Slope as the ratio of rise to run, the two-point formula and the consistent-order rule, what the sign of a slope says about a line, the slope conditions for parallel and perpendicular lines, and slope as an average rate of change with units.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 2 — Linear Equations and Functions
Find Slope and Rate of Change
Objectives
Five outcomes. The last two are what make slope useful outside a coordinate plane.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.2 Find Slope and Rate of Change §2.2, pp. 82-87 — the lesson these objectives are drawn from
Warm-up
In Lesson 2.1 you noticed that a linear equation's outputs changed by the same amount at every step. This lesson names that amount.
Discussion prompt
In the table for y equals negative two x minus one, the outputs went 3, 1, -1, -3, -5 as x went -2, -1, 0, 1, 2. How much did y change for each step of one in x, and where does that number appear in the equation?
Hint: Subtract consecutive outputs, then look at the equation.
Answer:
\[ \Delta y = -2 \text{ for each } \Delta x = 1 \]
\[ y = \underbrace{-2}_{\text{slope}}x - 1 \]
The constant change per step is the coefficient of x, and it has a name: the slope. Everything in this lesson is that observation made precise enough to use on two points rather than a whole table.
Concept
The slope of a nonvertical line is the ratio of vertical change to horizontal change — how far it climbs for each unit it moves sideways. Because a line is straight, that ratio is the same wherever on the line you measure it, which is what makes a single number able to describe the whole line.
slope — The ratio of vertical change, the rise, to horizontal change, the run, between any two points of a nonvertical line.
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{rise}}{\text{run}} \]
The formula is Pythagoras' triangle without the hypotenuse: draw the right triangle between two points and slope is the vertical leg over the horizontal leg.
Figure (svg): A line on a coordinate plane with a right triangle drawn between two points, its vertical leg labelled rise and its horizontal leg labelled run
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.2 Find Slope and Rate of Change §2.2, pp. 82-82 — Slope of a Line
Section
Section 1
Concept
Pick any two points on a line, subtract to get the rise and the run, and divide. Different pairs of points give different rises and runs but always the same ratio, which is why the slope belongs to the line rather than to the pair.
\[ m = \frac{y_2 - y_1}{x_2 - x_1}, \quad x_1 \neq x_2 \]
The condition that the two x values differ is what excludes vertical lines. Their run is zero, and dividing by zero is undefined — which is the honest reason a vertical line has no slope.
Figure (svg): A line on a coordinate plane with a right triangle drawn between two points, its vertical leg labelled rise and its horizontal leg labelled run
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.2 Find Slope and Rate of Change §2.2, pp. 82-82
Picture it
The rise and the run are the two legs of a right triangle whose hypotenuse lies along the line.
Figure (svg): A line on a coordinate plane with a right triangle drawn between two points, its vertical leg labelled rise and its horizontal leg labelled run
Move to a different pair of points and the triangle changes size but keeps its shape, so the ratio of its legs is unchanged. Similar triangles are the reason slope is well defined.
Worked example
Example 1. A physical rise and run, with no coordinate plane in sight.
\[ \text{A ramp rises } 15 \text{ inches over a run of } 54 \text{ inches. Find its slope.} \]
Write the ratio
Why: Rise over run, with the rise on top because slope measures climb per unit of travel.
\[ \frac{15}{54} \]
Reduce the fraction
Why: Three divides both, giving five over eighteen.
\[ \frac{5}{18} \]
Interpret it
Why: The ramp climbs five inches for every eighteen inches of horizontal travel.
\[ 5\text{ in up per } 18\text{ in across} \]
Figure (svg): The solution to Worked example a skateboard ramp shown as a ladder of expressions, one row per algebraic move
\[ m = \frac{15}{54} = \frac{5}{18} \]
Verify: scale the ratio back to the original numbers
Why: Multiplying five eighteenths by three gives fifteen fifty-fourths, the original ratio, so the reduction is faithful. As a decimal the slope is about 0.28, which means the ramp climbs a little over a quarter of an inch per inch — a gentle ramp, which matches a rise of just over a foot across four and a half feet.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.2 Find Slope and Rate of Change §2.2, pp. 82-82
Picture it
The textbook's Avoid Errors note, made into a picture.
Figure (svg): Two columns contrasting a consistent subtraction order with a mixed one, showing that mixing the order flips the sign of the slope
Either labelling works. What never works is one labelling on the top and the other on the bottom, because that negates exactly one of the two differences.
