Linear functions and the meaning of slope, polynomials and degree, the roots of a quadratic and what the discriminant predicts, power functions and end behaviour, the algebraic and transcendental families, piecewise-defined functions, and the four transformations of a graph — the classification that lets Chapter 3 give one differentiation rule per family.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 1 — Functions and Graphs
Basic Classes of Functions
Objectives
Five outcomes. The last two are the ones that keep paying: naming a family tells you which rule applies, and transformations let you graph a whole family from one parent.
OpenStax Calculus Volume 1, §1.2 Basic Classes of Functions §1.2, pp. 30-52 — the section these objectives are drawn from
Warm-up
Section 1.1 gave you the definition of a function. This section asks a different question: what kinds of function are there?
Discussion prompt
You are handed the four rules 3x minus 1, x squared over x plus 1, the square root of x squared plus 1, and the sine of x. Which of these could you have written down using only addition, multiplication and roots?
Hint: Ask whether a finite formula built from the usual operations produces the rule.
Answer:
\[ 3x - 1, \quad \frac{x^2}{x+1}, \quad \sqrt{x^2+1} \qquad \text{versus} \qquad \sin x \]
The first three are built from arithmetic and roots — they are called algebraic. Sine is not: no finite combination of those operations produces it, which is why it and its relatives are called transcendental, meaning they transcend algebra.
That divide is the top-level split in this section, and it survives all the way to Chapter 5, where the two halves need genuinely different integration techniques.
Concept
Rather than treating every formula as a new object, we sort functions into families by how they are built. Polynomials sit inside rational functions, which sit inside algebraic functions; the transcendental functions sit outside all of them. Each family has its own shape, its own domain habits, and later its own calculus.
algebraic function — A function that can be built from polynomials using addition, subtraction, multiplication, division and the taking of roots. A function that is not algebraic is called transcendental.
\[ \text{polynomial} \subset \text{rational} \subset \text{algebraic} \subset \text{all functions} \]
The containments are genuine, not merely conventional. A polynomial is a rational function whose denominator happens to be 1, and a rational function is an algebraic function that happens to use no roots. So a question about rational functions is automatically a question about polynomials too.
Figure (svg): The families of functions arranged from polynomial outward to transcendental, each with an example
OpenStax Calculus Volume 1, §1.2 Basic Classes of Functions §1.2, pp. 30-31
Section
Section 1
Concept
A linear function has the form f of x equals mx plus b. The number m is the slope, and it measures how much the output changes per unit change in the input. Its defining property is that this ratio is the same between any two points on the line.
slope — For a line through two distinct points, the change in output divided by the change in input. For a linear function written as mx plus b, the slope is m and the output at input zero is b.
\[ m = \frac{y_2 - y_1}{x_2 - x_1}, \qquad y - y_1 = m(x - x_1) \]
Point-slope form is the more useful of the two forms, because it needs only a point and a slope rather than the intercept. In Section 3.1 you will use it constantly: the tangent line to a curve is always written from a point on the curve and the slope there.
Figure (svg): A line with the rise and run drawn as a right triangle between two labelled points
OpenStax Calculus Volume 1, §1.2 Basic Classes of Functions §1.2, pp. 31-35 — linear functions and slope
Picture it
Rise over run between two points on a line.
Figure (svg): A line with the rise and run drawn as a right triangle between two labelled points
Sliding the triangle along the line changes both rise and run but never their ratio. That constancy is the whole content of linearity — and its failure is what makes every other function need calculus.
Worked example
Example 1.11. Two points are enough for everything.
\[ \text{Find an equation of the line through } (-2, 3) \text{ and } (1, -6). \]
Compute the slope from the two points
Why: Change in output over change in input, taking the points in a consistent order.
\[ m = \frac{-6 - 3}{1 - (-2)} \]
Simplify
Why: Negative 9 over 3.
\[ m = -3 \]
Write point-slope form using either point
Why: Using the second point keeps the arithmetic small.
\[ y - (-6) = -3(x - 1) \]
Rearrange to slope-intercept form
Why: Distribute and isolate y.
\[ y = -3 x - 3 \]
Figure (svg): The solution to Worked example slope and the equation of a line shown as a ladder of expressions, one row per legal move
\[ y = -3x - 3 \]
Verify: substitute the OTHER point
Why: Point-slope form used only the point (1, -6), so checking with (-2, 3) is a genuine test: negative 3 times negative 2 minus 3 is 6 minus 3, which is 3. It matches. Using one point to build and the other to check is the cheapest possible verification, and it catches a slope computed with the subtractions in inconsistent orders.
OpenStax Calculus Volume 1, §1.2 Basic Classes of Functions §1.2, pp. 33-34
Matching
Consistent order, top and bottom.
Match the pairs
Why: The last two are the instructive pair. A zero rise gives slope 0, a horizontal line, which IS a function. A zero run gives a zero denominator and no slope at all, a vertical line, which is NOT a function — the vertical line test's worst case from Section 1.1.
