Applies Chapter 1's transformations to the exponential parent. Establishes the three features to track — the point at height one, the horizontal asymptote, and the range — and works out which transformations move which. The vertical shift is the one that moves the asymptote, and forgetting that is what costs the range and the end behaviour together.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 4 — Exponential and Logarithmic Functions
§4.2 Graphs of Exponential Functions, pp. 482-500
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §4.2 Graphs of Exponential Functions §4.2, pp. 482-500 — the pages these objectives are drawn from
Warm-up
Three features, and one of them is new relative to Chapter 1.
Discussion prompt
Sketch 2 to the power x from memory. What point does it definitely pass through, and what does it do far to the left?
Hint: What is any base raised to the power zero?
Answer:
It passes through the point at height 1 when the input is zero, because any base to the power zero is 1. Every exponential with coefficient 1 does, whatever its base.
Far to the left the outputs get very small but stay positive — 2 to the power negative 10 is about a thousandth. The graph hugs the horizontal axis without touching it.
That last feature is the horizontal asymptote, and none of Chapter 1's toolkit parents had one. It is the feature to watch through every transformation in this section, because it is the one people forget to move.
Concept
Transforming an exponential moves the same things §1.5 said it would, plus one more: the horizontal asymptote. It moves under a vertical shift and stays put under everything else.
\[ f(x)=a\,b^{x-h}+k \;\Longrightarrow\; \text{asymptote } y=k \]
The reason the asymptote deserves separate attention is that it decides the range and the end behaviour together. Get it wrong and both are wrong, and the sketch is in the wrong place vertically even if every other feature was handled correctly.
Figure (svg): The parent exponential graph with its key features labelled: the point at height one on the vertical axis, the horizontal asymptote along the axis, and the domain and range marked
OpenStax, Precalculus, §4.2 Graphs of Exponential Functions §4.2, pp. 482-487
Section
Section 1
Concept
The exponential parent passes through the point at height one, has the horizontal axis as an asymptote, and produces only positive outputs.
The domain being everything is worth noticing, because it is unusual among the chapter's functions. There is no denominator and no even root, so §1.2's rules find no restriction at all — and the graph reflects that by extending forever in both directions.
Figure (svg): The parent exponential graph with its key features labelled: the point at height one on the vertical axis, the horizontal asymptote along the axis, and the domain and range marked
OpenStax, Precalculus, §4.2 Graphs of Exponential Functions §4.2, pp. 482-488
Picture it
One point, one asymptote, and a range that starts just above it.
Figure (svg): The parent exponential graph with its key features labelled: the point at height one on the vertical axis, the horizontal asymptote along the axis, and the domain and range marked
The shaded band is the range: everything above the asymptote, not including it. Every transformation in this section moves some of these three and leaves the others.
Worked example
Read off the three features without computing anything.
\[ \text{State the domain, range and asymptote of } f(x)=5^x. \]
Consider the domain
Why: Nothing restricts an exponent.
Consider the outputs
Why: A positive base to any power is positive.
Check whether zero is attained
Why: It is approached but not reached.
Name the asymptote
Why: The line the outputs approach.
\[ y = 0 \]
Figure (svg): The parent exponential graph with its key features labelled: the point at height one on the vertical axis, the horizontal asymptote along the axis, and the domain and range marked
\[ \text{domain } (-\infty,\infty), \; \text{range } (0,\infty), \; \text{asymptote } y=0 \]
Verify: test far to the left
Why: At x equal to negative 10 the output is 5 to the negative 10, which is about one over nine million — tiny, positive, and not zero. That is the asymptotic behaviour: arbitrarily close to the axis and never on it.
OpenStax, Precalculus, §4.2 Graphs of Exponential Functions §4.2, pp. 483-485
Prediction
Several exponentials with different bases and coefficient 1 are graphed together.
Predict first
What do they all have in common?
Correct: They all pass through the point at height 1 above the origin.
Why: Any base to the power zero is 1, so every such exponential gives 1 at input zero. They also share the asymptote along the horizontal axis. What differs is the steepness, which the base controls, and whether they climb to the right or the left.
