Educerie
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Educerie · IB Diploma · Mathematics: analysis and approaches

Topic 2 Functions · 2.4 Key features of graphs and intersections

Level
SL and HL. Nothing here is HL only, so every section is examinable for both.
Themes (key concepts)
representation, relationships, space. The features of a graph (its intercepts, turning points, symmetry and asymptotes) are the parameters of the function made visible, and the place where two graphs meet represents the solution of an equation.
The question this unit answers
once a function is on the screen, which points and lines describe it, and how do you find each of them, including where two graphs meet?
Where it is examined
mainly Paper 2, where the GDC is expected: "find the coordinates of the local minimum", "write down the equation of the vertical asymptote", "solve f(x) = g(x)", usually 1 to 3 marks each inside a longer question. Paper 1 asks for the same features without technology, for functions simple enough to read by hand: a factorised quadratic, a rational function (ax + b) ÷ (cx + d). Intersections are how Paper 2 solves the equations that algebra cannot.

What you must be able to do

You must be able toLevelWhat it looks like in the exam
Find the zeros of a function (the roots of f(x) = 0) and its x- and y-interceptsSL, HL"Find the x-intercepts of the graph of f" (2 marks); GDC on Paper 2, factorising on Paper 1
Find maximum and minimum points and values, local and on a restricted domainSL, HL"Find the coordinates of the local maximum" (2 marks); "Find the range of f" (2 to 3 marks)
Recognise and describe symmetry, including the axis of symmetry and vertex of a parabolaSL, HL"Write down the equation of the axis of symmetry" (1 mark)
Find vertical and horizontal asymptotes, using graphing technologySL, HL"Write down the equations of the asymptotes of the graph of f" (2 marks)
Find the points of intersection of two curves or lines using technologySL, HLPaper 2: "Solve f(x) = g(x)" or "Find the coordinates of the points where the graphs meet" (2 to 3 marks)
Give GDC answers to 3 significant figures and exact ones exactlySL, HLEvery Paper 2 question; an unrounded or over-rounded answer loses the accuracy mark

Before you start

You need the graph of a function and the window skills of 2.3, and the domain and range of 2.2. You should be comfortable sketching straight lines (2.1) and quadratics on your GDC, and able to solve a linear or factorised quadratic equation by hand. Rational functions of the form (ax + b) ÷ (cx + d) are studied in full in 2.8; here you meet their asymptotes as features to find.


1The idea in one paragraph

A graph is described by a short list of key features: where it crosses the axes, where it turns, whether it is symmetric, and which lines it approaches far out. Those features are what a sketch must show and what most questions ask for. On Paper 2 your GDC finds each one with a built-in tool (zero, maximum, minimum, intersect), and your job is to know which tool answers which question and to report the result properly. The most powerful of them is the intersection: two graphs meet exactly where their equations are equal, so any equation f(x) = g(x), however hard to solve by algebra, can be solved by looking for where y = f(x) and y = g(x) cross.

2The features, and their names

Figure 1 shows one graph with its features labelled: the cubic y = x³ − 3x² − 9x + 5.

Figure 1 · The key features of one graph Figure 1 · The key features of one graph x y local max (−1, 10) local min (3, −22) (0, 5) (−2.18, 0) (0.489, 0) (4.69, 0) y = x³ − 3x² − 9x + 5: three zeros, one y-intercept, a local maximum and a local minimum. The zeros of f, the roots of f(x) = 0 and the x-intercepts are the same three numbers.
Figure 1 · The key features of one graph
FeatureWhat it isHow to find it
y-interceptwhere the graph crosses the y-axisput x = 0: the point (0, f(0))
x-interceptswhere the graph crosses the x-axissolve f(x) = 0
Zeros of fthe x-values that make f(x) = 0the same numbers as the x-intercepts
Roots of an equationthe solutions of f(x) = 0the same numbers again
Local maximum / minimuma point where the graph turns, higher or lower than everything close to itGDC maximum and minimum tools; later, calculus
Vertexthe turning point of a parabolaGDC, or x = −b ÷ 2a from the formula booklet
Axis of symmetrya vertical line the graph is a mirror image inthrough the vertex of a parabola
Vertical asymptotea line x = a that the graph runs up or down alongsidewhere the function is undefined because of division by zero
Horizontal asymptotea line y = b that the graph settles towards as x → ±∞the value f(x) approaches for very large or very negative x

Three names for one idea deserve a sentence. The zeros of a function f, the roots of the equation f(x) = 0, and the x-intercepts of y = f(x) are the same numbers seen three ways: as inputs, as solutions, and as points on a graph. For Figure 1 they are x = −2.18, 0.489 and 4.69. A question asking for the zeros wants the x-values; a question asking for the x-intercepts wants points, (−2.18, 0) and so on.