Worked example
Example 1.12. The units are the point here.
\[ \text{A trucker's distance is } D(t) = 250 - 50t \text{ miles after } t \text{ hours. Interpret the numbers.} \]
Read the constant term as the value at input zero
Why: At t equal to 0 the output is 250.
\[ D(0) = 250\text{ miles} \]
Read the coefficient of t as the slope
Why: The slope is negative 50.
\[ m = -50 \]
Attach the units of output over input
Why: Miles divided by hours.
\[ -50\text{ miles per hour} \]
Interpret the sign
Why: A negative slope means the output falls as the input grows.
Figure (svg): The solution to Worked example slope as a rate of change shown as a ladder of expressions, one row per legal move
\[ D(0) = 250 \text{ mi}, \qquad m = -50 \text{ mi/h} \]
Verify: find where the output reaches zero
Why: Setting 250 minus 50t equal to 0 gives t equal to 5, so the trip takes 5 hours — which is exactly 250 miles at 50 miles per hour. The independent check agrees, and it confirms that the negative sign was interpreted as distance REMAINING rather than distance travelled.
OpenStax Calculus Volume 1, §1.2 Basic Classes of Functions §1.2, pp. 35-35
Trap
\[ (-2, 3) \text{ and } (1, -6) \]
Compute the slope with the orders mismatched
Why: Outputs taken second-minus-first, inputs taken first-minus-second.
\[ m = \frac{-6 - 3}{-2 - 1} = \frac{-9}{-3} = 3 \quad \text{(wrong sign)} \]
The line through these points falls, so a positive slope is visibly wrong — but nothing in the arithmetic complains.
\[ m = \frac{-6 - 3}{1 - (-2)} = \frac{-9}{3} = -3 \]
Fix an order and use it in BOTH subtractions
Why: Whichever point you call the second one, it must be second on the top and on the bottom.
Either order is fine — the two negatives cancel — but they must match. A quick guard: before computing, look at the points and decide whether the line rises or falls, then check your answer's sign against that.
Fill the middle
Distribute, then isolate the output.
Fill in the blanks
y + 6 = -3(x - 1) \;\Longrightarrow\; y = -3x + 3 - 6 = -3x - 3
Why: Distributing gives negative 3x plus 3, and subtracting the 6 leaves negative 3x minus 3. The sign slip to watch is the plus 3 from negative 3 times negative 1 — a negative times a negative, which is easy to lose when working quickly.
Prediction
Commit before reasoning.
Predict first
A tank drains according to V(t) = 400 - 12t litres after t minutes. What does the 12 mean?
Correct: The tank loses 12 litres every minute.
\[ V(t) = 400 - 12t: \quad V(0) = 400, \; m = -12 \text{ L/min}, \; V(t) = 0 \text{ at } t = \tfrac{100}{3} \]
Why: Slope always carries the units of output divided by input, here litres per minute, and the negative sign in front of it says the volume is falling. The tank holds 400 litres, and it empties after 400 divided by 12, about 33.3 minutes. The percentage reading would require the loss to depend on how much is left, which is exponential decay — a different family entirely, and the subject of Section 6.8.
Ranking
Given two points, produce an equation.
Put in order
Why: Steps a and c are the two people skip, and between them they eliminate the entire class of sign errors on this topic. Step e costs one substitution and independently confirms the whole answer.
Section
Section 2
Concept
A polynomial is a sum of terms, each a constant times a whole-number power of the input. The largest power is the degree and its coefficient is the leading coefficient. Together those two numbers decide entirely what the graph does far out on both sides.
degree and leading coefficient — For a polynomial written with its powers in descending order, the degree is the largest exponent and the leading coefficient is the constant multiplying that term. Together they determine the end behaviour of the graph.
\[ f(x) = a_n x^n + \cdots + a_1 x + a_0, \qquad a_n \ne 0 \]
An odd degree forces the two ends to disagree, so the graph must cross the horizontal axis at least once. An even degree forces them to agree, so a polynomial of even degree may miss the axis entirely — as x squared plus 1 does.
Figure (svg): Four polynomial graphs of degree one through four, showing how degree governs the end behaviour
OpenStax Calculus Volume 1, §1.2 Basic Classes of Functions §1.2, pp. 35-40 — polynomials, degree, and end behaviour
Picture it
The same question asked of degrees one through four.
Figure (svg): Four polynomial graphs of degree one through four, showing how degree governs the end behaviour
Notice that the wiggles in the middle vary from picture to picture, but the ends are forced. A degree-n polynomial has at most n minus 1 turning points and at most n roots — both facts Chapter 4 will prove with derivatives.
