The exponential parent function for a base between zero and one, the coefficient a and the y-intercept, translations and asymptotes, decay models built from a percent decrease, and the reflection relationship linking growth to decay.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 7 — Exponential and Logarithmic Functions
Graph Exponential Decay Functions
Objectives
Five outcomes. Everything is Lesson 7.1 with one condition changed.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.2 Graph Exponential Decay Functions §7.2, pp. 486-491 — the lesson these objectives are drawn from
Warm-up
Lesson 7.1 required a base greater than 1. This lesson asks what happens below it.
Discussion prompt
Tabulate y equals one half to the x at x equal to negative 3 through 2. Which way does the curve go, and why?
Hint: One half to a negative power is a reciprocal.
Answer:
\[ x: \; -3, -2, -1, 0, 1, 2 \qquad y: \; 8, 4, 2, 1, \tfrac{1}{2}, \tfrac{1}{4} \]
The values halve at each step to the right instead of doubling, so the curve falls rather than rises. Everything else — the point where y is 1, the asymptote along the horizontal axis, the domain and the range — is exactly as before.
Concept
An exponential function with a positive coefficient and a base between zero and one is an exponential decay function. Multiplying by a number less than 1 at each step makes the quantity shrink, so the graph falls to the right and approaches the horizontal axis there instead of on the left.
exponential decay function — A function y equals a times b to the x with a greater than zero and b between 0 and 1. The base b is called the decay factor.
\[ y = ab^x, \quad a > 0, \; 0 < b < 1 \]
The transformations, the asymptote rule and the modelling procedure are all unchanged from Lesson 7.1. Only the direction of the curve and the sign inside the factor differ.
Figure (svg): Two columns separating a growth factor from a decay factor
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.2 Graph Exponential Decay Functions §7.2, pp. 486-486
Section
Section 1
Concept
For a base between zero and one, the function b to the x falls from left to right, passing through the point where x is 0 and y is 1 and through the point where x is 1 and y is b. Its domain and range are unchanged from the growth case.
\[ f(x) = b^x, \quad 0 < b < 1 \]
The horizontal axis is still an asymptote, but the curve now approaches it going right rather than left, because the shrinking happens as x increases.
Figure (svg): The graph of one half to the x with its table of values and its horizontal asymptote
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.2 Graph Exponential Decay Functions §7.2, pp. 486-486 — Parent Function for Exponential Decay Functions
Picture it
Example 1: a table from negative 3 to 2, and the falling curve.
Figure (svg): The graph of one half to the x with its table of values and its horizontal asymptote
The point where y is 1 is shared with every exponential graph, and the point at x equal to 1 identifies the base. Both facts carry over from Lesson 7.1 unchanged.
Worked example
Example 1, in the book's three steps.
\[ \text{Graph } y = \left(\tfrac{1}{2}\right)^x. \]
Make a table
Why: From negative 3 to 2 the outputs are 8, 4, 2, 1, one half and one quarter.
Plot them
Why: The values fall steeply on the left and flatten on the right.
Draw the curve
Why: From right to left it begins just above the horizontal axis and rises to the left.
State the domain, range and asymptote
Why: Every real number is allowed and every output is positive.
\[ y > 0,\text{ asymptote } y = 0 \]
Figure (svg): The solution to Worked example graph the decay parent shown as a ladder of expressions, one row per algebraic move
\[ y = \left(\tfrac{1}{2}\right)^x, \quad y > 0 \]
Verify: check what happens far to the right
Why: At x equal to 10 the value is 1 over 1024, tiny but positive; at x equal to 100 it is smaller still and still positive. The asymptote is approached on the right in the decay case and on the left in the growth case, which is the only difference between the two pictures.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.2 Graph Exponential Decay Functions §7.2, pp. 486-486
Sorting
Compare the base with 1.
Sort into buckets
Sort each function, assuming a positive coefficient.
Only the base decides, and the comparison is always with 1 rather than with zero. A decimal such as 0.85 is a decay base just as three fifths is.
Worked example
Guided Practice 1, plus a comparison.
\[ \text{Graph } y = \left(\tfrac{2}{3}\right)^x \text{ and compare it with } y = \left(\tfrac{1}{2}\right)^x. \]
Tabulate the first
Why: At x equal to negative 1, 0, 1, 2 the outputs are three halves, 1, two thirds and four ninths.