One more distinction costs marks easily. The maximum value of a function is a y-value, such as 10. The maximum point is a pair of coordinates, (−1, 10). Answer the one you are asked for.

3Using the GDC: what each tool gives you

On Paper 2 you are expected to find these features with graphing technology, and a short note of the method you used helps the marker award method marks. The tools have different names on different calculators but the same jobs:

You wantToolThe GDC asks you for
a zero, root or x-interceptzero / roota left bound and a right bound either side of it
a local maximum or minimummaximum / minimumbounds either side of the turn
the y-interceptvalue at x = 0, or y-interceptnothing
where two graphs meetintersectthe two curves, and a guess near the point
an output for a given inputvalue, or the tablethe input

On a TI-84 these live under CALC; on a Casio under G-Solv; on a TI-Nspire under Analyze Graph. Whichever you use, three habits matter.

Find each feature separately. A cubic with three zeros needs the zero tool three times, with bounds around each one. The tool gives you one answer per use.

Round only at the end, and to 3 significant figures. The GDC shows 0.488872256…; you write 0.489. If that value feeds into a later calculation, use the stored value, not the rounded one, or the next answer drifts.

Write down what you did. A line such as "using GDC: local minimum at (3, −22)" is enough. An answer with no working and no mention of technology can still score, but only if it is exactly right.

On Paper 1, with no calculator, you find the same features by algebra: factorise to get zeros, substitute x = 0 for the y-intercept, read the vertex from a completed square (2.6), and read asymptotes from the form of the function. The features are the same; only the route changes.

4Maximum and minimum values

A local maximum is a point higher than every nearby point on the graph: the top of a hill. A local minimum is the bottom of a valley. For y = x³ − 3x² − 9x + 5 the GDC gives a local maximum at (−1, 10) and a local minimum at (3, −22).

"Local" is doing work there. The cubic in Figure 1 carries on upward forever on the right and downward forever on the left, so (−1, 10) is not the highest point on the graph; there is no highest point. It is only the highest point in its neighbourhood.

On a restricted domain the end points join the competition, as Figure 2 shows. Take the same cubic on −3 ≤ x ≤ 6. Now check four candidates: the two turning points and the two end points.

f(−3) = −27 − 27 + 27 + 5 = −22left end point
f(−1) = 10local maximum
f(3) = −22local minimum
f(6) = 216 − 108 − 54 + 5 = 59right end point

The greatest value on this domain is 59, at the end point (6, 59), not at the local maximum. The least value is −22, reached twice. So the range of f on this domain is −22 ≤ f(x) ≤ 59, the same method as the range questions in 2.2.

Figure 2 · On a restricted domain the end points compete Figure 2 · On a restricted domain the end points compete x y −3 3 6 −22 59 (−3, −22) (6, 59) greatest value local max (−1, 10) (3, −22) Same function on −3 ≤ x ≤ 6. The greatest value, 59, is at an end point, not at the local maximum. The least value, −22, is reached twice: at the end point x = −3 and at the local minimum.
Figure 2 · On a restricted domain the end points compete

On a restricted domain, the greatest and least values can be at the end points. Evaluate f at every end point as well as every turning point, then compare.

5Symmetry and the vertex

A graph has line symmetry in a vertical line x = h if folding along that line puts one half exactly onto the other. Every parabola has one, its axis of symmetry, and the turning point on it is the vertex. Figure 3(a) shows y = −2x² + 8x − 3. The formula booklet gives the axis of a quadratic ax² + bx + c as x = −b ÷ 2a:

x = −b / 2a = −8 / (2 × (−2)) = 2axis of symmetry: x = 2
y = −2(2)2 + 8(2) − 3 = −8 + 16 − 3 = 5vertex (2, 5)

The symmetry gives you points for free: since (0.5, 0.5) is on the curve, so is its mirror image (3.5, 0.5), the same distance on the other side of x = 2. It also means the two x-intercepts of a parabola sit equally either side of its axis, so the axis is halfway between them. For f(x) = (x − 1)(x + 3), the zeros are 1 and −3, so the axis is x = −1 without any formula.