Worked example
Example 1.15. Try factoring first; fall back to the formula.
\[ \text{Find the roots of } f(x) = x^2 - 3x + 1 \text{ and of } g(x) = x^2 - 5x + 6. \]
Try to factor the second one over the integers
Why: Two numbers multiplying to 6 and adding to negative 5.
\[ (x - 2) (x - 3) = 0 \]
Read off its roots
Why: The zero product property.
\[ x = 2, x = 3 \]
The first one does not factor over the integers, so use the formula
Why: Read a, b and c with their signs from standard form.
\[ a = 1, b = -3, c = 1 \]
Substitute and evaluate the discriminant
Why: Nine minus 4 is 5, which is positive but not a perfect square.
\[ x = \frac{3 + - \sqrt{5}}{2} \]
Figure (svg): A parabola with its two roots marked and the quadratic formula written beside it
\[ x = \frac{3 \pm \sqrt{5}}{2}; \qquad x = 2 \text{ and } x = 3 \]
Verify: use the sum and product of the roots
Why: For x squared plus bx plus c the roots sum to negative b and multiply to c. For the first: the two roots sum to 3 and multiply to (9 - 5)/4, which is 1 — matching b equal to negative 3 and c equal to 1. For the second: 2 plus 3 is 5 and 2 times 3 is 6. Both check without re-solving anything.
OpenStax Calculus Volume 1, §1.2 Basic Classes of Functions §1.2, pp. 38-39
Sorting
Ask how the rule is built.
Sort into buckets
Sort each function into the smallest family that contains it.
Smallest family is the point of the exercise: the first function is also rational and also algebraic, but calling it a polynomial says the most. In Chapter 3, the smaller the family, the simpler the differentiation rule.
Worked example
Example 1.17. Roots and denominators together.
\[ \text{Find the domain of } f(x) = \frac{\sqrt{x - 3}}{x^2 - 4}. \]
Demand the radicand be non-negative
Why: An even root needs a non-negative input.
\[ x - 3 \ge 0,\text{ so } x \ge 3 \]
Demand the denominator not vanish
Why: Factor to find where it is zero.
\[ x ^{2} - 4 = 0\text{ at } x = 2\text{ and } x = -2 \]
Combine the two conditions
Why: Both excluded values are below 3, so the first condition already removes them.
\[ x \ge 3 \]
Write in interval notation
Why: From 3 inclusive to infinity.
\[ D = [3, \infty] \]
Figure (svg): The solution to Worked example domain of an algebraic function shown as a ladder of expressions, one row per legal move
\[ D = [3, \infty) \]
Verify: check that the excluded points really are already gone
Why: The denominator vanishes at 2 and negative 2, and both are less than 3, so the root condition has already removed them — the second condition adds nothing here. That is worth noticing rather than assuming: had the radicand been x plus 3, the domain would have been x at least negative 3 with 2 and negative 2 punched out, a genuinely different answer.
OpenStax Calculus Volume 1, §1.2 Basic Classes of Functions §1.2, pp. 41-42
Error analysis
A student states the end behaviour of a polynomial written out of order.
Annotate
On: \( f(x) = 5x - 2x^3 + 7 \;\Longrightarrow\; \text{degree } 1, \text{ leading coefficient } 5 \)
Rewrite every polynomial in descending order before reading anything off it. The degree is a property of the polynomial, not of whichever term happens to be typed first.
Fill the middle
For the quadratic from the worked example, with a equal to 1, b equal to negative 3 and c equal to 1.
Fill in the blanks
b^2 - 4ac = (-3)^2 - 4(1)(1) = 5
Why: The discriminant is 5. It is positive, so there are two real roots; it is not a perfect square, so those roots are irrational and factoring over the integers was never going to work. Computing this one number first tells you which method to use.
Two truths and a lie
All three are about polynomials.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. A degree-n polynomial has AT MOST n real roots, and may have far fewer: x squared plus 1 has degree 2 and no real roots at all, while x to the fourth plus 1 has degree 4 and none either. The exact count of n holds only when complex roots are counted with multiplicity, which is a statement about a different number system.
Prediction
Commit before reasoning.
Predict first
For f(x) = -2x^3 + 5x + 7, what happens at the far left and far right?
Correct: Rises on the left and falls on the right.
\[ \text{as } x \to \infty, \; -2x^3 \text{ dominates} \;\Longrightarrow\; f(x) \to -\infty \]
Why: Odd degree makes the two ends disagree, and the negative leading coefficient flips the standard odd-degree picture. For large positive x the cubic term dominates everything else and is large and negative; for large negative x it is large and positive. The lower-degree terms are irrelevant out there, which is precisely the content of limits at infinity in Section 4.6.
Section
Section 3
Concept
A piecewise-defined function uses different formulas on different parts of its domain. It is still one function: each input gets exactly one output, because the pieces are assigned to non-overlapping sets of inputs.
piecewise-defined function — A function specified by different formulas on different intervals of its domain, with each input belonging to exactly one interval so that exactly one formula applies.
\[ f(x) = \begin{cases} x^2 & x \le 1 \\ 3 - x & x > 1 \end{cases} \]
The whole difficulty lives at the boundary. Exactly one piece owns the boundary input, and which one is decided by where the equals sign is written. A filled dot marks the piece that owns it, an open dot the piece that does not.