Note the shared point
Why: Both curves pass through the point where x is 0 and y is 1.
\[ (0, 1)\text{ on both} \]
Compare to the right
Why: At x equal to 1 the first gives two thirds and the second one half.
\[ \frac{2}{3}\text{ falls more slowly} \]
State the domain and range
Why: Unchanged for both.
Figure (svg): The solution to Worked example two more decay bases shown as a ladder of expressions, one row per algebraic move
\[ \left(\tfrac{2}{3}\right)^x \text{ decays more slowly than } \left(\tfrac{1}{2}\right)^x \]
Verify: check the ordering at a large input
Why: At x equal to 5, two thirds to the fifth is about 0.132 while one half to the fifth is 0.031 — more than four times smaller. A base closer to 1 means slower decay, exactly as a base closer to 1 from above means slower growth.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.2 Graph Exponential Decay Functions §7.2, pp. 487-487
Trap
\[ y = \left(\tfrac{1}{2}\right)^x \]
Take a base below 1 to mean negative outputs
Why: Small base, small values, so the curve is drawn below the axis.
\[ y < 0 \text{ for large } x \quad \text{(wrong)} \]
One half to the tenth is 1 over 1024, which is small but firmly positive. No power of a positive number is ever negative.
\[ \left(\tfrac{1}{2}\right)^x > 0 \text{ for every real } x \]
Keep the range as the positive numbers, whatever the base
Why: The base being small changes how fast the curve falls, not which side of the axis it sits on.
\[ \text{range } y > 0 \text{ for both growth and decay} \]
Only the sign of a can put a graph below the axis, and that makes it neither a growth nor a decay function by the textbook's definitions.
Fill the middle
Example 1, at x equal to negative 2.
Fill in the blanks
\left(\tfrac4___\right)^___ = \left(\tfrac______\right)^___ = ___
Why: A negative exponent inverts the base, so one half to the negative 2 is 2 squared, which is 4. That is why the decay curve rises on the left: negative inputs turn the small base into a large one.
Matching
Every graph passes through (1, b), growth or decay.
Match the pairs
Why: The rule is identical to the growth case: plot the point where y is 1, then the point at x equal to 1 whose height is the base. Since every base here is below 1, the second point is lower than the first and the curve falls.
Prediction
Commit before reasoning.
Predict first
The definition excludes b equal to 1. Why?
Correct: Because 1 to the x is always 1, giving a horizontal line.
\[ 1^x = 1 \text{ for every } x \;\Longrightarrow\; \text{a horizontal line} \]
Why: One raised to any power is 1, so the function would be the constant function 1 — a line, not an exponential curve, and neither growing nor decaying. The base is also required to be positive, since a negative base raised to a fractional exponent often has no real value. Both exclusions in the definition are there to keep the family well behaved.
Section
Section 2
Concept
The graph of a times b to the x is a vertical stretch or shrink of b to the x with its y-intercept at a. A negative a puts the curve below the horizontal axis, and the function is then not an exponential decay function.
\[ y = ab^x \;\Longrightarrow\; \text{y-intercept } (0, a) \]
The two conditions in the definition are independent: a must be positive and b must lie strictly between 0 and 1. Failing either one disqualifies the function.
Figure (svg): Two decay functions with different coefficients, one positive and one negative
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.2 Graph Exponential Decay Functions §7.2, pp. 487-487 — Graph y = ab^x for 0 < b < 1
Picture it
Example 2: 2 times one quarter to the x, and negative 3 times two fifths to the x.
Figure (svg): Two decay functions with different coefficients, one positive and one negative
The first falls toward the axis from above and the second rises toward it from below. Only the first is a decay function.
Worked example
Example 2, both parts.
\[ \text{Graph } y = 2\left(\tfrac{1}{4}\right)^x \text{ and } y = -3\left(\tfrac{2}{5}\right)^x. \]
First: find the anchor points
Why: At x equal to 0 the value is a, which is 2; at x equal to 1 it is a times b, which is one half.
\[ (0, 2)\text{ and } (1, \frac{1}{2}) \]
First: draw the curve
Why: From right to left it begins just above the axis and rises to the left.
Second: find the anchor points
Why: At x equal to 0 the value is negative 3; at x equal to 1 it is negative six fifths.
\[ (0, -3)\text{ and } (1, -\frac{6}{5}) \]
Second: draw and classify
Why: The curve lies below the axis and rises toward it; since a is negative it is not a decay function.