Worked example
The base changes the steepness and the direction, not the features.
\[ \text{Compare } 2^x \text{ with } 5^x. \]
Compare at input zero
Why: Both give 1.
Compare at input 1
Why: Two against five.
\[ 5 ^{x}\text{ is higher} \]
Compare at input negative 1
Why: One half against one fifth.
\[ 5 ^{x}\text{ is lower} \]
Describe the difference
Why: A larger base is steeper both ways.
Figure (svg): The solution to Worked example compare two bases shown as a ladder of expressions, one row per legal move
\[ \text{both through } (0,1), \text{ both asymptote } y=0; \; 5^x \text{ steeper} \]
Verify: check the crossing point is shared
Why: Every exponential with coefficient 1 passes through the point at height 1, because any base to the power zero is 1. So all of them cross at that single point and fan out from it, steeper for larger bases. The three features are identical; only the steepness differs.
OpenStax, Precalculus, §4.2 Graphs of Exponential Functions §4.2, pp. 485-488
Trap
\[ 2^x \text{ reaches } 0 \text{ eventually as } x \text{ goes to } -\infty \]
Observe that the outputs become extremely small
Why: At negative 100 the output is far below any measurable value.
The graph is described as reaching zero and continuing along the axis.
It never reaches zero. A positive base raised to any power is positive, however large the negative exponent is.
The outputs become arbitrarily small, which is a different claim: they get closer to zero than any number you name, without ever equalling it.
That is exactly what an asymptote means, and it is why the range is written with a round bracket at zero. The distinction matters for the range and for §3.7's asymptote language alike.
Sorting
Compare against the toolkit functions from §1.2.
Sort into buckets
Sort each statement about the exponential parent.
Faded example
For the function 7 to the power x.
Fill in the blanks
\text0; \quad \text0 y > ___; \quad \text___ y = ___
Why: Both are zero for the parent, which is why they are easy to conflate. The range starts just above zero and the asymptote sits at zero, and a vertical shift later will move both together — which is when keeping them distinct starts to matter.
Explain it to yourself
The exponential parent has a feature none of Chapter 1's toolkit functions had.
Discussion prompt
Explain why the horizontal axis is an asymptote, using what a negative exponent means.
Hint: What is 2 to the power negative n?
Answer:
A negative exponent is a reciprocal: 2 to the power negative n is 1 over 2 to the n. As n grows, the denominator grows without bound.
So the output becomes arbitrarily small and stays positive, since a reciprocal of a positive number is positive. It gets closer to zero than any number you care to name and never arrives.
That is precisely the definition of a horizontal asymptote from §3.7, arrived at from a completely different family. Approaching without reaching is what both rational and exponential functions do, though for different reasons — one from a growing denominator in the formula and one from a growing power.
Section
Section 2
Concept
A vertical shift moves the whole graph, asymptote included. A horizontal shift slides the graph along without touching the asymptote at all.
The last point is a real change in character. The parent never crosses the horizontal axis, but shifting it down moves part of the curve below the axis, so a crossing appears. Finding it requires solving an exponential equation, which is §4.6's business.
Figure (svg): An exponential shifted vertically, with the horizontal asymptote moving with it, drawn against the parent whose asymptote stays on the axis
OpenStax, Precalculus, §4.2 Graphs of Exponential Functions §4.2, pp. 488-494
Picture it
The dashed curve is the parent and the solid one has been shifted down 3.
Figure (svg): An exponential shifted vertically, with the horizontal asymptote moving with it, drawn against the parent whose asymptote stays on the axis
The asymptote came down with the graph, so the range now starts at negative 3 rather than zero. The curve also crosses the horizontal axis, which the parent never did.