Figure (svg): A piecewise function graphed with a filled dot where the piece is included and an open dot where it is not
OpenStax Calculus Volume 1, §1.2 Basic Classes of Functions §1.2, pp. 43-46 — piecewise-defined functions
Picture it
Two rules meeting at an input where they disagree.
Figure (svg): A piecewise function graphed with a filled dot where the piece is included and an open dot where it is not
At the boundary the two rules give 1 and 2 respectively, and the definition awards the input to the first. The gap that leaves is a discontinuity, and Section 2.4 is entirely about when such gaps appear.
Worked example
Example 1.18. Choose the piece first, then substitute.
\[ \text{For } f(x) = \begin{cases} x^2 & x \le 1 \\ 3 - x & x > 1 \end{cases} \text{ find } f(-1), f(1) \text{ and } f(4). \]
Decide which piece owns each input
Why: Compare the input with the boundary value 1.
Evaluate the first piece at negative 1
Why: Squaring.
\[ f(-1) = 1 \]
Evaluate at the boundary, using the piece with the equals sign
Why: The condition x at most 1 includes 1 itself.
\[ f(1) = 1 \]
Evaluate the second piece at 4
Why: Three minus 4.
\[ f(4) = -1 \]
Figure (svg): The solution to Worked example evaluating a piecewise function shown as a ladder of expressions, one row per legal move
\[ f(-1) = 1, \quad f(1) = 1, \quad f(4) = -1 \]
Verify: check the boundary against the wrong piece
Why: Using the second rule at the input 1 would give 3 minus 1, which is 2 — a different number. The definition assigns 1 to the first piece, so 1 is correct and 2 is not. The fact that the two rules disagree at the boundary is exactly why the equals sign has to be somewhere, and only in one place.
OpenStax Calculus Volume 1, §1.2 Basic Classes of Functions §1.2, pp. 44-44
Matching
The function is x squared for inputs at most 1, and 3 minus x above that.
Match the pairs
Why: Inputs 1 and 2 both give output 1, by different rules — a reminder that two inputs sharing an output is perfectly legal. The boundary input 1 uses the first piece because its condition carries the equals sign.
Worked example
Example 1.19. A real charging structure with a break in it.
\[ \text{A garage charges } \$10 \text{ for the first hour and } \$5 \text{ for each hour after. Model the cost.} \]
Describe the first interval
Why: For any positive time up to 1 hour the cost is flat.
\[ C = 10\text{ for } 0 < t \le 1 \]
Describe the later intervals
Why: After the first hour, each additional hour adds 5.
\[ C = 10 + 5(t - 1)\text{ for } t > 1 \]
Simplify the second rule
Why: Distribute and collect.
\[ C = 5 t + 5\text{ for } t > 1 \]
Assemble, giving the boundary to exactly one piece
Why: The first hour includes t equal to 1.
\[ \text{piecewise with boundary at } t = 1 \]
Figure (svg): The solution to Worked example building a piecewise model shown as a ladder of expressions, one row per legal move
\[ C(t) = \begin{cases} 10 & 0 < t \le 1 \\ 5t + 5 & t > 1 \end{cases} \]
Verify: check the two rules agree at the boundary
Why: The first rule gives 10 at t equal to 1, and the second rule approaches 5 times 1 plus 5, which is also 10. They agree, so this model has no jump — unlike the previous example. That agreement is worth checking because it distinguishes a model that is merely defined in pieces from one that actually breaks, and Chapter 2 will call the first kind continuous.
OpenStax Calculus Volume 1, §1.2 Basic Classes of Functions §1.2, pp. 45-46
Trap
\[ f(x) = \begin{cases} x^2 & x \le 1 \\ 3 - x & x > 1 \end{cases} \]
Evaluate at the boundary with both rules
Why: The student reports two values for the input 1.
\[ f(1) = 1 \text{ and } f(1) = 2 \quad \text{(wrong)} \]
This would make f fail the definition of a function: one input has been given two outputs.
\[ x \le 1 \text{ includes } 1 \;\Longrightarrow\; f(1) = 1^2 = 1 \]
Find the equals sign and let it decide
Why: Exactly one piece's condition includes the boundary, and only that piece applies there.
The other rule is not wrong — it simply does not apply at that input. Its value there is the one-sided limit, which Section 2.2 will name and which is genuinely useful; but it is not f of 1, and keeping the two apart is what makes continuity a meaningful question later.
Fill the middle
The boundary of the garage-charge model from the worked example.
Fill in the blanks
C(1) = 10 \text___ 0 < t \le 1
Why: The first piece owns t equal to 1 and gives a flat 10 dollars. The second piece would also give 10 there, so this model happens to have no jump — but the reason C(1) is 10 is the condition, not the coincidence.
Prediction
Commit before reasoning.
Predict first
Does defining a function by two different formulas violate the definition from Section 1.1?