Worked example
Example 2. The formula, with the labelling written out.
\[ \text{Find the slope of the line through } (-2, 1) \text{ and } (3, 5). \]
Label the two points
Why: Call the first one point one and the second point two. Which is which is a free choice, but it must then be used consistently.
\[ (x 1, y 1) = (-2, 1), (x 2, y 2) = (3, 5) \]
Compute the rise
Why: Second y minus first y: five minus one.
\[ \text{rise } = 4 \]
Compute the run
Why: Second x minus first x: three minus negative two, which is five.
\[ r u n = 5 \]
Divide
Why: Four fifths, and it is positive, so the line rises.
\[ m = \frac{4}{5} \]
Figure (svg): The solution to Worked example slope from two points shown as a ladder of expressions, one row per algebraic move
\[ m = \frac{5 - 1}{3 - (-2)} = \frac{4}{5} \]
Verify: swap the labels and recompute
Why: Calling (3,5) the first point gives a rise of one minus five, which is negative four, and a run of negative two minus three, which is negative five. Negative four over negative five is four fifths — the same slope. The labelling really is free, provided it is consistent.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.2 Find Slope and Rate of Change §2.2, pp. 82-82
Trap
\[ \text{Slope through } (-2, 1) \text{ and } (3, 5). \]
Take the y difference one way and the x difference the other
Why: The two subtractions are done independently rather than as one consistent choice.
\[ m = \frac{5 - 1}{-2 - 3} = \frac{4}{-5} = -\frac{4}{5} \]
The magnitude is right and the sign is wrong, which is exactly what makes this error survive a glance at the answer.
\[ \text{Slope through } (-2, 1) \text{ and } (3, 5). \]
Choose which point is first, then use it first in BOTH subtractions
Why: The formula is second minus first on top and second minus first on the bottom.
\[ m = \frac{5 - 1}{3 - (-2)} = \frac{4}{5} \]
Sketching the two points settles it independently: the second point is up and to the right of the first, so the line rises and the slope must be positive.
Fill the middle
Guided Practice 2, from the exercise set.
Fill in the blanks
m = \frac-8 - (-4)___} = \frac______ = \frac______
Why: The numerator used the second point's y first, so the denominator must use the second point's x first: negative eight minus negative four, which is negative four. Negative six over negative four reduces to three halves. Note that both differences came out negative and the two negatives cancelled — a sign that the order really was consistent.
Elimination
Guided Practice 2, as a multiple-choice question.
Eliminate the wrong options
What is the slope of the line through (-4, 9) and (-8, 3)?
Survives elimination: A
Why: Second y minus first y is 3 minus 9, or -6. Second x minus first x is -8 minus -4, or -4. The quotient is positive three halves. A sketch confirms the sign: moving from (-8,3) to (-4,9) goes right and up, so the line rises and its slope is positive.
Explain it to yourself
Slope is defined using two points, but it is a property of the whole line.
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
Discussion prompt
Explain why choosing a different pair of points on the same line gives the same slope. What geometric fact is doing the work, and what would go wrong if the graph were a curve instead of a line?
Hint: Draw two different right triangles under the same line and compare their shapes.
Answer:
Two right triangles drawn under the same line have equal angles, so they are similar, and similar triangles have proportional sides. The ratio of the legs is therefore the same for both, however different their sizes.
On a curve this fails: the steepness genuinely changes from place to place, so different pairs of points give different ratios. That is precisely why a curve needs calculus and a line does not — the single number that describes a line has to be replaced by a different number at every point.
Section
Section 2
Concept
A positive slope means the line rises left to right, a negative slope means it falls, a zero slope means it is horizontal, and an undefined slope means it is vertical. You can classify a line from two points without drawing anything.
\[ m > 0 \text{ rises}, \quad m < 0 \text{ falls}, \quad m = 0 \text{ horizontal}, \quad m \text{ undefined: vertical} \]
Zero and undefined are the two cases students swap. Zero slope has zero on top; undefined slope has zero on the bottom.
Figure (svg): Four small coordinate planes showing a rising line, a falling line, a horizontal line and a vertical line, each labelled with the sign of its slope
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.2 Find Slope and Rate of Change §2.2, pp. 83-83 — Classify lines using slope
Picture it
Example 3 produced one of each.
Figure (svg): Four small coordinate planes showing a rising line, a falling line, a horizontal line and a vertical line, each labelled with the sign of its slope
The horizontal line has a rise of zero, so its slope is zero over something, which is zero. The vertical line has a run of zero, so its slope would be something over zero, which is not a number at all.
Worked example
Example 3, all four parts. No graphing is needed.
\[ \text{Classify the line through each pair: } (-5,1),(3,1); \; (-6,0),(2,-4); \; (-1,3),(5,8); \; (4,6),(4,-1). \]
First pair: compute the slope
Why: One minus one is zero on top, and three minus negative five is eight on the bottom.
\[ m = \frac{0}{8} = 0 \]
Second pair
Why: Negative four minus zero is negative four; two minus negative six is eight. The slope is negative.
\[ m = -\frac{4}{8} = -\frac{1}{2} \]
Third pair
Why: Eight minus three is five; five minus negative one is six. The slope is positive.
\[ m = \frac{5}{6} \]
Fourth pair
Why: Negative one minus six is negative seven; four minus four is zero. Division by zero is undefined.
\[ m = -\frac{7}{0},\text{ undefined} \]
Read off the four classifications
Why: Zero is horizontal, negative falls, positive rises, undefined is vertical.