Figure (svg): The solution to Worked example two coefficients shown as a ladder of expressions, one row per algebraic move
\[ (0, 2), (1, \tfrac{1}{2}); \qquad (0, -3), (1, -\tfrac{6}{5}) \]
Verify: check the second's direction
Why: Negative 3 times two fifths is negative 1.2, which is above negative 3 on the number line — so the curve is rising as x increases, even though the base is a decay base. Multiplying by a fraction moves a negative number toward zero, which is upward. The sign of a reverses the visual direction as well as the side.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.2 Graph Exponential Decay Functions §7.2, pp. 487-487
Sorting
Check the sign of a and the size of b.
Sort into buckets
Sort each function.
Every base here is a decay base, so once again the sign of the coefficient is doing all the sorting.
Worked example
Guided Practice 2 and 3.
\[ \text{Graph } y = -2\left(\tfrac{3}{4}\right)^x \text{ and } f(x) = 4\left(\tfrac{1}{5}\right)^x. \]
First: anchor points
Why: At x equal to 0 the value is negative 2; at x equal to 1 it is negative three halves.
\[ (0, -2), (1, -1.5) \]
First: classify
Why: The coefficient is negative, so this is not a decay function.
Second: anchor points
Why: At x equal to 0 the value is 4; at x equal to 1 it is four fifths.
\[ (0, 4), (1, 0.8) \]
Second: classify
Why: Positive coefficient and a base between 0 and 1, so it is a decay function.
Figure (svg): The solution to Worked example two more shown as a ladder of expressions, one row per algebraic move
\[ -2\left(\tfrac{3}{4}\right)^x: \text{ no}; \qquad 4\left(\tfrac{1}{5}\right)^x: \text{ yes} \]
Verify: compare the two decay rates
Why: The second has base one fifth and drops from 4 to 0.8 in one step, a loss of 80 percent. A base of three quarters would lose only 25 percent per step. The smaller the base, the faster the decay — and a base very close to 1 decays very slowly indeed.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.2 Graph Exponential Decay Functions §7.2, pp. 487-487
Error analysis
A student classifies an exponential function.
Annotate
On: \( y = -3\left(\tfrac{2}{5}\right)^x: \text{ the base } \tfrac{2}{5} \text{ is between } 0 \text{ and } 1, \text{ so it is exponential decay} \)
Two conditions, both required. The textbook flags this exact case in a Classify Functions note, as it did for growth in Lesson 7.1.
Fill the middle
Example 2a.
Fill in the blanks
y = 2\left(\tfrac1/2___\right)^x: \; \text___ x=1, \; y = 2\cdot\tfrac______ = ___
Why: Two times one quarter is one half, so the second anchor point sits at half the height. In general the two anchor points are at heights a and a times b, and for decay the second is always lower than the first.
Comparison
Fill the blanks. One condition apart.
Comparison matrix
| Question | Growth, Lesson 7.1 | Decay, here |
|---|---|---|
| Base condition | b > 1 | 0 < b < 1 |
| Coefficient condition | a > 0 | a > 0 |
| Direction | rises to the right | falls to the right |
| Where the asymptote is approached | on the left | on the right |
The middle row is the same in both, which is why a negative coefficient disqualifies a function from either family.
Prediction
Commit before reasoning.
Predict first
Among bases one fifth, one half and three quarters, which gives the fastest decay?
Correct: One fifth, since it is smallest.
\[ \left(\tfrac{1}{5}\right)^3 = 0.008, \; \left(\tfrac{1}{2}\right)^3 = 0.125, \; \left(\tfrac{3}{4}\right)^3 \approx 0.42 \]
Why: Each step multiplies by the base, so a smaller base removes more at every step: one fifth keeps 20 percent, one half keeps 50 percent and three quarters keeps 75 percent. The closer the base is to 1 the slower the decay, which mirrors the growth case where a base close to 1 grows slowly.
Section
Section 3
Concept
To graph a times b to the x minus h, plus k, sketch a times b to the x and shift it h units horizontally and k units vertically. The asymptote moves to the line y equals k and the range becomes everything above it for a positive a.
\[ y = ab^{x-h}+k \;\Longrightarrow\; \text{asymptote } y = k \]
The procedure is word for word the one from Lesson 7.1. Only the direction the curve falls has changed, and translations do not affect that.
