Everything from the last three lessons assembles into one procedure. Work through domain, symmetry, intercepts, stationary points, concavity, inflections and end behaviour, and the graph draws itself. Then run the process backwards: read $f'$ and $f''$ straight off a picture of $f$.
Commit to a prediction about reading one graph from another.
You are shown only the graph of $f'$, and it crosses the $x$-axis at $x = -1$ going from positive to negative, and again at $x = 3$ going from negative to positive. Without looking ahead say what $f$ does at each of those points, and whether you can find the $y$-values.
Learn the seven-item sketching checklist.
Seven checks · in this order, every time
We just saw stationary points, concavity and inflections handled separately. That raises a question: a "sketch the curve" question wants all of it at once, so in what order do you work and what must appear on the page? This card answers it → a seven-item checklist, each item cheap on its own, run in an order where each one narrows the picture.
| check | what you do |
|---|---|
| 1. Domain | Where is $f$ defined? Exclude zeros of denominators and negatives under even roots. |
| 2. Symmetry | $f(-x) = f(x)$ is even (reflect in the $y$-axis); $f(-x) = -f(x)$ is odd (rotate about the origin). Either one halves the work. |
| 3. Intercepts | $y$-intercept is $f(0)$; $x$-intercepts solve $f(x) = 0$. Skip if the solving is unreasonable and say so. |
| 4. Stationary points | Solve $f'(x) = 0$ and classify, by the sign of $f'$ or by $f''$. |
| 5. Concavity and inflections | Solve $f''(x) = 0$, test the sign either side, keep only genuine changes. |
| 6. Asymptotes and discontinuities | Vertical asymptotes at excluded values where $f$ grows without bound; note any holes. |
| 7. End behaviour | What happens as $x \to \infty$ and $x \to -\infty$? |
Sketching checklist: domain, symmetry, intercepts, stationary points, concavity and inflections, asymptotes and discontinuities, end behaviour.; Marks come from LABELLED features: coordinates on every turning point and inflection, dashed asymptotes, intercepts shown.
Pause, copy the seven-item sketching checklist in order, and the note that marks are awarded for labelled features rather than for the quality of the drawing, into your book.
Quick check: Which function is odd, so that its graph has rotational symmetry about the origin?
Run the checklist end to end on two curves.
Worked examples · reveal as you go
Sketch $f(x) = x^3 - 3x$, showing all significant features.
Sketch $f(x) = x + \dfrac{1}{x}$, showing all significant features.
Fill the blanks: For $f(x) = x^3 - 3x$.
number of stationary points $=$
$x$-coordinate of the inflection $=$
$f(-1) =$
Read f' and f'' off a graph of f, and back again.
Three graphs · one function
We just saw a curve built from its formula. That raises a question: the syllabus also asks you to graph $f'$ and $f''$ given only a graph of $f$, with no formula at all, so what carries across? This card answers it → a fixed dictionary between features of one graph and features of the next.
| on the graph of $f$ | on the graph of $f'$ |
|---|---|
| turning point (max or min) | crosses the $x$-axis |
| horizontal point of inflection | touches the $x$-axis without crossing |
| increasing | above the $x$-axis |
| decreasing | below the $x$-axis |
| point of inflection (any) | turning point of $f'$ |
| steepest section | maximum or minimum of $f'$ |
The same table applied a second time relates $f'$ to $f''$. So an inflection of $f$ is a turning point of $f'$ and a zero-crossing of $f''$ — three descriptions of one event.
Line the two graphs up vertically and work column by column. Every feature of $f$ sits directly above its counterpart on $f'$, which is why exam sketches are always drawn one under the other.
Turning point of $f$ = zero CROSSING of $f'$; horizontal inflection of $f$ = $f'$ TOUCHES zero without crossing; inflection of $f$ = turning point of $f'$ = zero crossing of $f''$.; From $f'$ alone you cannot recover $y$-values on $f$: every vertical shift has the same derivative.
Pause, copy the translation table between features of $f$ and features of $f'$, and the note that a vertical shift is unrecoverable from $f'$ alone, into your book.