Worked example
Move the asymptote first, then everything follows.
\[ \text{Describe } f(x)=2^x-3. \]
Identify the transformation
Why: Minus 3 outside.
\[ \text{shift down } 3 \]
Move the asymptote
Why: It goes down with the graph.
\[ \text{asymptote } y = -3 \]
State the range
Why: Everything above the new asymptote.
\[ y > -3 \]
Find the point that was at height 1
Why: It moves down 3 too.
\[ (0, -2) \]
Figure (svg): An exponential shifted vertically, with the horizontal asymptote moving with it, drawn against the parent whose asymptote stays on the axis
\[ \text{asymptote } y=-3, \; \text{range } (-3,\infty) \]
Verify: check for a new x-intercept
Why: The graph now dips below the horizontal axis on the left, so it must cross it. Setting the rule to zero gives 2 to the x equal to 3, which happens somewhere between 1 and 2. The parent had no x-intercept and this one does, which is the shift's most visible consequence.
OpenStax, Precalculus, §4.2 Graphs of Exponential Functions §4.2, pp. 489-491
Sorting
Only vertical shifts move it.
Sort into buckets
Sort each transformation of an exponential.
Worked example
The asymptote does not move.
\[ \text{Describe } f(x)=2^{x-4}. \]
Identify the transformation
Why: Minus 4 inside the exponent.
\[ \text{shift right } 4 \]
Check the asymptote
Why: Nothing was done to the output.
\[ \text{still } y = 0 \]
State the range
Why: Unchanged.
\[ y > 0 \]
Find where the height-1 point went
Why: It moves right 4.
\[ (4, 1) \]
Figure (svg): The solution to Worked example a horizontal shift shown as a ladder of expressions, one row per legal move
\[ \text{right } 4; \; \text{asymptote } y=0, \; \text{range } (0,\infty) \]
Verify: confirm with a substitution
Why: At x equal to 4 the exponent is zero, so the output is 1 — the height-1 point has moved to input 4, as predicted. And the outputs are still all positive, since shifting the input horizontally cannot change what values come out. Only the vertical shift touches the range.
OpenStax, Precalculus, §4.2 Graphs of Exponential Functions §4.2, pp. 492-494
Error analysis
A student describes a shifted exponential.
Annotate
On: \( f(x)=3^x+5: \quad \text{asymptote } y=0, \; \text{range } y>0 \)
A vertical shift moves everything vertical, and the asymptote is a vertical feature. Checking the behaviour far to the left is the quickest test: whatever the outputs approach there is the asymptote's height.
Faded example
For the function 4 to the power x, plus 6.
Fill in the blanks
\text6 y = 7, \qquad \text___ y > ___, \qquad \text___ (0, ___)
Why: The shift up 6 moves the asymptote to height 6 and the range to everything above it. The point that was at height 1 moves to height 7, since it too is shifted up 6. All three features move together under a vertical shift.
Prediction
An exponential with a positive coefficient is shifted vertically.
Predict first
When does the graph acquire an x-intercept?
Correct: When it is shifted down, so part of it drops below the axis.
Why: The parent stays entirely above the axis, so no crossing exists. Shifting down moves the asymptote below the axis, and the curve then rises from below it through the axis, producing exactly one crossing. Shifting up moves it further away and horizontal shifts do not change which side of the axis the graph is on.
Explain it to yourself
Three of the four transformations leave the asymptote alone.
Discussion prompt
Explain why a horizontal shift and a vertical stretch both leave the asymptote at zero.
Hint: What is a stretched version of something approaching zero?
Answer:
A horizontal shift slides the graph sideways, and the asymptote is a horizontal line extending forever in both directions. Sliding a curve along beside such a line leaves the line exactly where it was.
A vertical stretch multiplies every output by a constant. Something approaching zero, multiplied by any constant, still approaches zero — so the asymptote does not move, though the graph gets taller everywhere else.
Only adding a constant shifts the limiting value itself, from zero to that constant. So the rule is simple: whatever is added outside is where the asymptote goes, and nothing else affects it.
Section
Section 3
Concept
A coefficient outside stretches the graph vertically; a negative outside reflects it below the asymptote. A negative inside the exponent turns growth into decay.