Figure (svg): The solution to Worked example classify four lines shown as a ladder of expressions, one row per algebraic move
\[ m = 0; \;\; m = -\tfrac{1}{2}; \;\; m = \tfrac{5}{6}; \;\; m \text{ undefined} \]
Verify: check the two degenerate cases against their coordinates
Why: The horizontal pair shares the y value 1, so the line is level — as a zero rise requires. The vertical pair shares the x value 4, so the line is straight up — as a zero run requires. Reading the shared coordinate is a faster classification than computing the slope at all.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.2 Find Slope and Rate of Change §2.2, pp. 83-83
Sorting
Classify each from its two points, without computing more than you need to.
Sort into buckets
Sort each pair of points by the kind of line through it.
Scanning for a shared coordinate first turns two of these six into one-second answers.
Worked example
Guided Practice 3 and the pattern behind Example 3's first and last parts.
\[ \text{Classify the line through } (0,3),(4,8), \text{ then through } (7,-2),(7,5). \]
First pair: compute the slope
Why: Eight minus three is five; four minus zero is four.
\[ m = \frac{5}{4} \]
Classify it
Why: Positive, so the line rises left to right.
Second pair: look before computing
Why: Both points have x equal to seven, so the run will be zero and the slope undefined.
State the shortcut
Why: Equal y values means horizontal; equal x values means vertical. Neither needs the formula.
Figure (svg): The solution to Worked example two more, and the shortcut shown as a ladder of expressions, one row per algebraic move
\[ m = \tfrac{5}{4} \text{ (rises)}; \qquad \text{undefined (vertical)} \]
Verify: compute the second one anyway
Why: Five minus negative two is seven; seven minus seven is zero. Seven over zero is undefined, confirming the shortcut. Notice this is the same fact that made a vertical line fail the vertical line test in Lesson 2.1 — one input with many outputs, and a run of zero, are the same situation described two ways.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.2 Find Slope and Rate of Change §2.2, pp. 83-83
Trap
\[ \text{The line through } (4,6) \text{ and } (4,-1). \]
Compute the slope and call it zero
Why: A zero appears somewhere in the fraction, so the slope is reported as zero.
\[ m = \frac{-7}{0} = 0 \quad \text{(wrong)} \]
Zero divided by something is zero. Something divided by zero is not a number at all.
\[ \text{The line through } (4,6) \text{ and } (4,-1). \]
Look at WHERE the zero is
Why: Zero on top gives a slope of zero, which is a horizontal line. Zero on the bottom gives an undefined slope, which is a vertical line.
\[ m = \frac{-7}{0} \quad \text{is undefined; the line is vertical} \]
\[ \text{compare } m = \frac{0}{8} = 0 \quad \text{: horizontal} \]
Prediction
Commit before you compute.
Predict first
Which line is steeper: the one through (0,0) and (4,3), or the one through (0,0) and (2,3)?
Correct: The second — it climbs 3 over a run of 2 rather than over a run of 4.
\[ m_1 = \tfrac{3}{4} = 0.75 \qquad m_2 = \tfrac{3}{2} = 1.5 \]
Why: Steepness is the ratio, not the total climb. Both lines rise by three, but one takes four units of run to do it and the other takes two, so the slopes are three quarters and three halves. The second is twice as steep. Comparing total climbs rather than ratios is the usual error, and it is the same mistake as comparing distances travelled instead of speeds.
Edge cases
Take the line through the origin and the point (h, 3), and shrink h.
Discussion prompt
Compute the slope for h equal to 3, then 1, then 0.1, then 0.01. Describe what happens to the slope as h approaches zero, and say what the line looks like in the limit. Why is the slope not simply a very large number at h equal to zero?
Hint: Compute the four slopes before reasoning about the limit.
Answer:
\[ h = 3: \; m = 1 \qquad h = 1: \; m = 3 \qquad h = 0.1: \; m = 30 \qquad h = 0.01: \; m = 300 \]
The slope grows without bound as the run shrinks, and the line stands up ever steeper. At h exactly zero the two points share an x value, the line is vertical, and there is no number the slope could be — not a very large one, because for any candidate you could name, a smaller h gives a larger slope.
Undefined is the honest word. It records that no number works, rather than that the number is infinite.
Comparison
Fill the blanks. These two are swapped more often than any other pair in the chapter.