Figure (svg): A decay function translated left and down, with its asymptote at y equals negative two
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.2 Graph Exponential Decay Functions §7.2, pp. 487-487 — Graph y = ab^(x-h) + k for 0 < b < 1
Picture it
Example 3: 3 times one half to the x plus 1, minus 2.
Figure (svg): A decay function translated left and down, with its asymptote at y equals negative two
The anchor points moved left 1 and down 2, and so did the asymptote — from the horizontal axis to the line two units below it.
Worked example
Example 3, with domain and range.
\[ \text{Graph } y = 3\left(\tfrac{1}{2}\right)^{x+1}-2 \text{ and state the domain and range.} \]
Sketch the untranslated version
Why: Three times one half to the x passes through the points at heights 3 and three halves.
\[ (0, 3)\text{ and } (1, \frac{3}{2}) \]
Read h and k
Why: X plus 1 gives h equal to negative 1; the constant outside is negative 2.
\[ h = -1, k = -2 \]
Shift the graph
Why: Left 1 and down 2, carrying the anchor points.
\[ (-1, 1)\text{ and } (0, -\frac{1}{2}) \]
State the asymptote, domain and range
Why: The asymptote moves to the line y equals negative 2.
Figure (svg): The solution to Worked example translate a decay function shown as a ladder of expressions, one row per algebraic move
\[ y > -2, \quad \text{asymptote } y = -2 \]
Verify: check a shifted anchor point
Why: The point at height 3 should move to x equal to negative 1 and y equal to 1. Substituting x equal to negative 1 gives 3 times one half to the zero, minus 2, which is 3 minus 2, or 1. The shift was applied correctly.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.2 Graph Exponential Decay Functions §7.2, pp. 487-487
Matching
The asymptote is the line y equals k.
Match the pairs
Why: Two of these share an asymptote despite having different bases, coefficients and horizontal shifts — because only k matters. The last has no added constant, so k is zero and the asymptote is the horizontal axis.
Worked example
Guided Practice 4 and 5.
\[ \text{Graph } y = \left(\tfrac{1}{4}\right)^{x-1}+1 \text{ and } y = 5\left(\tfrac{2}{3}\right)^{x+1}-2. \]
First: read h and k
Why: X minus 1 gives h equal to 1, and the constant outside is 1.
\[ h = 1, k = 1 \]
First: state the asymptote and range
Why: The asymptote moves up to the line y equals 1.
\[ y > 1 \]
Second: read h and k
Why: X plus 1 gives h equal to negative 1, and the constant is negative 2.
\[ h = -1, k = -2 \]
Second: state the asymptote and range
Why: The asymptote is the line y equals negative 2.
\[ y > -2 \]
Figure (svg): The solution to Worked example two more translations shown as a ladder of expressions, one row per algebraic move
\[ y > 1; \qquad y > -2 \]
Verify: check the first at its shifted anchor
Why: At x equal to 1 the exponent is zero, so the value is 1 plus 1, which is 2 — the point where the parent's y-intercept of 1 has landed. Every translated exponential graph has that point at the coordinates h and a plus k.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.2 Graph Exponential Decay Functions §7.2, pp. 488-488
Trap
\[ y = 3\left(\tfrac{1}{2}\right)^{x+1}-2 \]
Take the left shift to reverse the fall
Why: Shifting left is treated as flipping the graph.
\[ \text{the curve now rises to the right} \quad \text{(wrong)} \]
At x equal to 0 the value is negative one half and at x equal to 1 it is negative one and a quarter — still falling.
\[ \text{translations move a graph; they never flip it} \]
Keep the shape and move it
Why: Only a negative coefficient reverses direction, and h and k do not touch the coefficient.
\[ \text{base } \tfrac{1}{2} < 1 \;\Longrightarrow\; \text{falls to the right, wherever it sits} \]
The direction is decided by the base and the sign of a alone. Both are read before any shifting is considered.
Fill the middle
Example 3.
Fill in the blanks
3\left(\tfrac-2___\right)^___ > 0 \;\Longrightarrow\; 3\left(\tfrac______\right)^___-2 > ___
Why: The exponential part is always positive, so subtracting 2 always leaves something above negative 2. The range follows from an inequality rather than from the picture, which makes it easy to state without drawing anything.