True or false: Given only the graph of $f'$, you can determine the $y$-coordinates of the turning points of $f$.
Odd one out: Three of these correctly translate a feature of $f$ into a feature of $f'$. Which one is wrong?
Read the traps, work the quick-fire set, then revisit.
Common errors · traps that cost marks
Quick-fire practice · sketch and translate
Determine whether $f(x) = x^4 - 4x^2$ is odd, even or neither, and say what that tells you about the sketch.
Find the domain and any vertical asymptotes of $f(x) = \dfrac{2}{x - 3}$, and describe the end behaviour.
A graph of $f'$ is a parabola opening upwards with $x$-intercepts at $x = 0$ and $x = 4$. Describe everything you can about $f$.
The graph of $f'$ touches the $x$-axis at $x = 2$ without crossing. What does $f$ do at $x = 2$?
Run all seven checks on $f(x) = x^3 - 3x^2$ and list what you would put on the sketch.
Match each feature of $f$ to what appears on the graph of $f'$:
Earlier you were told only that $f'$ crosses zero at $x = -1$ from positive to negative, and at $x = 3$ from negative to positive. That is enough to say $f$ has a local maximum at $x = -1$ and a local minimum at $x = 3$, that $f$ is increasing before $-1$, decreasing between, and increasing after $3$, and that $f$ has a point of inflection somewhere between them, where $f'$ bottoms out. What you cannot get is either $y$-value: every vertical shift of $f$ has exactly the same $f'$, so the heights are undetermined until you are given one point on $f$ itself.
Sketching finds every local feature. The last lesson of this strand asks a different question: across a whole domain, where is the function actually largest? That means comparing local maxima against the values at endpoints and at any discontinuity, and then applying the whole method to problems where you must build the function yourself from a described situation.
Apply the lesson methods, then compare each response with its comprehensive answer.
Use the visible short-answer practice below during this lesson. After completing the focus area, use Checkpoint 1 for the checkpoint question bank.
Q1. Sketch $f(x) = x^3 - 3x^2 - 9x + 5$.
(a) Find the stationary points and classify them. (b) Find the point of inflection. (c) Describe the end behaviour and list every feature you would label on the sketch. (3 marks)
Q2. The graph of $f'$ is a parabola opening downwards, crossing the $x$-axis at $x = -2$ and $x = 4$, with its maximum at $x = 1$.
(a) State where $f$ is increasing and where it is decreasing. (b) Classify the stationary points of $f$. (c) State the $x$-coordinate of the point of inflection of $f$, and explain how you know from the graph of $f'$ alone. (3 marks)
Q3. Two students are given the same graph of $f'$ and asked to sketch $f$. Student A draws a curve passing through the origin; Student B draws an identical curve shifted three units up. Each insists the other is wrong.
(a) Explain who is right. (b) State precisely what a graph of $f'$ does and does not determine about $f$. (c) Describe the single smallest piece of extra information that would settle the disagreement, and explain why nothing less would do. (3 marks)
Drill 1: $f(-x) = (-x)^4 - 4(-x)^2 = x^4 - 4x^2 = f(x)$, so $f$ is even. The graph is symmetric in the $y$-axis, so everything found for $x > 0$ is mirrored to $x < 0$ and only half the analysis is needed.
Drill 2: Domain is $x \neq 3$. As $x \to 3$ the denominator tends to 0 while the numerator stays at 2, so $|f| \to \infty$: a vertical asymptote at $x = 3$ (from the right $f \to +\infty$, from the left $f \to -\infty$). As $x \to \pm\infty$, $f \to 0$, so $y = 0$ is a horizontal asymptote.
Drill 3: An upward parabola with roots 0 and 4 is negative between them and positive outside. So $f$ is increasing for $x < 0$ and for $x > 4$, and decreasing on $0 < x < 4$. At $x = 0$ the sign of $f'$ goes $+$ to $-$: a local maximum. At $x = 4$ it goes $-$ to $+$: a local minimum. The inflection of $f$ is at the turning point of $f'$, which by symmetry of the parabola is $x = 2$.