Comparison matrix
| Feature | Slope is zero | Slope is undefined |
|---|---|---|
| Where the zero sits | in the numerator | in the denominator |
| What the two points share | the same y value | the same x value |
| Direction of the line | horizontal | vertical |
| Is it a function? | yes | no - fails the vertical line test |
The last row ties this back to Lesson 2.1: a vertical line is exactly the graph that fails the vertical line test, and it is exactly the one whose slope is undefined. Those are the same fact.
Section
Section 3
Concept
Two different nonvertical lines are parallel if and only if their slopes are equal. They are perpendicular if and only if their slopes are negative reciprocals, which is the same as saying the slopes multiply to negative one.
\[ \text{parallel: } m_1 = m_2 \qquad \text{perpendicular: } m_1 m_2 = -1 \]
The if-and-only-if runs both ways: equal slopes prove parallel, and parallel proves equal slopes. The same is true of the perpendicular condition.
Figure (svg): Two panels: one showing two parallel lines with equal slopes, one showing two perpendicular lines whose slopes are negative reciprocals
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.2 Find Slope and Rate of Change §2.2, pp. 84-84 — Slopes of Parallel and Perpendicular Lines
Picture it
Example 4 produced one of each.
Figure (svg): Two panels: one showing two parallel lines with equal slopes, one showing two perpendicular lines whose slopes are negative reciprocals
Negative reciprocal means two changes at once: flip the fraction, and change the sign. Doing only one of the two is the standard error.
Worked example
Example 4a. Two slopes, and a product that settles it.
\[ \text{Line 1 through } (-2,2),(0,-1). \text{ Line 2 through } (-4,-1),(2,3). \text{ Parallel, perpendicular, or neither?} \]
Find the slope of line 1
Why: Negative one minus two is negative three; zero minus negative two is two.
\[ m 1 = -\frac{3}{2} \]
Find the slope of line 2
Why: Three minus negative one is four; two minus negative four is six.
\[ m 2 = \frac{4}{6} = \frac{2}{3} \]
Multiply the two slopes
Why: Negative three halves times two thirds is negative one, since the threes and the twos cancel.
\[ m 1 \times m 2 = -1 \]
Conclude
Why: A product of negative one is the perpendicular condition.
Figure (svg): The solution to Worked example perpendicular lines shown as a ladder of expressions, one row per algebraic move
\[ m_1 m_2 = -\tfrac{3}{2} \cdot \tfrac{2}{3} = -1 \;\Longrightarrow\; \text{perpendicular} \]
Verify: check the negative-reciprocal description separately
Why: The reciprocal of negative three halves is negative two thirds, and its negative is positive two thirds — which is exactly m2. Both descriptions of the condition agree, so the verdict does not depend on which form you remember.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.2 Find Slope and Rate of Change §2.2, pp. 84-84
Matching
Flip the fraction and change the sign — both, every time.
Match the pairs
Why: A whole number is a fraction over one, so the perpendicular of 4 is negative one quarter. The pair with slopes -1 and 1 is the familiar pair of diagonals meeting at right angles, and it is the one case where the negative reciprocal is easiest to see. Note that the first two rows are each other's partners, which is what makes the relationship symmetric.
Worked example
Example 4b. Equal slopes, and one thing worth checking afterwards.
\[ \text{Line 1 through } (1,2),(4,-3). \text{ Line 2 through } (-4,3),(-1,-2). \text{ Classify them.} \]
Find the slope of line 1
Why: Negative three minus two is negative five; four minus one is three.
\[ m 1 = -\frac{5}{3} \]
Find the slope of line 2
Why: Negative two minus three is negative five; negative one minus negative four is three.
\[ m 2 = -\frac{5}{3} \]
Compare
Why: The slopes are equal.
\[ m 1 = m 2 \]
Check the lines are actually different
Why: Two points of line 2 do not lie on line 1, so they are distinct lines rather than the same line written twice.
Conclude
Why: Equal slopes and different lines means parallel.
Figure (svg): The solution to Worked example parallel lines shown as a ladder of expressions, one row per algebraic move
\[ m_1 = m_2 = -\tfrac{5}{3} \;\Longrightarrow\; \text{parallel} \]
Verify: confirm the lines really are distinct
Why: Line 1 passes through (1,2). Does line 2? Starting from (-4,3) and moving three right and five down gives (-1,-2), and three right again gives (2,-7) — never (1,2). So the two lines are genuinely different, which the definition of parallel requires. Identical lines have equal slopes too, and are not called parallel.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.2 Find Slope and Rate of Change §2.2, pp. 84-84
Error analysis
A student checks whether two lines are perpendicular.
Annotate
On: \( m_1 = \tfrac{3}{4}, \quad m_2 = \tfrac{4}{3} \;\Longrightarrow\; \text{perpendicular, because the slopes are reciprocals} \)
Use the product test rather than the reciprocal description. Multiplying the two slopes is one step and it cannot be half-applied.
Discrimination
You are given the two slopes. Do not compute anything else.