Sorting
The sign of a decides.
Sort into buckets
Sort each function by where its graph lies relative to its asymptote.
Identical to the growth case and to the radical functions of Lesson 6.5: k is the boundary and a says which side.
Prediction
Commit before reasoning.
Predict first
Two functions have k equal to 3 but bases of one half and 5. Where are their asymptotes?
Correct: Both at y equals 3.
\[ ab^{x-h} \to 0 \;\Longrightarrow\; ab^{x-h}+k \to k \]
Why: The exponential part shrinks toward zero — on the right for a decay base and on the left for a growth base — so both functions approach k. The base decides which end the flattening happens at and how quickly, but not which line is approached. Only k determines the asymptote.
Section
Section 4
Concept
When a quantity decreases by the same percent each period, its amount after t periods is the initial amount times the decay factor raised to t, where the decay factor is 1 minus the percent written as a decimal.
decay factor — The base of an exponential decay function. For a percent decrease r written as a decimal, the decay factor is 1 minus r.
\[ y = a(1-r)^t \]
The percent decrease says how much is lost each period; the decay factor says what fraction remains. A 10 percent loss and a factor of 0.90 are the same statement.
Figure (svg): A depreciation model for a snowmobile, with the value after three years and the year it reaches twenty-five hundred dollars
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.2 Graph Exponential Decay Functions §7.2, pp. 488-488 — Exponential decay models
Picture it
Example 4: 4200 dollars losing 10 percent of its value each year.
Figure (svg): A depreciation model for a snowmobile, with the value after three years and the year it reaches twenty-five hundred dollars
After three years the value is 3061.80 dollars, and the graph reaches 2500 dollars after about five years. The curve flattens because each year's loss is 10 percent of a smaller amount.
Worked example
Example 4, in the book's three steps.
\[ \text{A } 4200 \text{ dollar snowmobile loses } 10 \text{ percent of its value a year. Model } y \text{ and find the value after } 3 \text{ years.} \]
Identify a and r
Why: The initial amount is 4200 and the percent decrease as a decimal is 0.10.
\[ a = 4200, r = 0.10 \]
Write the model
Why: The decay factor is 1 minus 0.10, which is 0.90.
\[ y = 4200(0.90) ^{t} \]
Evaluate at three years
Why: Point nine cubed is 0.729.
\[ 4200(0.729) \]
Compute
Why: The value is 3061.80 dollars.
\[ 3061.80\text{ dollars} \]
Figure (svg): The solution to Worked example build and use a decay model shown as a ladder of expressions, one row per algebraic move
\[ y = 4200(0.90)^t, \quad y(3) = 3061.80 \]
Verify: check the first year by hand
Why: Ten percent of 4200 is 420, so after one year the value should be 3780 — and the model gives 4200 times 0.9, which is 3780. After the second year the loss is 10 percent of 3780, which is 378, not 420 again: each year's loss is smaller, which is why the graph flattens.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.2 Graph Exponential Decay Functions §7.2, pp. 488-488
Matching
Subtract the decimal from 1.
Match the pairs
Why: The last is the clearest check: losing half leaves half, so the factor is 0.5. The factor and the percent always add to 1, which is what makes the arithmetic a subtraction rather than anything more complicated.
Worked example
Example 4, step 3, plus Exercise 1's model.
\[ \text{Estimate when the snowmobile is worth } 2500 \text{ dollars, and describe } y = 1250(0.85)^t. \]
Use the graph
Why: The curve crosses the line at 2500 dollars at about t equal to 5.
\[ \text{about } 5\text{ years} \]
Confirm with the model
Why: Point nine to the fifth is about 0.590, and 4200 times that is about 2480.
\[ \text{close to } 2500 \]
Read the second model's initial amount
Why: The coefficient is the value at t equal to zero.
\[ a = 1250 \]
Read its decay factor and percent decrease
Why: The base 0.85 is the decay factor, so r is 0.15.
\[ 15 \% \]
Figure (svg): The solution to Worked example read a time from the graph shown as a ladder of expressions, one row per algebraic move
\[ 5 \text{ years}; \quad 1250, \; 0.85, \; 15\% \]
Verify: separate the decay factor from the percent
Why: The decay factor 0.85 and the percent decrease 15 are complements: what remains plus what is lost is the whole. Reporting 0.85 as the percent decrease would claim an 85 percent loss, which would leave almost nothing after two years rather than most of it.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.2 Graph Exponential Decay Functions §7.2, pp. 488-489
Error analysis
A student writes a decay model for a 10 percent annual loss.