Correct: No — the pieces cover non-overlapping sets of inputs, so each input gets exactly one rule.
\[ x \le 1 \text{ and } x > 1 \text{ are disjoint and cover } \mathbb{R} \]
Why: The definition demands one output per input, and says nothing at all about how many formulas were used to describe the rule. As long as every input falls under exactly one condition, the definition is satisfied. The graph may have a visible jump, and it may fail the vertical line test nowhere at all — a jump is a break in the curve, not a doubling of outputs.
Sorting
Compare the two rules' values at the boundary.
Sort into buckets
Sort each piecewise function by whether the pieces meet.
The fourth is the one to check by hand: at 2, the first rule gives 3 and the second approaches 6 minus 3, which is also 3. They meet. This question — do the pieces agree at the seam — is exactly the definition of continuity that Section 2.4 will state formally.
Section
Section 4
Concept
Adding a constant outside the function shifts the graph vertically. Adding a constant inside shifts it horizontally — and in the direction opposite to the sign, because the input must compensate. Multiplying outside stretches vertically; multiplying inside compresses horizontally.
transformation of a graph — A change to a function's formula that moves, reflects or scales its graph without changing its shape. Changes applied outside the function affect outputs directly; changes applied inside affect inputs, and therefore act on the graph in the opposite sense.
\[ y = a f(b(x - h)) + k \]
The reason the horizontal shift runs backwards is worth having rather than memorising. In f of x minus 2, the output that f used to produce at input 0 now appears when x minus 2 equals 0 — that is, at x equal to 2. The graph therefore moves right, even though the formula says minus.
Figure (svg): The four transformations applied to one parent parabola, each labelled with the change to the formula
OpenStax Calculus Volume 1, §1.2 Basic Classes of Functions §1.2, pp. 46-50 — transformations of functions
Picture it
The squaring function, shifted up, shifted right, and flipped.
Figure (svg): The four transformations applied to one parent parabola, each labelled with the change to the formula
Every one of these is the same curve in a different place. Recognising that saves you from plotting points for any function that is a transformed version of one you already know — which, in this chapter, is nearly all of them.
Worked example
Example 1.20. Read the formula from the inside out.
\[ \text{Describe how the graph of } y = -|x + 2| - 3 \text{ comes from } y = |x|. \]
Look inside the function first
Why: The input has 2 added to it, which shifts the graph in the opposite direction.
\[ \text{shift LEFT } 2 \]
Look at the sign outside the function
Why: A minus in front reflects outputs across the horizontal axis.
Look at the constant added outside
Why: Subtracting 3 lowers every output.
\[ \text{shift DOWN } 3 \]
Apply them in that order and locate the vertex
Why: The corner of the absolute value moves from the origin.
\[ \text{vertex at } (-2, -3) \]
Figure (svg): The solution to Worked example describing a transformation shown as a ladder of expressions, one row per legal move
\[ \text{vertex } (-2, -3), \text{ opening downward} \]
Verify: evaluate at the claimed vertex
Why: At x equal to negative 2 the inside is 0, so the absolute value is 0 and y is negative 3. That confirms the vertex is at (-2, -3). Checking a second point: at x equal to 0 the formula gives negative 2 minus 3, which is negative 5, and the point (0, -5) is indeed 2 units right of the vertex and 2 units below it, as a downward-opening absolute value should be.
OpenStax Calculus Volume 1, §1.2 Basic Classes of Functions §1.2, pp. 49-50
Matching
Read inside and outside separately.
Match the pairs
Why: The pairs differ only in whether the 3 or the minus sits inside or outside, and that single distinction changes the axis and, for the shift, the direction. Note that the last two are precisely the even and odd tests from Section 1.1: an even function is unchanged by the fourth, an odd function is turned into the third.
Worked example
The two scalings act on different variables, and it shows.
\[ \text{Compare } y = 2f(x) \text{ with } y = f(2x) \text{ for } f(x) = x^2 \text{ at } x = 3. \]
Apply the outside factor
Why: The output of f is doubled.
\[ 2 f(3) = 2(9) = 18 \]
Apply the inside factor
Why: The input is doubled before f acts.
\[ f(2 \cdot 3) = f(6) = 36 \]
Describe each geometrically
Why: Outside stretches vertically; inside compresses horizontally.
\[ \text{stretch by } 2\text{ versus compress by } \frac{1}{2} \]
Note why the numbers differ so much here
Why: Squaring turns a doubled input into a quadrupled output.
\[ 18\text{ versus } 36 \]
Figure (svg): The solution to Worked example a horizontal versus a vertical scaling shown as a ladder of expressions, one row per legal move
\[ 2f(x) \text{ stretches vertically}; \quad f(2x) \text{ compresses horizontally} \]
Verify: check the horizontal compression at a landmark
Why: The parabola y equals x squared passes through (2, 4). Under y equals f(2x), the same output 4 now appears when 2x equals 2, that is at x equal to 1 — half the distance from the axis, confirming a compression by a factor of one half rather than a stretch by 2. Tracking one landmark point is the fastest way to settle which way an inside change goes.
OpenStax Calculus Volume 1, §1.2 Basic Classes of Functions §1.2, pp. 47-49
Error analysis
A student describes the graph of y equals the square of x minus 3.
