points (2, 7) and (8, 19), the average rate of change is (19-7)/(8-2) = 10/6 = 5/3.

Problem 145

Correct the average-rate error in From x=0 to x=10, f(0)=5 and f(10)=25. The average rate is (25-5)/(10-0) = 20/5 = 4.

Problem 146

Correct the average-rate error in For f(x) from x=1 to x=3, with f(1)=5 and f(3)=9, the average rate is (5+9)/(3-1) = 14/2 = 7.

Problem 147

Correct the average-rate error in The average rate of change of f(x) from x=2 to x=8, with f(2)=10 and f(8)=30, is (10+30)/2 = 20.

Problem 148

Correct the average-rate error in If f(x) goes from 5 to 15 as x goes from 2 to 7, the average rate is (15-5)/(15-5) = 1.

Problem 149

Correct the average-rate error in A car travels 150 miles in 3 hours and 350 miles in 7 hours. The average speed is (350-150)/3 = 200/3 mph.

Problem 150

Correct the average-rate error in If the average rate of change of a stock price is $0 over a month, it means the price never changed.