Sort into buckets
Sort each pair of slopes.
Fill the middle
Guided Practice 11, one slope given.
Fill in the blanks
m_1 = \frac\frac{2 - 1}{-2 - (-5)}___ = -3, \quad m_2 = ___ = \tfrac______ \;\Longrightarrow\; \text___
Why: Two minus one is 1 on top, and negative two minus negative five is 3 on the bottom, giving one third. The product of negative three and one third is negative one, so the lines are perpendicular. Note that negative three is really negative three over one, and its negative reciprocal is one third — the same answer by the other description.
Counterexample
A classmate offers a rule about perpendicular lines.
\[ \text{two lines are perpendicular whenever their slopes have opposite signs} \]
Discussion prompt
Find two lines with opposite-signed slopes that are clearly not perpendicular, then state what the condition actually requires beyond opposite signs.
Hint: Pick a very shallow negative slope against a very shallow positive one.
Answer:
\[ m_1 = 1, \quad m_2 = -0.1 \;\Longrightarrow\; m_1 m_2 = -0.1 \neq -1 \]
Opposite signs are necessary but nowhere near sufficient: one line rises gently and the other falls almost flat, so they cross at a shallow angle, not a right angle.
The condition needs the magnitudes to be reciprocals as well, which is what the product being exactly negative one encodes. Opposite signs alone gets you a negative product, not a product of negative one.
Section
Section 4
Concept
When the two quantities have units, the slope becomes an average rate of change: how much one quantity changes, on average, per unit change in the other. The unit of the rate is the output unit divided by the input unit.
average rate of change — The change in one quantity divided by the change in another. It is a slope carrying units, such as miles per hour or inches per year.
\[ \text{rate} = \frac{\text{change in output}}{\text{change in input}} \]
The word average matters. The rate describes the whole interval and says nothing about what happened inside it.
Figure (svg): A sequoia trunk diameter growing from 137 inches in 1965 to 141 inches in 2005, with the average rate of change computed beside it
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.2 Find Slope and Rate of Change §2.2, pp. 85-85 — Rate of change
Picture it
Example 5: a giant sequoia measured in 1965 and again in 2005.
Figure (svg): A sequoia trunk diameter growing from 137 inches in 1965 to 141 inches in 2005, with the average rate of change computed beside it
Four inches over forty years is a tenth of an inch per year. The units divide exactly as the numbers do, which is the unit analysis of Lesson 1.1 doing familiar work.
Worked example
Example 5, step one. The diameter grew from 137 inches to 141 inches.
\[ \text{Diameter } 137 \text{ in in } 1965 \text{ and } 141 \text{ in in } 2005. \text{ Find the average rate of change.} \]
Compute the change in the output quantity
Why: One hundred and forty-one minus 137 is four inches.
\[ \text{change in diameter } = 4\text{ in} \]
Compute the change in the input quantity
Why: Two thousand and five minus 1965 is forty years.
\[ \text{change in time } = 40\text{ yr} \]
Divide, keeping the units
Why: Four inches over forty years, and inches divided by years gives inches per year.
\[ 4\text{ in } / 40\text{ yr} \]
Simplify
Why: Four fortieths is one tenth.
\[ 0.1\text{ in per year} \]
Figure (svg): The solution to Worked example find the rate of change shown as a ladder of expressions, one row per algebraic move
\[ \text{rate} = \frac{141 - 137}{2005 - 1965} = \frac{4 \text{ in}}{40 \text{ yr}} = 0.1 \; \text{in/yr} \]
Verify: scale the rate back over the interval
Why: A tenth of an inch per year for forty years is four inches, which is exactly the growth observed. The unit also survives: inches per year times years gives inches, which is what a change in diameter should be.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.2 Find Slope and Rate of Change §2.2, pp. 85-85
Matching
The unit of a rate is the output unit over the input unit, always.
Match the pairs
Why: In every case the unit is read straight off the division, with no rule to remember: the quantity on the vertical axis divided by the quantity on the horizontal one. This is why identifying which variable is the input, as Lesson 2.1 insisted, decides the unit of every slope you ever compute.
Worked example
Guided Practice 13. The same two steps with different numbers.
\[ \text{Diameter } 248 \text{ in in } 1965 \text{ and } 251 \text{ in in } 2005. \text{ Find the rate.} \]
Compute the change in diameter
Why: Two hundred and fifty-one minus 248 is three inches.
\[ \text{change } = 3\text{ in} \]
Compute the change in time
Why: The same forty-year interval as before.
\[ \text{change } = 40\text{ yr} \]
Divide
Why: Three over forty is 0.075.
\[ 0.075\text{ in per year} \]
Compare with the first tree
Why: This tree is much thicker but growing more slowly, which the two rates make plain.
\[ 0.075\text{ against } 0.1 \]
Figure (svg): The solution to Worked example a different tree shown as a ladder of expressions, one row per algebraic move
\[ \text{rate} = \frac{3 \text{ in}}{40 \text{ yr}} = 0.075 \; \text{in/yr} \]
Verify: check the size against the other tree
Why: Three inches of growth over the same forty years must give a smaller rate than four inches did, and 0.075 is indeed less than 0.1. Comparing two rates computed over the same interval is a check that needs no arithmetic at all.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.2 Find Slope and Rate of Change §2.2, pp. 85-85
Error analysis
A student computes the growth rate of the sequoia and reports a number without units.