That last point is a genuine peculiarity of this family. Replacing x by 2x in 2 to the x gives 4 to the x, so a horizontal compression is indistinguishable from a change of base. No other family in the course has that overlap, and it is why horizontal scalings of exponentials are usually absorbed into the base instead.
Figure (svg): Two reflections of the exponential parent shown side by side: one over the horizontal axis flipping the graph below it, and one over the vertical axis turning growth into decay
OpenStax, Precalculus, §4.2 Graphs of Exponential Functions §4.2, pp. 494-498
Picture it
Outside on the left, inside on the right.
Figure (svg): Two reflections of the exponential parent shown side by side: one over the horizontal axis flipping the graph below it, and one over the vertical axis turning growth into decay
The outside minus puts the graph below the asymptote and reverses the range. The inside minus leaves the range alone and reverses the direction of growth.
Worked example
Read the coefficient's size and its sign separately.
\[ \text{Describe } f(x)=-3(2^x). \]
Read the coefficient's size
Why: Three, greater than one.
\[ \text{stretch by } 3 \]
Read its sign
Why: Negative, acting on the output.
\[ \text{reflect over } y = 0 \]
Check the asymptote
Why: A stretch and a reflection leave it.
\[ \text{still } y = 0 \]
State the range
Why: The graph is now below the asymptote.
\[ y < 0 \]
Figure (svg): Two reflections of the exponential parent shown side by side: one over the horizontal axis flipping the graph below it, and one over the vertical axis turning growth into decay
\[ \text{asymptote } y=0, \; \text{range } (-\infty,0) \]
Verify: check the value at zero
Why: At input zero the rule gives negative 3 times 1, which is negative 3 — the coefficient, as always. The graph passes through that point and lies entirely below the axis, so the range is the negatives. The asymptote did not move, but the graph swapped sides of it.
OpenStax, Precalculus, §4.2 Graphs of Exponential Functions §4.2, pp. 495-496
Matching
Inside is horizontal, outside is vertical, as always.
Match the pairs
Why: Only the last one moves the asymptote. The first flips which side of it the graph is on, the second reverses the direction of growth without touching the range, and the third scales the outputs without changing what they approach.
Worked example
An inside minus turns growth into decay.
\[ \text{Show that } 2^{-x} \text{ is the same as } (1/2)^x. \]
Use the negative exponent rule
Why: A negative exponent is a reciprocal.
\[ \frac{1}{2 ^{x}} \]
Rewrite the fraction
Why: One over a power is that reciprocal to the power.
\[ (\frac{1}{2}) ^{x} \]
Interpret the graph
Why: The reflection turned growth into decay.
Check the asymptote
Why: Still along the horizontal axis.
\[ y = 0 \]
Figure (svg): The solution to Worked example a negative exponent shown as a ladder of expressions, one row per legal move
\[ 2^{-x}=\Bigl(\tfrac{1}{2}\Bigr)^x \]
Verify: test one input
Why: At x equal to 3: 2 to the negative 3 is one eighth, and one half cubed is also one eighth. The two agree, so reflecting a growth exponential horizontally produces a decay one — which is why growth and decay graphs are always mirror images in the vertical axis.
OpenStax, Precalculus, §4.2 Graphs of Exponential Functions §4.2, pp. 496-498
Trap
\[ f(x)=-2^x \;\Longrightarrow\; \text{asymptote } y=0 \text{ becomes } y=0 \text{ reflected, so } y=-0 \]
Apply the reflection to every feature including the asymptote
Why: The graph flips, so the asymptote is taken to flip too.
Confusion follows about where the asymptote has gone.
The asymptote is at zero and reflecting zero gives zero, so it does not move. The line stays exactly where it was.
What the reflection changes is which side of it the graph occupies: the range flips from the positives to the negatives.
A reflection over the horizontal axis moves the range and not the asymptote. The two are different features, and the parent's asymptote sitting at zero is what makes them easy to conflate — which is why a shifted case is the better test of understanding.
Prediction
An exponential has a negative coefficient and is shifted up 4.