Annotate
On: \( y = 4200(0.10)^t \)
Test the model after one step: it should give the original minus 10 percent of it. The textbook flags this in an Avoid Errors note beside Example 4.
Fill the middle
Example 4, at t equal to 3.
Fill in the blanks
y = 4200(0.90)^3 = 4200 \times 0.729 = 3061.80
Why: Point nine cubed is 0.729, so nearly three quarters of the value remains after three years. Losing 10 percent three times leaves 72.9 percent rather than 70 percent, because each loss is taken from a smaller amount.
Prediction
Commit before reasoning.
Predict first
At 10 percent depreciation a year, when is the snowmobile worth nothing?
Correct: Never, according to the model.
\[ 4200(0.9)^{50} \approx 22 \text{ dollars, still not zero} \]
Why: Each year the value is multiplied by 0.9, and multiplying a positive number by 0.9 always leaves something positive. The graph approaches the horizontal axis asymptotically without reaching it. That is a limitation of the model rather than a fact about snowmobiles: in reality a vehicle eventually reaches scrap value, and no exponential model captures that.
Comparison
Fill the blanks. One sign apart.
Comparison matrix
| Question | Growth, 7.1 | Decay, here |
|---|---|---|
| The model | y = a(1 + r)^t | y = a(1 - r)^t |
| The base | greater than 1 | between 0 and 1 |
| A 10 percent change gives | 1.10 | 0.90 |
| The graph | rises to the right | falls to the right |
One formula with a plus or a minus, and everything else about the two situations follows from which sign was used.
Section
Section 5
Concept
Because one half to the x equals 2 to the negative x, a decay function with base b is the same as a growth function with base 1 over b and the input negated. The two graphs are reflections of each other across the vertical axis.
\[ \left(\tfrac{1}{b}\right)^x = b^{-x} \]
That is why the two lessons need only one set of rules. Everything about transformations, asymptotes, domains and models is shared, and only the direction differs.
Figure (svg): A growth curve and a decay curve on the same axes, mirror images across the vertical axis
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.2 Graph Exponential Decay Functions §7.2, pp. 486-489 — Exercise 2
Picture it
The growth and decay parents on the same axes.
Figure (svg): A growth curve and a decay curve on the same axes, mirror images across the vertical axis
They meet at the point where y is 1 and are mirror images either side of the vertical axis. Each is the other read backwards.
Worked example
The reciprocal identity, applied.
\[ \text{Show that } y = \left(\tfrac{1}{2}\right)^x \text{ and } y = 2^{-x} \text{ are the same function.} \]
Write the fraction as a negative power
Why: One half is 2 to the negative 1, by Lesson 5.1.
\[ \frac{1}{2} = 2 ^{-1} \]
Apply the power of a power rule
Why: Two to the negative 1, all to the x, multiplies the exponents.
\[ (2 ^{-1}) ^{x} = 2 ^{-x} \]
Check a value
Why: At x equal to 3 both give one eighth.
Interpret geometrically
Why: Replacing x by negative x reflects a graph across the vertical axis.
Figure (svg): The solution to Worked example rewrite a decay function as a growth function shown as a ladder of expressions, one row per algebraic move
\[ \left(\tfrac{1}{2}\right)^x = 2^{-x} \]
Verify: check a second value
Why: At x equal to negative 2 both give 4: one half to the negative 2 is 2 squared, and 2 to the positive 2 is also 4. The identity holds at every input, which is what it means for two expressions to define the same function.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.2 Graph Exponential Decay Functions §7.2, pp. 486-486
Matching
Take the reciprocal and negate the exponent.
Match the pairs
Why: Each pair is the same function written two ways, since a reciprocal base is a negative exponent. The third shows that the rule works for any fraction, not only for unit fractions: the reciprocal of two thirds is three halves.