Annotate
On: \( \frac{141 - 137}{2005 - 1965} = \frac{4}{40} = 0.1 \)
A rate without its unit is not an answer to a rate question. The unit is what makes it possible to use the number in the next calculation.
Estimation
A tree grows from 248 inches to 251 inches over forty years.
Predict first
Roughly what is the annual growth rate?
Correct: About 0.075 inch per year.
\[ \frac{3}{40} = 0.075 \quad \text{compare} \quad \frac{4}{40} = 0.1 \]
Why: Three inches spread over forty years must be far less than a tenth of an inch each year, since a tenth for forty years would give four inches. The exact value is 0.075. Estimating first catches the two most common slips here — reporting the total change of 3 as if it were the rate, and slipping the decimal by a factor of ten.
Socratic
One deep question, and nothing else on this slide.
\[ \text{rate} = 0.1 \text{ in/yr over } 1965\text{-}2005 \]
Discussion prompt
The sequoia's average growth rate over forty years is a tenth of an inch per year. Does that mean it grew a tenth of an inch in 1983? What would you need to know to answer that, and what does this tell you about the word average in average rate of change?
Hint: Think about a drought year, or a year with unusually good conditions.
Answer:
No. The average describes the whole forty-year interval and says nothing about any single year within it. The tree might have grown half an inch in one wet year and nothing at all in a drought.
To answer for 1983 you would need measurements close to that year — the shorter the interval, the more local the rate. This is exactly the idea calculus formalises: shrinking the interval until the average rate becomes an instantaneous one.
For a straight line the distinction vanishes, because the rate is genuinely the same everywhere. For anything else, average is a real qualification and not a hedge.
Fill the middle
A car's odometer reads 12,480 miles on Monday and 12,900 miles on Friday.
Fill in the blanks
\text4 = \frac______} = 105 \; \text___
Why: The change in distance is 420 miles, and Monday to Friday is four days of driving, so the rate is 105 miles per day. The subtlety is the four rather than five: the interval between Monday and Friday is four days, even though five days are named — the same off-by-one that Lesson 1.5 met with fence posts.
Section
Section 5
Concept
Once you have an average rate of change, multiplying it by an interval predicts how much the output will change over that interval. Adding that to the last known value predicts the future value.
\[ \text{future} = \text{current} + \text{rate} \times \text{elapsed} \]
The prediction assumes the rate continues. That assumption is the whole strength and the whole weakness of the method, and it should be stated whenever the prediction is.
Figure (svg): A sequoia trunk diameter growing from 137 inches in 1965 to 141 inches in 2005, with the average rate of change computed beside it
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.2 Find Slope and Rate of Change §2.2, pp. 85-85
Picture it
The rate was measured over 1965 to 2005 and is now used beyond 2005.
Figure (svg): A sequoia trunk diameter growing from 137 inches in 1965 to 141 inches in 2005, with the average rate of change computed beside it
Note what the prediction extends: a pattern observed over forty years, projected sixty years forward. That is a longer extrapolation than the data behind it, which is worth saying out loud.
Worked example
Example 5, step two. The rate from step one is now used.
\[ \text{The sequoia was } 141 \text{ in in } 2005 \text{ and grows } 0.1 \text{ in/yr. Predict its diameter in } 2065. \]
Find the number of years elapsed
Why: Two thousand and sixty-five minus 2005 is sixty years.
\[ 60\text{ years} \]
Multiply the rate by the elapsed time
Why: Sixty years times a tenth of an inch per year, and the years cancel to leave inches.
\[ 60 \times 0.1 = 6\text{ in} \]
Add the increase to the last known diameter
Why: One hundred and forty-one plus six.
\[ 141 + 6 \]
State the prediction
Why: About 147 inches, with the assumption that the growth rate holds.
\[ \text{about } 147\text{ in} \]
Figure (svg): The solution to Worked example predict the diameter in 2065 shown as a ladder of expressions, one row per algebraic move
\[ 141 + (60)(0.1) = 147 \text{ inches} \]
Verify: check the unit cancellation and the plausibility
Why: Years times inches per year gives inches, so the six really is a change in diameter. The size is plausible too: the tree grew four inches in the previous forty years, so six inches in sixty years continues the same pace rather than assuming a sudden change.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.2 Find Slope and Rate of Change §2.2, pp. 85-85
Real world
A town's population was 8,400 in 2010 and 10,200 in 2020.