Predict first
What is its range?
Correct: Everything below 4.
Why: The shift puts the asymptote at height 4, and the negative coefficient puts the graph below it. So the outputs approach 4 from below and extend downward without bound. Getting either the asymptote or the side wrong produces a range that is wrong in a different way, which is why both have to be tracked.
Faded example
Express 5 to the power negative x as a decay with a fractional base.
Fill in the blanks
5^5 = \fracx___ = \Bigl(\frac______}\Bigr)^___}
Why: A negative exponent is a reciprocal, and one over 5 to the x is the same as one fifth to the x. So a horizontal reflection of a growth exponential is a decay exponential, which is why the two graphs are mirror images in the vertical axis.
Counterexample
A classmate says every transformation of a graph moves its asymptote.
Discussion prompt
Give two transformations that leave an exponential's asymptote exactly where it is.
Hint: What happens to something approaching zero when you stretch it or slide it sideways?
Answer:
A horizontal shift leaves it: the asymptote is a horizontal line extending forever, so sliding the curve along beside it changes nothing about the line.
A vertical stretch leaves it too: multiplying something approaching zero by any constant still gives something approaching zero. The graph gets taller but its limiting value does not move.
Only adding a constant outside moves it, because that changes the limiting value itself. So the rule is unusually clean: whatever is added outside is the asymptote's height, and nothing else has any effect on it.
Section
Section 4
Concept
No transformation restricts an exponential's domain, since nothing about an exponent can be illegal. The range is decided by the asymptote's position and which side the graph occupies.
The domain being immune is worth stating explicitly, because it is unusual. Every other family in Chapter 3 could have its domain restricted by a transformation; an exponential cannot, because the only operations involved are a power and some arithmetic, and none of them refuses an input.
Figure (svg): A table showing which transformations move the horizontal asymptote and which leave it in place, with the effect on the range noted for each
OpenStax, Precalculus, §4.2 Graphs of Exponential Functions §4.2, pp. 490-496
Picture it
Four transformations and their effect on the asymptote and the range.
Figure (svg): A table showing which transformations move the horizontal asymptote and which leave it in place, with the effect on the range noted for each
Only the first row moves the asymptote. The last row changes the range without moving the asymptote, which is the case most often got wrong.
Worked example
Asymptote first, then the range, then the intercepts.
\[ \text{For } f(x)=4(3^x)-8, \text{ give the domain, range, asymptote and intercepts.} \]
State the domain
Why: Never restricted.
Find the asymptote
Why: The constant added outside.
\[ y = -8 \]
State the range
Why: Positive coefficient, so above it.
\[ y > -8 \]
Find the y-intercept
Why: Substitute zero.
\[ 4 - 8 = -4 \]
Figure (svg): The solution to Worked example state everything shown as a ladder of expressions, one row per legal move
\[ \text{range } (-8,\infty), \; \text{asymptote } y=-8, \; (0,-4) \]
Verify: check whether an x-intercept exists
Why: The graph runs from just above negative 8 on the left up to infinity on the right, so it must cross zero somewhere. Setting the rule to zero gives 3 to the x equal to 2, which happens between 0 and 1. So there is exactly one x-intercept, and finding it exactly needs logarithms.
OpenStax, Precalculus, §4.2 Graphs of Exponential Functions §4.2, pp. 491-493
Faded example
For the function that subtracts 5 from twice 3 to the power x.
Fill in the blanks
\text-5 y = ___; \text___ y > ___
Why: The constant subtracted outside puts the asymptote at negative 5, and the positive coefficient keeps the graph above it. So the range is everything above negative 5. The coefficient 2 stretches the graph but does not affect either the asymptote or which side of it the graph sits on.
Worked example
The graph must cross the axis for one to exist.
\[ \text{Does } f(x)=2^x+3 \text{ have an x-intercept?} \]
Find the asymptote
Why: The constant added outside.
\[ y = 3 \]
State the range
Why: Positive coefficient, above the asymptote.
\[ y > 3 \]
Compare with zero
Why: Every output exceeds 3.