Annotate
On: \( y = (x - 3)^2 \;\Longrightarrow\; \text{shift the parabola LEFT by } 3 \)
The reliable fix is never to memorise the direction but to ask where the inside expression equals zero. That input is where the parent's landmark has moved to, and it settles the direction in one step every time.
Fill the middle
The vertex of a shifted absolute value.
Fill in the blanks
y = -|x + 2| - 3: \;\text-2 x = ___, \text___ (-2, -3)
Why: Setting the inside expression to zero locates where the parent's corner has moved to. Adding 2 inside moves the graph two units LEFT, which is why the vertex sits at negative 2 rather than positive 2.
Two truths and a lie
All three are about transformations.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Vertical and horizontal changes are independent of each other and may indeed be interleaved — but within one direction, order matters. Stretching vertically by 2 and then shifting up 3 gives 2f(x) + 3; shifting up 3 first and then stretching gives 2(f(x) + 3), which is 2f(x) + 6. Those are different graphs.
Prediction
Commit before reasoning.
Predict first
Why does y = f(x - 2) shift the graph right rather than left?
Correct: Because the input must be 2 larger to produce the output f used to give.
\[ f(x - 2) \text{ at } x = 2 \;\text{ equals }\; f(0) \]
\[ f(2x - 6) = f(2(x - 3)) \;\Longrightarrow\; \text{compress by } \tfrac{1}{2}, \text{ then shift right } 3 \]
Why: Whatever output f produced at input 0, the new function produces when x minus 2 equals 0, that is at x equal to 2. Every feature of the graph therefore appears 2 units further along. Nothing has been reversed arbitrarily; the input is compensating for the subtraction. Holding onto this reasoning rather than the rule keeps it right when the inside becomes something more complicated, such as f of 2x minus 6, where the shift is 3 rather than 6.
Section
Section 5
Concept
A power function is a constant times x raised to a fixed power. In a polynomial, the highest-power term eventually overwhelms all the others, so the whole polynomial behaves like that single term far from the origin.
end behaviour — The behaviour of a function's outputs as the inputs grow without bound in the positive or negative direction. For a polynomial, it is determined entirely by the degree and the leading coefficient.
\[ f(x) = a x^n, \qquad n \text{ a positive integer} \]
The rate at which powers grow separates them decisively. At x equal to 10, x squared is 100 while x cubed is 1000; at x equal to 100 the gap is a factor of 100. That is why the leading term wins, and it is the intuition behind every limit at infinity in Section 4.6.
Figure (svg): Four polynomial graphs of degree one through four, showing how degree governs the end behaviour
OpenStax Calculus Volume 1, §1.2 Basic Classes of Functions §1.2, pp. 36-38 — power functions and behaviour at infinity
Picture it
Degrees one through four side by side.
Figure (svg): Four polynomial graphs of degree one through four, showing how degree governs the end behaviour
The even-degree graphs open the same way on both sides; the odd-degree ones go opposite ways. That single distinction is what guarantees an odd-degree polynomial has a real root and lets an even-degree one avoid the axis entirely.
Worked example
Two numbers decide the whole picture.
\[ \text{Describe the end behaviour of } f(x) = -3x^4 + 2x^3 - 7x + 1. \]
Identify the degree and leading coefficient
Why: Highest power 4, its coefficient negative 3.
\[ n = 4, a = -3 \]
Even degree means the ends agree
Why: Both sides do the same thing.
The negative coefficient flips the standard picture
Why: An even power is always non-negative, and negative 3 times it is non-positive.
Sanity-check with a large input
Why: At x equal to 10 the leading term is negative 30000 against a next term of only 2000.
\[ f(10)\text{ is large and negative} \]
Figure (svg): The solution to Worked example end behaviour from the leading term shown as a ladder of expressions, one row per legal move
\[ \text{as } x \to \pm\infty, \; f(x) \to -\infty \]
Verify: compare term sizes at a large input
Why: At x equal to 100: the leading term is negative 300 million, while 2x cubed is 2 million, negative 7x is negative 700, and the constant is 1. The leading term is over a hundred times the next one and grows further ahead as x grows. That is what 'the leading term dominates' means quantitatively, and it is why the other three terms cannot rescue the sign.
OpenStax Calculus Volume 1, §1.2 Basic Classes of Functions §1.2, pp. 37-38
Sorting
Degree parity and the leading coefficient's sign, together.
Sort into buckets
Sort each polynomial by its end behaviour.
Notice how little of the polynomial you needed. Every lower-degree term was irrelevant, which is the practical payoff of end behaviour and the reason Section 4.6 can compute limits at infinity by inspection.