Discussion prompt
Find the average rate of change with its units, predict the population in 2035, and then say what has to remain true for that prediction to be worth anything. Give one concrete thing that could make it badly wrong.
Hint: Compute the rate first, then the elapsed time from the LATER measurement.
Answer:
\[ \text{rate} = \frac{10200 - 8400}{2020 - 2010} = 180 \text{ people per year} \]
\[ 10200 + (15)(180) = 12\,900 \text{ people in } 2035 \]
The prediction assumes the town keeps growing at a constant 180 people a year for fifteen more years. A new factory, a closed factory, or a housing development would break it, and so would the simple fact that towns often grow proportionally rather than by a fixed number — which is Chapter 7's exponential model rather than this one.
Worked example
Guided Practice 13, second part. The interval is now a full century.
\[ \text{The tree was } 251 \text{ in in } 2005 \text{ and grows } 0.075 \text{ in/yr. Predict its diameter in } 2105. \]
Find the elapsed time
Why: Twenty-one hundred and five minus 2005 is one hundred years.
\[ 100\text{ years} \]
Multiply the rate by the elapsed time
Why: A hundred years times 0.075 inch per year is 7.5 inches.
\[ 100 \times 0.075 = 7.5\text{ in} \]
Add to the last known value
Why: Two hundred and fifty-one plus 7.5.
\[ 251 + 7.5 \]
State the prediction and its assumption
Why: About 258.5 inches, assuming the growth rate is unchanged for a century.
\[ \text{about } 258.5\text{ in} \]
Figure (svg): The solution to Worked example a longer prediction shown as a ladder of expressions, one row per algebraic move
\[ 251 + (100)(0.075) = 258.5 \text{ inches} \]
Verify: compare the prediction interval with the data interval
Why: The rate came from forty years of data and is being projected a hundred years forward — two and a half times as far as the evidence reaches. The arithmetic is exact, but the confidence should not be, and saying so is part of a complete answer.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.2 Find Slope and Rate of Change §2.2, pp. 85-85
Error analysis
A student predicts the sequoia's diameter in 2065.
Annotate
On: \( 141 + (2065 - 1965)(0.1) = 141 + 10 = 151 \text{ inches} \)
Always pair the starting value with the year it was measured. Mixing a later measurement with an earlier start double-counts the interval between them.
Ranking
Predicting a future value from two measurements.
Put in order
Why: The two subtractions can happen in either order but both must precede the division. What cannot move is step four: the elapsed time must be measured from the later of the two known values, because that value already contains all the growth up to its own date. Adding growth measured from the earlier date would count the middle interval twice.
Commit first
Answer, then rate your confidence honestly.
Predict first
You compute a rate from two measurements forty years apart and use it to predict two hundred years ahead. Is the arithmetic wrong?
Correct: No — the arithmetic is fine, but the assumption behind it may not be.
The professional habit is to state both: here is the number, and here is what it assumes. Chapter 11 will give you tools for saying how much confidence a prediction deserves.
Why: Multiplying a rate by an interval is exact arithmetic however long the interval. What degrades is the modelling assumption that the rate stays constant, and it degrades with distance from the data. Separating these two things — the calculation, which is either right or wrong, and the assumption, which is either reasonable or not — is what lets you state a prediction honestly instead of either over-trusting or refusing to make one.
Reverse engineer
A prediction was made and the working has been lost.
Fill in the blanks
\text0.1 = 141 + 60 \cdot ___ = 147
Why: The predicted increase is six inches over sixty years, so the rate is a tenth of an inch per year. Working backwards from a prediction to the rate that produced it is a useful check on somebody else's claim: if the implied rate is wildly different from what the data supports, the prediction is not built on the measurements it cites.
Comparison
Fill the blanks. The same ratio does four different jobs.
Comparison matrix
| Situation | What the slope is | What its sign or value tells you |
|---|---|---|
| A ramp with rise and run | rise divided by run | how steep the climb is |
| Two points on a plane | (y2 - y1) / (x2 - x1) | whether the line rises or falls |
| Two lines compared | each line's own slope | equal means parallel; product -1 means perpendicular |
| Two quantities with units | change in output over change in input | the average rate of change |
| A vertical line | run is zero | undefined, not zero |
The last row is the only one where the answer is not a number, and it is the one worth over-learning.
Pattern
One routine covers every slope question in this lesson.
Step one turns two of the four classification cases into one-second answers and removes the risk of dividing by zero without noticing.
OpenStax Algebra and Trigonometry 2e, §3.3 Rates of Change and Behavior of Graphs §3.3
Check
The formula, with negatives in the coordinates.
Check your understanding
What is the slope of the line through (-4, 9) and (-8, 3)?