Conclude
Why: The graph never reaches zero.
Figure (svg): The solution to Worked example when there is no x-intercept shown as a ladder of expressions, one row per legal move
\[ \text{No: the range is } (3,\infty), \text{ which excludes } 0. \]
Verify: confirm algebraically
Why: Setting the rule to zero gives 2 to the x equal to negative 3, and an exponential is never negative, so there is no solution. The range argument and the algebra agree, and the range argument is faster — checking whether zero lies in the range settles the question before any solving is attempted.
OpenStax, Precalculus, §4.2 Graphs of Exponential Functions §4.2, pp. 493-496
Trap
\[ f(x)=2^{x-5}: \quad \text{domain } x\ge 5, \text{ since the exponent must be nonnegative} \]
Apply the radical-style domain reasoning to the exponent
Why: The exponent is treated as needing to be nonnegative, by analogy with a square root.
The domain is restricted to inputs at or above 5.
An exponent may be negative. A negative exponent means a reciprocal, which is perfectly well defined for a positive base.
At x equal to 3 the rule gives 2 to the negative 2, which is one quarter — a legitimate output. The domain is every real number.
Only denominators and even roots restrict a domain, per §1.2, and an exponent is neither. Exponentials are one of the few families in the course whose domain is never restricted by anything.
Sorting
It has one exactly when zero lies in the range.
Sort into buckets
Sort each function.
Prediction
An exponential is stretched, reflected, and shifted in both directions.
Predict first
What is its domain?
Correct: All real numbers, whatever the transformations.
Why: None of these operations can make an input illegal: an exponent may be any real number, and stretching, reflecting and shifting are all arithmetic on legal outputs. Exponentials are unusual in this respect, and the domain question for them is answered before it is asked.
Step zero
You are asked for the range of a transformed exponential.
Discussion prompt
What do you find first, and why does the range follow from it?
Hint: What line does the graph approach?
Answer:
Find the asymptote first, which is whatever constant is added outside. It is the one number the range depends on.
Then decide which side of it the graph occupies, from the sign of the coefficient: positive puts it above and negative below.
The range is then everything on that side, with a round bracket at the asymptote since it is approached and never reached. Two facts, one number and one sign, and the range follows without any further work.
Section
Section 5
Concept
A fully transformed exponential has four parameters. Reading them in the right order gives the asymptote, the range, the shape and the intercepts without plotting anything.
\[ f(x)=a\,b^{x-h}+k \]
Reading k first is the efficient order because everything else is described relative to the asymptote. The range, the end behaviour and whether an x-intercept exists all depend on where that line is, and none of them can be settled until it is known.
Figure (svg): A table showing which transformations move the horizontal asymptote and which leave it in place, with the effect on the range noted for each
OpenStax, Precalculus, §4.2 Graphs of Exponential Functions §4.2, pp. 496-500
Picture it
Four parameters, and only one of them touches the asymptote.
Figure (svg): A table showing which transformations move the horizontal asymptote and which leave it in place, with the effect on the range noted for each
Finding k first and the sign of a second settles the range immediately, which is the fact everything else in a sketch is positioned against.
Worked example
Asymptote first, then the sign, then the rest.
\[ \text{Describe } f(x)=-2(3^{x+1})+4 \text{ completely.} \]
Read k
Why: The constant added outside.
\[ \text{asymptote } y = 4 \]
Read the sign of a
Why: Negative.
State the range
Why: Below 4.
\[ y < 4 \]
Read h and b
Why: Plus 1 inside shifts left 1; base 3 grows.
\[ \text{left } 1,\text{ growth} \]
Figure (svg): The solution to Worked example read all four parameters shown as a ladder of expressions, one row per legal move
\[ \text{asymptote } y=4, \; \text{range } (-\infty,4), \; \text{left } 1 \]
Verify: check the y-intercept and the ends
Why: At input zero: negative 2 times 3 to the first, plus 4, which is negative 6 plus 4, giving negative 2. Far to the left the exponential shrinks to nothing and the output approaches 4 from below, confirming both the asymptote and which side the graph is on. Far to the right the output falls without bound, since the negative coefficient reverses the growth.