Figure 3 · Two kinds of symmetry Figure 3 · Two kinds of symmetry (a) y = −2x² + 8x − 3 x y 1 2 3 4 −2 2 4 x = 2 vertex (2, 5) (b) y = x³ − 4x x y −2 −1 1 2 −3 3 (1, −3) (−1, 3) (a) Line symmetry: fold along x = 2. (b) Rotational symmetry: turn half a turn about the origin.
Figure 3 · Two kinds of symmetry

The special case of line symmetry in the y-axis (x = 0) is worth knowing by its algebra: it happens exactly when f(−x) = f(x) for every x. Even powers do this: x², x⁴, and 10 ÷ (x² + 1).

A graph has rotational symmetry about the origin if turning it through half a turn (180°) about (0, 0) leaves it unchanged. Figure 3(b) shows y = x³ − 4x: the point (1, −3) turns onto (−1, 3), and every point on one side matches one on the other, with the origin exactly halfway. The algebra is f(−x) = −f(x). Odd powers do this: x, x³, and 4x ÷ (x² + 1).

When a question asks you to describe a symmetry, name the line (x = 2) or the centre (the origin) precisely.

6Asymptotes

An asymptote is a straight line that a graph gets closer and closer to as it heads off towards infinity. At standard level you need two kinds, and the guide expects you to find them using graphing technology. Figure 4 shows both for y = (2x + 1) ÷ (x − 3).

Figure 4 · Vertical and horizontal asymptotes Figure 4 · Vertical and horizontal asymptotes x y −6 3 6 9 −6 −3 3 6 9 x = 3 y = 2 (−0.5, 0) (0, −⅓) y = (2x + 1)/(x − 3). Near x = 3 the curve shoots off; far to either side it settles towards y = 2.
Figure 4 · Vertical and horizontal asymptotes

A vertical asymptote, x = a, is where the function is undefined because it would divide by zero, and nearby the outputs grow without limit, up or down. Here the denominator x − 3 is zero at x = 3. The GDC shows the curve shooting off the top and bottom of the screen either side of x = 3, and the table shows why:

x2.92.9933.013.1
y−68−698undefined70272

A horizontal asymptote, y = b, is the value the outputs settle towards as x becomes very large, positive or negative. Scroll the table far out:

x1001000−1000
y2.072.0071.993

The outputs settle on 2, so the horizontal asymptote is y = 2. On Paper 1 you can see it without a table: for very large x the +1 and the −3 hardly matter, so (2x + 1) ÷ (x − 3) behaves like 2x ÷ x = 2. The other features follow in the usual way: the y-intercept is (0, 1 ÷ (−3)) = (0, −1/3), and the x-intercept is where the top is zero, 2x + 1 = 0, at (−0.5, 0).

Always write an asymptote as the equation of a line: x = 3 and y = 2, never just "3" and "2". On a sketch, draw it dashed and label it with that equation.

A horizontal asymptote is about what happens far away, not close up, so a graph can cross its horizontal asymptote. Figure 5 shows y = 4x ÷ (x² + 1), whose horizontal asymptote is y = 0: it passes straight through the origin, turns at (1, 2) and (−1, −2), and only then settles towards y = 0 on both sides. A graph can never cross a vertical asymptote, because the function has no value there.

Figure 5 · A curve may cross its horizontal asymptote Figure 5 · A curve may cross its horizontal asymptote x y −6 −3 3 6 −2 2 max (1, 2) min (−1, −2) approaches y = 0 approaches y = 0 y = 4x/(x² + 1) passes through the origin on y = 0, then approaches y = 0 again as x → ±∞.
Figure 5 · A curve may cross its horizontal asymptote

Two warnings about technology. A GDC in connected mode sometimes draws a steep near-vertical line joining the two branches at x = 3. That line is not part of the graph; the table proves it. And a GDC window that stops at x = 10 can make an approach to an asymptote look like a curve that is still falling. Check the table at large x before you claim an asymptote.

7Where two graphs meet

Two graphs meet where they have the same y-value at the same x. So the x-coordinates of the points of intersection of y = f(x) and y = g(x) are exactly the solutions of f(x) = g(x). That turns an equation into a picture, and a picture is something the GDC can read.