Worked example
Exercises 3 to 6, plus two of the guided practice items.
\[ \text{Classify } 3\left(\tfrac{3}{4}\right)^x, \; 4\left(\tfrac{5}{2}\right)^x, \; \tfrac{2}{7}\cdot 4^x, \; -5(0.25)^x, \; 4\left(\tfrac{1}{5}\right)^x, \; -2\left(\tfrac{3}{4}\right)^x. \]
Check each coefficient first
Why: Four are positive and two are negative.
\[ a > 0\text{ for four} \]
Check each base against 1
Why: Three quarters, one quarter, one fifth and three quarters are below 1; five halves and 4 are above.
\[ \text{compare with } 1 \]
Classify the four with positive coefficients
Why: Decay, growth, growth, decay in order.
Classify the two with negative coefficients
Why: Neither is a growth or a decay function, whatever its base.
Figure (svg): The solution to Worked example classify six functions shown as a ladder of expressions, one row per algebraic move
\[ \text{decay, growth, growth, neither, decay, neither} \]
Verify: check the trickiest one
Why: The function with coefficient two sevenths and base 4 has a small coefficient, which might suggest decay — but the coefficient only scales the graph vertically. The base 4 exceeds 1, so the function grows, from two sevenths to eight sevenths to thirty-two sevenths. Only the base decides the direction.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.2 Graph Exponential Decay Functions §7.2, pp. 489-489
Trap
\[ y = \tfrac{2}{7}\cdot 4^x \]
Call it decay because the coefficient is a small fraction
Why: The two sevenths is taken as the shrinking factor.
\[ \text{exponential decay} \quad \text{(wrong)} \]
The values are two sevenths, eight sevenths, thirty-two sevenths — quadrupling every step. This grows very fast.
\[ \text{base } 4 > 1 \;\Longrightarrow\; \text{growth} \]
Read the base for the direction and the coefficient for the starting height
Why: They answer two different questions and neither substitutes for the other.
\[ a \text{ sets the y-intercept}; \; b \text{ sets the direction} \]
A small coefficient with a large base starts low and climbs fast. A large coefficient with a small base starts high and falls. The two readings are independent.
Fill the middle
The reciprocal identity.
Fill in the blanks
\left(\tfrac-x___\right)^x = (3^___)^x = 3^___}
Why: Negative 1 times x is negative x, so one third to the x is 3 to the negative x. Replacing x by negative x reflects a graph across the vertical axis, which is exactly why the growth and decay parents are mirror images.
Sorting
Two readings, two questions.
Sort into buckets
Sort each feature of a graph by which letter controls it.
Three independent readings from three parts of the equation, and none of them substitutes for another.
Prediction
Commit before reasoning.
Predict first
Why do growth and decay functions need only one set of transformation rules?
Correct: Because a decay function is a growth function with the input negated.
\[ ab^{x-h}+k \text{ with } b<1 \;\Longleftrightarrow\; a\left(\tfrac{1}{b}\right)^{-(x-h)}+k \]
Why: One half to the x is 2 to the negative x, so every decay function can be rewritten as a growth function of negative x. A transformation that works for one therefore works for the other with the horizontal direction reversed — and since h and k act on position rather than direction, the rules come out identical. The two lessons really are one family.
Comparison
Fill the blanks. Four cases, one form.
Comparison matrix
| a | b | Behaviour |
|---|---|---|
| positive | greater than 1 | exponential growth |
| positive | between 0 and 1 | exponential decay |
| negative | greater than 1 | falls, below the axis; neither |
| negative | between 0 and 1 | rises toward the axis from below; neither |
Only the top two rows have names, and both require a positive coefficient. The bottom two are still exponential functions, just not growth or decay ones.
Pattern
One routine, identical to Lesson 7.1's.
A percent decrease of r gives a decay factor of 1 minus r. Using r itself as the base models a far steeper loss than intended.
OpenStax Algebra and Trigonometry 2e, §6.2 Graphs of Exponential Functions §6.2
Check
Classifying. Both conditions.
Check your understanding
Which of these is an exponential decay function?
Answer: A
Why: The coefficient is positive and the base lies between 0 and 1, so both conditions hold.
Check
A translation. Find the asymptote.
Check your understanding
What are the asymptote and range of y = 3(1/2)^(x+1) - 2?
Answer: A
Why: The constant k = -2 moves the asymptote, and the positive coefficient keeps the graph above it.
Check
A decay model. Watch the base.
Check your understanding
A 4200 dollar snowmobile loses 10 percent of its value a year. What is the model?