OpenStax, Precalculus, §4.2 Graphs of Exponential Functions §4.2, pp. 497-499
Ranking
The efficient order for describing a transformed exponential.
Put in order
Why: The asymptote comes first because everything vertical is described relative to it. Its position plus the coefficient's sign gives the range immediately. The y-intercept is computed separately and last, since no parameter gives it directly once a shift is present.
Worked example
Asymptote, then a point, then the direction.
\[ \text{Sketch } f(x)=2^{x-3}+1. \]
Draw the asymptote
Why: At the height added outside.
\[ \text{dashed line } y = 1 \]
Find a convenient point
Why: Where the exponent is zero.
\[ (3, 2) \]
Determine the direction
Why: Base above 1, positive coefficient.
Sketch
Why: Approaching the asymptote on the left.
\[ \text{curve above } y = 1 \]
Figure (svg): The solution to Worked example sketch from the parameters shown as a ladder of expressions, one row per legal move
\[ \text{asymptote } y=1, \text{ through } (3,2), \text{ growing right} \]
Verify: check the y-intercept is consistent
Why: At input zero the exponent is negative 3, giving one eighth, plus 1, which is 1.125 — just above the asymptote, as the sketch should show. Finding the input that makes the exponent zero is the fastest way to get an exact point on a transformed exponential, since the power is then 1.
OpenStax, Precalculus, §4.2 Graphs of Exponential Functions §4.2, pp. 499-500
Error analysis
A student states the y-intercept of a transformed exponential.
Annotate
On: \( f(x)=5(2^{x-3})+1: \quad y\text{-intercept} = 5 \)
The y-intercept must be computed by substituting zero, not read off a parameter. That shortcut works only for the simplest form, and any shift breaks it.
Faded example
For the function 3 to the power x minus 5, plus 2, find the input making the exponent zero.
Fill in the blanks
x - 5 = 0 \;\Longrightarrow\; x = 5, \text3 1 + 2 = ___
Why: The exponent vanishes at x equal to 5, where the power is 1, so the output is 1 plus 2, which is 3. This is the fastest exact point to find on any transformed exponential, and it is more useful than the y-intercept when the horizontal shift is large.
Prediction
An exponential has a negative coefficient, base above 1, and is shifted up 6.
Predict first
What happens far to the left and far to the right?
Correct: Approaches 6 from below on the left; falls without bound on the right.
Why: Far left the exponential part shrinks to nothing, so the output approaches the asymptote at 6 — from below, because the negative coefficient puts the graph there. Far right the exponential grows and the negative coefficient sends the output downward without bound.
Explain it
Four parameters, and the order they are read in matters for efficiency.
Discussion prompt
Explain to a classmate why finding the asymptote first makes everything else easier.
Hint: What is the range described relative to?
Answer:
The range is described relative to the asymptote — everything above it or everything below it. So the range cannot be stated at all until the asymptote's height is known.
The end behaviour is too: on one side the graph approaches the asymptote and on the other it runs off. Both halves of that description reference the same line.
And whether an x-intercept exists depends on whether zero lies in the range, which again needs the asymptote. So one number unlocks three separate questions, which is why it is worth finding before anything else — and it is the easiest parameter to read, being simply whatever is added outside.
Comparison
Fill the blanks from memory. The asymptote is the column that catches people out.
Comparison matrix
| moves the asymptote | changes the range | changes the domain | |
|---|---|---|---|
| vertical shift | yes | yes, it moves with it | no |
| horizontal shift | no | no | no |
| vertical stretch | no | no | no |
| reflection over the x axis | no | yes, it flips sides | no |
The last column is entirely 'no', which is unusual and worth remembering: nothing ever restricts an exponential's domain.
Pattern
Five steps, and the first one unlocks three of the others.