That matters because most equations cannot be solved by algebra. There is no algebraic method on the syllabus for 2ˣ = x + 3. The GDC solves it in seconds, two ways, as Figure 6 shows.

Figure 6 · Solving 2ˣ = x + 3 with technology, two ways Figure 6 · Solving 2ˣ = x + 3 with technology, two ways (a) y = 2ˣ and y = x + 3 x y −4 2 4 2 4 6 8 (−2.86, 0.137) (2.44, 5.44) y = 2ˣ y = x + 3 (b) y = 2ˣ − x − 3 x y −4 −2 2 4 2 4 x = −2.86 x = 2.44 (a) Intersect the two graphs. (b) Or graph the difference and find its zeros. The x-values are the same.
Figure 6 · Solving 2ˣ = x + 3 with technology, two ways

Method 1: intersect. Graph y = 2ˣ and y = x + 3 and use the intersect tool at each crossing. The graphs meet at (−2.86, 0.137) and (2.44, 5.44), so the solutions are x = −2.86 and x = 2.44.

Method 2: zeros of the difference. Rearrange to 2ˣ − x − 3 = 0, graph y = 2ˣ − x − 3, and use the zero tool twice. You get the same x-values. This is the method to reach for when one of the two graphs is awkward to enter, or when you want to be sure you have found every solution: count the zeros in a wide window.

Two checks before you write the answer. Have you found them all? Zoom out; an exponential and a line, or two curves, can meet more than once, and the second meeting is often off the default screen. Does the question want x-values or points? "Solve" wants x = …; "find the coordinates" wants (x, y).

Intersections have a real-world meaning whenever the two functions do. In economics, a market equilibrium is where a demand curve meets a supply curve: the price at which the quantity buyers want equals the quantity sellers offer. In 2.3, the break-even points were where revenue met cost. In each case, name the meaning in your answer, with units.

8Where marks are lost

Giving a y-value when a point was asked for, or the reverse. "The maximum value" is a number; "the maximum point" is a pair of coordinates. Read the question's words.

Calling a local maximum the greatest value on a restricted domain. The end points may be higher or lower. Evaluate them and compare.

Writing asymptotes as numbers. "Vertical asymptote 3" is not an equation. Write x = 3 and y = 2.

Believing a graph cannot cross a horizontal asymptote. It can, at finite x. The asymptote only describes what happens as x → ±∞.

Stopping at the first intersection. Many pairs of graphs meet twice or three times. Look at a wide window and count.

Rounding too early. Using 2.44 instead of the stored 2.4449… in a later part drifts the next answer off the mark scheme. Keep full values in the calculator; round what you write to 3 s.f.

Mixing up zeros and the y-intercept. Zeros come from f(x) = 0. The y-intercept comes from x = 0. They are different equations.

Trusting a steep line the GDC draws at a vertical asymptote. It is the calculator joining two branches, not part of the graph. Leave it out of your sketch.

9Work it right

  1. Read what is asked for: a value, an x-value, a point, or an equation of a line.
  2. On Paper 2, graph the function in a window that shows every feature before using any tool.
  3. Use one tool per feature: zero for each x-intercept, maximum or minimum for each turn, intersect for each meeting point.
  4. On a restricted domain, evaluate the end points too, and compare them with the turning points.
  5. For asymptotes, look for division by zero (vertical) and check the table at very large positive and negative x (horizontal). Write both as equations.
  6. For f(x) = g(x), either intersect the two graphs or find the zeros of f(x) − g(x); zoom out to be sure you have every solution.
  7. Write "using GDC" with the result, and give answers to 3 s.f. unless they are exact.

10Try it

Marks in brackets. Q1 and Q2 are Paper 1 style, no calculator. Q3 to Q5 are Paper 2 style, with a GDC.

Q1. Let f(x) = (x − 1)(x + 3).

(a) Write down the zeros of f. 2 marks

(b) Find the y-intercept of the graph of f. 1 mark

(c) Write down the equation of the axis of symmetry of the graph. 1 mark

(d) Find the coordinates of the vertex. 2 marks

Q2. Let g(x) = (3x − 6) ÷ (x + 2), for x ≠ −2. Write down the equations of the vertical and horizontal asymptotes of the graph of g, and find the coordinates of its intercepts with the axes. 4 marks

Q3. Let q(x) = 2x³ − 9x² + 7, for −1 ≤ x ≤ 5.