Answer: A
Why: Three minus nine is -6 on top, and -8 minus -4 is -4 on the bottom. A negative divided by a negative is positive, giving 3/2. A sketch confirms it: moving from (-8,3) to (-4,9) goes right and up.
Check
Classifying two lines. Compute both slopes first.
Check your understanding
Line 1 passes through (-2, 8) and (2, -4). Line 2 passes through (-5, 1) and (-2, 2). Are they parallel, perpendicular, or neither?
Answer: A
Why: Line 1 has slope -12/4, which is -3. Line 2 has slope 1/3. Their product is -1, so the lines meet at a right angle.
Check
A rate of change and a prediction. Watch which year you start from.
Check your understanding
A sequoia's diameter was 248 inches in 1965 and 251 inches in 2005. Predict its diameter in 2105.
Answer: A
Why: The rate is 3 inches over 40 years, or 0.075 inch per year. From 2005 to 2105 is 100 years, giving 7.5 inches of growth, and 251 plus 7.5 is 258.5 inches.
Real world
A wheelchair ramp must have a slope no steeper than 1 in 12 to meet accessibility guidance. You need to reach a doorway 30 inches above the pavement.
Discussion prompt
What is the shortest run the ramp may have? Then say what happens to the required run if the doorway is twice as high, and why that relationship is a consequence of slope being a ratio.
Hint: Set up the slope as rise over run and compare it with the limit.
Answer:
\[ \frac{30}{\text{run}} \leq \frac{1}{12} \;\Longrightarrow\; \text{run} \geq 360 \text{ inches, or } 30 \text{ feet} \]
Thirty feet of ramp for two and a half feet of rise. Doubling the height to sixty inches doubles the required run to sixty feet, because the ratio must stay fixed and the rise is on top.
That proportionality is why accessible entrances need so much space, and why switchback ramps exist: the run cannot be shortened without breaking the ratio, so it gets folded instead.
Commit first
Answer, then rate your confidence honestly.
Predict first
A horizontal line and a vertical line are perpendicular. Does their pair of slopes satisfy the product test?
Correct: No — the test cannot be applied, because one slope is not a number.
This is a good example of a formula having a domain. The product test is stated for two nonvertical lines, and reading that condition rather than skipping it is what keeps you out of trouble.
Why: They genuinely are perpendicular; that is a geometric fact about the axes. But a vertical line has no slope at all, so there is nothing to put into the product, and zero times undefined is not an arithmetic statement. This is the single exception to the product test, and the honest response is to fall back on geometry: a horizontal and a vertical line meet at a right angle by definition.
Explain it
They can plot points but have never computed a slope.
Discussion prompt
In four sentences or fewer, explain what slope measures, how to compute it from two points, and the one bookkeeping rule that prevents most sign errors. Give them a way to check their sign without the formula.
Hint: The check is one sketch.
Answer:
Slope measures how much a line climbs for each unit it travels sideways: the vertical change divided by the horizontal change. To compute it, pick one point as first and the other as second, then subtract second minus first on the top and on the bottom.
The bookkeeping rule is that the same point must be first in both subtractions. To check the sign without the formula, sketch the two points: if the line goes up as you read left to right, the slope is positive.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For order, write the labels next to the points before you subtract anything. For zero against undefined, ask where the zero sits — on top gives zero, on the bottom gives undefined. For negative reciprocals, do both operations and then check the product is negative one. For units, write the output unit over the input unit as part of the division. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Draw one coordinate plane in the middle of a page with a line through two points you choose, and draw the right triangle between them with its legs labelled rise and run. Write the slope formula beside it and compute the slope both ways, once with each point labelled first, showing that the answer is the same. Around the outside of the page, sketch four small lines showing a positive slope, a negative slope, a zero slope and an undefined slope, writing beside each what the two points would share. In one corner draw two parallel lines and write their equal slopes; in another corner draw two perpendicular lines and write the product of their slopes. At the bottom, take two real measurements you can find — anything with a before and an after — compute the rate of change with its units, and write one sentence predicting a future value together with the assumption that prediction rests on.
If the assumption sentence was hard to write, reread Section 5. A prediction without its assumption stated is a number pretending to be a fact.
Recap
Five things, and the last two are the reason slope appears outside mathematics at all.
| If you see | Then |
|---|---|
| Two points sharing a y value | Horizontal line, slope zero |
| Two points sharing an x value | Vertical line, slope undefined |
| Two equal slopes | Parallel lines |
| Two slopes multiplying to -1 | Perpendicular lines |
| Quantities with units | The slope is a rate; attach the units |
Lesson 2.3 puts the slope and one point together to draw the line, and shows why the constant term of an equation is where the line crosses the vertical axis.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.2 Find Slope and Rate of Change §2.2, pp. 82-87 — everything on these slides traces back here
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