Answer: A
Why: The decay factor is 1 - r = 1 - 0.10 = 0.90.
Real world
A medicine leaves the bloodstream at 25 percent per hour. A patient is given 400 milligrams, and the drug is considered ineffective below 50 milligrams.
Discussion prompt
Write the decay model, find the amount after 4 hours, and estimate when a second dose is needed. Then explain why halving the elimination rate does not double the effective time.
Hint: The decay factor is 1 minus 0.25.
Answer:
\[ A = 400(0.75)^t: \; A(4) = 400(0.3164) \approx 127 \text{ mg} \]
\[ 400(0.75)^t = 50 \;\Longrightarrow\; (0.75)^t = 0.125 \;\Longrightarrow\; t \approx 7.2 \text{ hours} \]
After 4 hours about 127 milligrams remain, and the level falls to 50 milligrams after about 7.2 hours.
At an elimination rate of 12.5 percent per hour the factor becomes 0.875, and the level reaches 50 milligrams after about 15.6 hours — slightly more than double, not exactly double. The reason is that halving the rate does not halve the factor: 0.75 becomes 0.875 rather than 0.375. Exponential decay is governed by the factor, and the relationship between the percent and the factor is a subtraction rather than a proportion.
Commit first
Answer, then rate your confidence honestly.
Predict first
Is y equals two sevenths times 4 to the x a growth function or a decay function?
Correct: Growth — the base 4 exceeds 1.
\[ \tfrac{2}{7}\cdot 4^x: \; \tfrac{2}{7}, \tfrac{8}{7}, \tfrac{32}{7}, \tfrac{128}{7}, \dots \]
Why: The coefficient sets the starting height and nothing else: the function's values are two sevenths, eight sevenths, thirty-two sevenths, quadrupling at every step. The base is what decides direction, and 4 is well above 1. A small coefficient with a large base starts low and climbs fast, which is a very common shape in real growth models — a population starting small and growing quickly is exactly this.
Explain it
They can graph exponential growth and have just met a fractional base.
Discussion prompt
In four sentences or fewer, explain what changes and what stays the same when the base drops below 1.
Hint: Only one thing changes.
Answer:
The curve falls instead of rising, because each step to the right now multiplies by a number less than 1 rather than more than 1. Everything else is the same: it still passes through the point where the output is 1, its domain is still every real number, its outputs are still positive, and the horizontal axis is still an asymptote.
The only other difference is which end flattens: a growth curve hugs the axis on the left and a decay curve hugs it on the right. In fact the two are mirror images, because one half to the x is the same as 2 to the negative x.
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 direction, compare the base with 1 and ignore everything else. For the coefficient, check its sign before naming the function. For decay factors, subtract the decimal from 1 and test one step by hand. For asymptotes, only k matters: the line is y equals k. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Take y equals 3 times one half to the x plus 1, minus 2, and build the page around it. Top left: sketch y equals one half to the x with a table from negative 3 to 2, marking the two anchor points and the asymptote. Top right: sketch y equals 3 times one half to the x, then the fully translated graph, marking where each anchor point and the asymptote moved to. Bottom left: on one set of axes, sketch y equals 2 to the x and y equals one half to the x together, mark their shared point, and write the identity that makes them mirror images. Bottom right: write a decay model of your own from an initial amount and a percent decrease you invent, evaluate it at t equal to 0, 1 and 5, and check the first two by subtracting the percent by hand. In a margin, write the four-row table of what a and b together produce.
If your bottom-right check disagrees at t equal to 1, you have almost certainly used the percent rather than one minus the percent as the base.
Recap
Five things, and four of them are Lesson 7.1's with one condition changed.
| If you see | Then |
|---|---|
| A base between 0 and 1 | Exponential decay, if a is positive |
| A base greater than 1 | Exponential growth, if a is positive |
| A negative coefficient | Neither; the graph is below the axis |
| A percent decrease r | The decay factor is 1 - r |
| A constant k added outside | The asymptote is y = k |
| A reciprocal base | The same as a negative exponent |
Lesson 7.3 introduces one particular base, the number e, which arises as the limit of the compound interest formula from Lesson 7.1 and turns out to be the natural base for every continuous growth and decay process.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.2 Graph Exponential Decay Functions §7.2, pp. 486-491 — everything on these slides traces back here
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