The domain step takes no work at all and is worth including anyway, because a question asking for both domain and range expects both to be stated.
OpenStax Algebra and Trigonometry 2e, §6.2 Graphs of Exponential Functions §6.2
Check
The asymptote moves with a vertical shift.
Check your understanding
What is the horizontal asymptote of the function 3 to the power x, minus 7?
Answer: A
Why: Subtracting 7 outside shifts the whole graph down 7, carrying the asymptote from height 0 to height negative 7. Far to the left the outputs approach negative 7 rather than zero.
Check
Asymptote plus the sign of the coefficient.
Check your understanding
What is the range of the function that takes negative 4 times 2 to the power x, plus 1?
Answer: A
Why: The asymptote sits at height 1 from the constant added outside, and the negative coefficient puts the graph below it. So the outputs approach 1 from below and extend downward without bound.
Check
Nothing restricts an exponent.
Check your understanding
What is the domain of the function 5 to the power x minus 2, plus 9?
Answer: A
Why: An exponent may be any real number, including negative ones, which simply produce reciprocals. No transformation of an exponential ever restricts its domain.
Real world
A shifted asymptote is what distinguishes a cooling model from a pure decay model.
Discussion prompt
A cup of coffee cools towards room temperature rather than towards zero. How does that appear in the model?
Hint: What is the asymptote, and what has been added?
Answer:
The asymptote sits at room temperature rather than at zero, so the model is an exponential decay with a vertical shift equal to the surroundings' temperature.
The decaying part describes the difference between the coffee and the room, and that difference does decay towards zero. Adding the room temperature back converts it into the actual temperature.
This is Newton's law of cooling, and it is exactly this section's transformation. Modelling the temperature directly as a pure decay would predict the coffee freezing, which is why the shift is the physical content rather than a mathematical detail.
Commit first
State your confidence along with your answer.
Predict first
Which transformation of an exponential can create an x-intercept where there was none?
Correct: A vertical shift that moves the asymptote past zero.
Why: The parent stays entirely on one side of the horizontal axis. Shifting it vertically so that the asymptote ends up on the opposite side of zero from the graph's growth means the curve must cross the axis somewhere. Horizontal shifts and stretches never change which side of the axis the graph occupies.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate why the horizontal asymptote needs tracking separately from everything else.
Hint: What three things depend on where it is?
Answer:
Because three separate questions depend on it: the range is described relative to it, the end behaviour on one side is 'approaches it', and whether an x-intercept exists depends on whether zero is on the occupied side.
Get the asymptote wrong and all three are wrong together, even if every transformation was applied correctly. That is why one small oversight costs so much here.
The rule itself is simple — whatever is added outside is the asymptote's height — so the cost of tracking it is almost nothing. The reason to emphasise it is that Chapter 1's parents had no asymptote, so there is no habit of looking for one.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The second is the section's central point and the source of most errors. The third follows immediately from it, so understanding the second usually fixes the third at no extra cost.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Sketch the exponential parent with its asymptote, its point at height one, and its range marked. Then sketch a fully transformed version beside it — reflected, stretched, and shifted both ways — and label its new asymptote, its range, and the point where its exponent is zero. Beneath both, write which of the four transformations moved the asymptote and which did not.
If your transformed sketch has its asymptote at the constant added outside, and its range on the side the coefficient's sign dictates, both of the section's decisions are on the page.
Recap
Five things, and the second is what Chapter 1 had no occasion to teach.
| if you remember one thing | it should be this |
|---|---|
| about the asymptote | it sits at whatever is added outside, and nothing else moves it |
| about the range | the asymptote's height plus the coefficient's sign gives it |
| about the domain | always every real number, whatever the transformations |
| about the y-intercept | compute it; do not read it off a parameter |
Section 4.3 introduces the inverse of the exponential — the logarithm — which is the tool needed to solve for an exponent and therefore to find the x-intercepts this section could only locate approximately.
OpenStax, Precalculus, §4.2 Graphs of Exponential Functions §4.2, pp. 482-500 — everything on these slides traces back here
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