(a) Find the coordinates of the local maximum point and the local minimum point. 2 marks

(b) Find the zeros of q. 2 marks

(c) Find the range of q. 2 marks

Q4. Let f(x) = 10 ÷ (x² + 1) and g(x) = x + 2.

(a) Write down the equation of the horizontal asymptote of the graph of f. 1 mark

(b) Describe the symmetry of the graph of f. 1 mark

(c) Find the coordinates of the point where the graphs of f and g intersect. 2 marks

Q5. In a market for a new board game, the demand is modelled by p = 120 ÷ (q + 2) and the supply by p = 0.5q + 3, where q is the quantity in thousands and p is the price in euros. Find the equilibrium quantity and price, and state what they mean in context. 3 marks

11In one breath

A graph is described by its intercepts, turning points, symmetry and asymptotes. Zeros, roots and x-intercepts are the same numbers; the y-intercept is f(0). A local maximum or minimum is a turn, and on a restricted domain the end points can beat it, so evaluate them too. A parabola has an axis of symmetry through its vertex, halfway between its zeros; f(−x) = f(x) means symmetry in the y-axis, f(−x) = −f(x) means rotational symmetry about the origin. A vertical asymptote x = a sits where the function would divide by zero; a horizontal asymptote y = b is where outputs settle as x → ±∞, and a graph may cross it. Write asymptotes as equations. On Paper 2 the GDC finds every feature with its zero, maximum, minimum and intersect tools, one feature per use, answers to 3 s.f. And any equation f(x) = g(x) is solved where the two graphs cross, or where f(x) − g(x) is zero, so zoom out and find every crossing.


Answers

Q1. (a) x = 1 and x = −3. A1 for each. Writing (1, 0) and (−3, 0) is accepted for zeros only if the x-values are clear.

(b) f(0) = (−1)(3) = −3, so the y-intercept is (0, −3). A1.

(c) Halfway between the zeros: x = −1. A1. The answer must be an equation; "−1" alone scores 0.

(d) f(−1) = (−2)(2) = −4, so the vertex is (−1, −4). M1 for substituting their axis value into f, A1 for (−1, −4).

Q2. Vertical asymptote x = −2, since the denominator is zero there. Horizontal asymptote y = 3, since for large x the function behaves like 3x ÷ x. The y-intercept is g(0) = −6 ÷ 2, giving (0, −3). The x-intercept is where 3x − 6 = 0, giving (2, 0). A1 for each asymptote as an equation, A1 for each intercept as a point. Asymptotes given as numbers alone score A0.

Q3. (a) Using the GDC: local maximum (0, 7), local minimum (3, −20). A1 for each point.

(b) Using the GDC: x = −0.812, x = 1, x = 4.31 (3 s.f.). A2 for all three; A1 for two of them.

(c) The end points are q(−1) = −4 and q(5) = 32. Comparing with the turning points, the least value is −20 and the greatest is 32, at the end point. The range is −20 ≤ q(x) ≤ 32. M1 for evaluating both end points and comparing with the turning points, A1 for the range. "−20 ≤ q(x) ≤ 7", using the local maximum, scores M0 A0.

Q4. (a) y = 0. As x gets large, x² + 1 grows without limit, so 10 ÷ (x² + 1) approaches 0. A1 for the equation.

(b) The graph is symmetric in the y-axis (line symmetry in x = 0), since f(−x) = f(x). A1. "It is symmetrical", with no line named, scores 0.

(c) Using the GDC to intersect the graphs: (1.39, 3.39) (3 s.f.). There is only one point of intersection. A1 for x = 1.39, A1 for y = 3.39. Follow-through is not available for y if x is wrong.

Q5. Using the GDC, the graphs of p = 120 ÷ (q + 2) and p = 0.5q + 3 intersect at q = 11.6 and p = 8.81 (3 s.f.). The equilibrium is about 11 600 games at a price of €8.81: at that price, the number of games buyers want equals the number sellers are willing to supply. A1 for q = 11.6, A1 for p = 8.81, R1 for interpreting both in context with units. Leaving q as 11.6 with no reference to thousands scores R0.


Educerie · written from the published IB Diploma Programme Mathematics: analysis and approaches guide, first assessment 2021, section 2.4 Key features of graphs and intersections. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.

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