Worked example
Which of two powers eventually wins is never in doubt.
\[ \text{For large } x, \text{ compare } x^2 \text{ and } x^3, \text{ and find where } x^3 \text{ overtakes } 100x^2. \]
Set the two expressions equal
Why: The crossing point is where neither is ahead.
\[ x ^{3} = 100 x ^{2} \]
Divide by the common factor, noting x is positive
Why: Dividing by x squared is legal for positive x.
\[ x = 100 \]
Interpret
Why: Below 100 the squared term is bigger; above it the cubed term is.
\[ x ^{3}\text{ wins for } x > 100 \]
Note that the constant only delays the outcome
Why: Any constant multiple is eventually beaten by the higher power.
Figure (svg): The solution to Worked example comparing growth rates shown as a ladder of expressions, one row per legal move
\[ x^3 > 100x^2 \;\text{ for }\; x > 100 \]
Verify: test either side of the crossing
Why: At x equal to 50: x cubed is 125000 while 100 x squared is 250000, so the squared term is still ahead. At x equal to 200: x cubed is 8 million while 100 x squared is 4 million, so the cubed term has taken over. The crossing at 100 is confirmed from both sides, and the lesson is that a large constant delays the higher power but never defeats it.
OpenStax Calculus Volume 1, §1.2 Basic Classes of Functions §1.2, pp. 37-37
Trap
\[ f(x) = x^3 - 400x \]
Graph it from -5 to 5 and describe the ends
Why: In that window the cubic term is tiny and the linear term dominates.
\[ \text{on } [-5, 5]: \; f \text{ looks like the falling line } -400x \quad \text{(misleading)} \]
Concluding that the graph falls on the right would be exactly wrong: it eventually rises steeply.
\[ \text{degree } 3, \text{ leading coefficient } +1 \;\Longrightarrow\; \text{falls left, rises right} \]
Read the ends from the leading term, not from a window
Why: End behaviour is a statement about arbitrarily large inputs, which no finite window shows.
The crossover here is late: x cubed only overtakes 400x at x equal to 20. A window of plus or minus 5 shows none of it. This is a genuine hazard with graphing technology — the picture on the screen is always a finite window, and the algebra is what tells you what lies beyond it.
Fill the middle
Comparing a cube with a hundred times a square.
Fill in the blanks
x^3 = 100x^2 \;\Longrightarrow\; x = 100 \text___ x > 0
Why: Dividing by x squared leaves x equal to 100. Below that the squared term is larger; above it the cube pulls ahead permanently. A bigger constant would push the crossing further out but could never prevent it.
Prediction
Commit before reasoning.
Predict first
Is there a constant c large enough that c times x squared exceeds x cubed for every positive x?
Correct: No. Whatever c is, the cube overtakes at x equal to c and never falls behind again.
\[ c x^2 = x^3 \;\Longrightarrow\; x = c, \quad \text{and } x^3 > cx^2 \text{ for all } x > c \]
Why: Setting c x squared equal to x cubed gives x equal to c, so a larger constant merely postpones the crossing. This is the qualitative fact that makes end behaviour depend only on the degree: coefficients decide where things happen, degree decides what happens eventually. The same reasoning in Section 4.6 lets you evaluate a limit at infinity by looking only at the highest powers.
Ranking
Which is largest for very large positive x?
Put in order
Why: Degree beats every coefficient in the end, so the order is simply by degree: 1, 2, 3, 4, 5. The constants shift the crossings around considerably — x to the fourth over 100 does not overtake x cubed until x equals 100 — but they never change the final ordering.
Comparison
Fill the blanks. This table is the reason the section exists.
Comparison matrix
| Family | Built from | Domain habit |
|---|---|---|
| Polynomial | sums of whole-number powers | all real numbers, always |
| Rational | one polynomial over another | all reals except the denominator's zeros |
| Algebraic | arithmetic and roots | roots and denominators both restrict it |
| Transcendental | sine, exponential, logarithm | varies: all reals for sin and exp, positives for log |
The third column is what you actually use day to day. Naming the family is a fast route to the domain, and from Chapter 3 onward it is also a fast route to the derivative.
Pattern
Given an unfamiliar formula, place it and describe it.
Step five is the one that saves the most time and is used the least. A great many functions in this course are a parent graph moved, flipped or scaled, and recognising that turns a plotting exercise into a one-line description.
Stewart, Calculus: Early Transcendentals 8e, §1.2 Mathematical Models: A Catalog of Essential Functions §1.2, pp. 23-35
Check
Slope. Consistent order.
Check your understanding
Find the slope of the line through (-2, 3) and (1, -6).
Answer: A
Why: Negative 6 minus 3 is negative 9, over 1 minus negative 2, which is 3. That gives negative 3.
Check
End behaviour. Two numbers only.
Check your understanding
Describe the end behaviour of f(x) = -3x^4 + 2x^3 - 7x + 1.
Answer: A
Why: Even degree makes the ends agree; the negative leading coefficient sends both downward.
Check
Transformations. Inside runs backwards.
Check your understanding
Where is the vertex of y = -|x + 2| - 3?
Answer: A
Why: The inside is zero at x equal to negative 2, and the constant outside puts the output at negative 3.
Real world
A company's revenue from selling x units is 60 dollars per unit. Its cost, in dollars, is four thousand plus ten per unit plus two hundredths times the square of the number of units, for production up to 2000 units.