Drill 4: Touching without crossing means $f'$ does not change sign at $x = 2$, though $f'(2) = 0$. So $f$ has a stationary point that is not a turning point: a horizontal point of inflection at $x = 2$.
Drill 5: $f(x) = x^3 - 3x^2$. Domain all reals; neither odd nor even (since $f(-x) = -x^3 - 3x^2$). Intercepts: $x^2(x - 3) = 0$ gives $x = 0$ (a double root, so the curve touches) and $x = 3$; $y$-intercept 0. $f'(x) = 3x^2 - 6x = 3x(x-2)$, stationary at $x = 0$ and $x = 2$; $f''(x) = 6x - 6$, so $f''(0) = -6$ gives a maximum at $(0, 0)$ and $f''(2) = 6$ gives a minimum at $(2, -4)$. Inflection where $f'' = 0$, at $x = 1$, and $f(1) = -2$, so $(1, -2)$; the sign of $f''$ changes there. No asymptotes. As $x \to \infty$, $f \to \infty$; as $x \to -\infty$, $f \to -\infty$.
Q1 (3 marks): (a) $f'(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x - 3)(x + 1)$, so stationary at $x = 3$ and $x = -1$. $f''(x) = 6x - 6$, so $f''(3) = 12 > 0$ giving a local minimum at $(3, -22)$, and $f''(-1) = -12 < 0$ giving a local maximum at $(-1, 10)$. (Values: $f(3) = 27 - 27 - 27 + 5 = -22$; $f(-1) = -1 - 3 + 9 + 5 = 10$.) [1.5] (b) $f''(x) = 6x - 6 = 0$ at $x = 1$, and $f''$ changes from negative to positive there, so $(1, -6)$ is a point of inflection. ($f(1) = 1 - 3 - 9 + 5 = -6$.) [1] (c) As $x \to \infty$, $f \to \infty$; as $x \to -\infty$, $f \to -\infty$. On the sketch label: the maximum $(-1, 10)$, the minimum $(3, -22)$, the inflection $(1, -6)$, the $y$-intercept $(0, 5)$, and the arrows showing end behaviour. There are no asymptotes to draw [0.5].
Q2 (3 marks): (a) A downward parabola with roots $-2$ and 4 is positive between the roots and negative outside them. So $f$ is increasing on $-2 < x < 4$ and decreasing for $x < -2$ and for $x > 4$ [1]. (b) At $x = -2$ the sign of $f'$ goes from negative to positive, so $f$ has a local minimum. At $x = 4$ it goes from positive to negative, so $f$ has a local maximum [1]. (c) A point of inflection of $f$ occurs where $f'$ has a turning point, because that is where $f''$ (the gradient of $f'$) crosses zero. The graph of $f'$ has its maximum at $x = 1$, so $f$ has its point of inflection at $x = 1$. It is also where $f$ is steepest, being the largest value of $f'$ [1].
Q3 (3 marks): (a) Both are right, and neither has grounds to correct the other. Any two functions differing by a constant have identical derivatives, so both curves are legitimate answers to "sketch $f$" from this information [1]. (b) A graph of $f'$ determines the shape of $f$ completely: where it increases and decreases, the $x$-coordinates of its stationary points and their classification, the $x$-coordinates of its inflections, and where it is steepest. It does not determine the vertical position, because $f$ and $f + c$ share a derivative for every constant $c$. So the family of possible answers is a set of identical curves stacked vertically [1]. (c) One point on $f$ itself — a single value $f(a) = b$ for any $a$ in the domain — is enough, and it is the smallest thing that works: it fixes $c$ uniquely and therefore pins the whole curve. Nothing purely about $f'$ can do it, no matter how much of $f'$ you are given, because every member of the family produces exactly the same $f'$. In integral terms, the point supplies the constant of integration [1].
Retrieve the central ideas, then mark the lesson complete or continue to the module quiz.
Answer from memory first, then return to the matching Learn checkpoint to check and correct your response.
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
Start the module quiz →Tick when you've finished the practice and review.
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