Discussion prompt
Name the family of each function, write the profit function, and say what its degree and leading coefficient force to happen for large x. Then find the break-even points.
Hint: Profit is revenue minus cost, and the sign of the leading coefficient decides the long-run story.
Answer:
\[ R(x) = 60x, \qquad C(x) = 0.02x^2 + 10x + 4000, \qquad 0 \le x \le 2000 \]
Revenue is linear; cost is a quadratic polynomial. Profit is their difference, so it is also a quadratic polynomial.
\[ P(x) = 60x - (0.02x^2 + 10x + 4000) = -0.02x^2 + 50x - 4000 \]
Degree 2 with a negative leading coefficient, so both ends fall: profit rises, peaks, and then falls again. That is not an artefact of the model but its central claim — selling more is not indefinitely better, because cost grows quadratically while revenue grows only linearly.
\[ -0.02x^2 + 50x - 4000 = 0 \;\Longrightarrow\; x^2 - 2500x + 200000 = 0 \]
\[ x = \frac{2500 \pm \sqrt{6250000 - 800000}}{2} = \frac{2500 \pm \sqrt{5450000}}{2} \approx 85 \text{ or } 2415 \]
The discriminant is positive, so there are two break-even points. Only the first, about 85 units, lies inside the stated domain of 0 to 2000 — a reminder that a model's domain can discard a mathematically valid root.
Commit first
Answer, then rate your confidence honestly.
Predict first
A polynomial has degree 5 and a negative leading coefficient. What is guaranteed?
Correct: It has at least one real root.
\[ \text{as } x \to -\infty, f(x) \to +\infty; \quad \text{as } x \to +\infty, f(x) \to -\infty \]
A graph that is eventually positive and eventually negative, and has no breaks, must pass through zero.
Why: Odd degree forces the two ends to go opposite ways — here rising on the left and falling on the right — so the graph must cross the horizontal axis somewhere along the way. It need not cross five times: a degree-5 polynomial can have as few as one real root. The last option describes even-degree behaviour, which odd degree rules out. The guarantee here is exactly the Intermediate Value Theorem, which Section 2.4 will state and which this is the most common application of.
Explain it
They can graph y equals x squared but insist that y equals the square of x minus 3 should move left, because of the minus sign.
Discussion prompt
In four sentences or fewer, explain why the graph moves right, without asking them to memorise a rule.
Hint: Ask them where the vertex is, and make them solve for it.
Answer:
Ask where the squared quantity equals zero, because that is where the parabola's lowest point sits. For y equals x squared it is at x equal to 0; for y equals the square of x minus 3 you must solve x minus 3 equals 0, which gives x equal to 3.
So the vertex has moved from 0 to 3 — to the right. The input has to be 3 bigger to give the function the same thing it used to receive, and every point on the graph moves along with it. Nothing was reversed by convention; the input is compensating for the subtraction.
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 shifts, never memorise a direction — solve for where the inside expression is zero. For end behaviour, rewrite in descending order first, then look only at the leading term. For algebraic domains, write the root condition and the denominator condition on separate lines before combining them. For piecewise boundaries, find the equals sign and let it decide. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Across the top of the page, draw the four-family diagram: polynomial inside rational inside algebraic, with transcendental outside all three, and write one example in each region. Below it, take the polynomial -3x^4 + 2x^3 - 7x + 1 and record its degree, leading coefficient, end behaviour at both ends, and the value of the leading term against the next term at x = 100. In the middle of the page, take y = x^2 as a parent and draw four transformed copies in one set of axes: up 2, right 2, flipped vertically, and compressed horizontally by a half, labelling each with its formula and its moved vertex. At the bottom, write the piecewise function that is x^2 for inputs at most 1 and 3 - x above that, graph it with the correct filled and open dots, and write one sentence saying which piece owns the input 1 and how you know. In a margin, write the quadratic formula and beside it the three things the discriminant can tell you.
If your four transformed parabolas all look congruent to the parent, three of them should be — a shift or a reflection does not change shape. Only the horizontal compression should look genuinely narrower, and if it does not, check whether you changed the input or the output.
Recap
Five things, and the first is the one Chapter 3 will lean on hardest.
| If you see | Then |
|---|---|
| A ratio of polynomials | Rational: exclude the denominator's zeros |
| A root of a polynomial | Algebraic: demand a non-negative radicand |
| sin, exp or log | Transcendental: no algebraic formula exists |
| Odd degree | The ends disagree, so there is a real root |
| Even degree | The ends agree, and there may be no real root |
| A change inside f | Horizontal, and in the opposite direction |
| A change outside f | Vertical, and in the direction written |
Section 1.3 takes the first of the transcendental families seriously: the trigonometric functions, measured in radians, which is the only unit in which their calculus comes out clean.
OpenStax Calculus Volume 1, §1.2 Basic Classes of Functions §1.2, pp. 30-52 — everything on these slides traces back here
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