2.4 Key features of graphs

Introduction

A graph makes visible facts about a function that are hard to see in its formula: the sales level at which a company stops making a loss, the time at which a drug in the bloodstream reaches its highest level, or the price at which buyers and sellers agree. Each of these is a particular point or line on a graph, such as a crossing of an axis, the top of a hump, a level that the curve approaches, or the place where two graphs meet. These are the key features of the graph, and reading them precisely with technology is a skill used throughout applied mathematics.

Population data give a real example. For decades China had the largest population in the world, but India's population was growing faster, and the two population curves met. The point where they cross marks the year India became the most populous country, a fact read straight from the intersection of two graphs.

Reading features with technology

Subtopic 2.3, Graphing functions with technology, showed how to graph a function, choose a viewing window and transfer the result to paper. Reading a feature precisely needs the GDC's analysis tools, which solve for the feature numerically instead of estimating it from the pixels. On a TI-84 Plus CE they are in the CALC menu (2nd TRACE); on a Casio fx-CG50 in G-SOLVE (SHIFT F5). The names vary a little, but every GDC offers the same set.

Each tool works on a stretch of curve chosen by the user. A TI-84 Plus CE asks for a "left bound" and a "right bound" on either side of the feature and then a guess; a Casio fx-CG50 moves from one feature to the next with the arrow keys. Choosing bounds that contain only the feature wanted prevents the calculator from finding a different one nearby.

Each kind of feature has its own written form. A point has two coordinates and is written in brackets, $(2.88, 4.33)$. A line, such as an asymptote or an axis of symmetry, is written as an equation, $x = 1$ or $y = 3$, because the single number 1 does not say which line is meant. A solution of an equation is a value of $x$, written $x = -1.43$, not as a point.

Intercepts and zeros

The points where a graph crosses the axes are usually the first features to find, because they are the easiest to interpret: the starting value of a model, and the inputs at which its output is zero.

The y-intercept

The $y$-intercept is where the graph crosses the vertical axis. Every point on that axis has $x = 0$, so the $y$-intercept is the point $(0, f(0))$. It exists only if 0 is in the domain, and a function has at most one $y$-intercept, because it gives only one output for the input 0. In a model with time as the input, $f(0)$ is the starting value: the initial temperature, the fixed cost when nothing is produced, the height at launch.

Zeros of a function and roots of an equation

The $x$-intercepts are where the graph meets the horizontal axis, so their $y$-coordinates are 0. The inputs at which this happens are the zeros of the function, and they are exactly the roots of the equation $f(x) = 0$.

The three descriptions are one fact seen in three representations: a value of $x$ (zero), a solution (root) and a point on the graph (intercept). A function can have no zeros, one, or many; a horizontal line $y = 5$ never meets the $x$-axis, and a sinusoidal model can cross it again and again.

Maximum and minimum points

A graph that rises and then falls has a highest point between the two, and one that falls and then rises has a lowest point. These points matter in almost every application: the greatest profit, the lowest temperature, the peak concentration of a drug.

Local maxima and minima

The highest point of a hump is called a local maximum because it is highest only compared with the points near it; a taller part of the graph may lie elsewhere.

The same idea upside down describes the bottom of a dip, the lowest point among its neighbours.

On a smooth curve, a local maximum or minimum is where the graph changes direction. Moving from left to right, a rising graph that starts to fall has just passed a local maximum, and a falling graph that starts to rise has just passed a local minimum.

The cubic $f(x) = x^3 - 6x^2 + 9x + 1$ has two turning points. The GDC's maximum tool, with bounds either side of the hump, gives $(1, 5)$; the minimum tool gives $(3, 1)$. Between them the graph falls, and outside them it rises. A description of where a graph rises and falls is written with inequalities: $f$ is increasing for $x < 1$ and $x > 3$, and decreasing for $1 < x < 3$. The formal terms increasing and decreasing and the use of calculus to find turning points exactly belong to subtopics 5.2 and 5.6.

Greatest and least values on a domain

A local maximum need not be the highest point of the whole graph. The cubic's local maximum has $y = 5$, but the graph rises beyond 5 when $x > 4$, and on the domain $0 \le x \le 5$ its greatest value is $f(5) = 125 - 150 + 45 + 1 = 21$, at an end point. The greatest and least outputs on a domain are the maximum and minimum values of the function there, and they occur either at turning points or at the ends of the domain, so both must be checked.

So on $0 \le x \le 5$ the greatest value is 21 (at $x = 5$) and the least value is 1 (at both $x = 0$ and $x = 3$). The range of $f$ on this domain is therefore $1 \le f(x) \le 21$, which is how the range of a function is found from its graph (subtopic 2.2).

Symmetry

Many graphs are symmetrical: one half is the mirror image of the other in a line. A vertical line of symmetry $x = a$ means that inputs the same distance either side of $a$ give the same output. Once one side of a symmetrical graph is known, the other side follows by reflection.

Every parabola has a vertical axis of symmetry through its vertex, its turning point (subtopic 2.5 gives the formula $x = -\frac{b}{2a}$). With technology, the axis is found from the vertex: the maximum or minimum tool gives the vertex, and its $x$-coordinate gives the equation of the axis. Symmetry also locates the vertex from two points at the same height: if a parabola passes through $(1, 7)$ and $(5, 7)$, its axis is halfway between them, $x = 3$.

Other graphs have symmetry too. The function $h(x) = \frac{8}{x^2 + 2}$ gives the same output for $x$ and $-x$, so its graph is symmetrical in the $y$-axis, the line $x = 0$; its maximum point is $(0, 4)$. A sinusoidal model such as $y = 80 - 35\cos(90t°)$ has a line of symmetry through every maximum and minimum, and is also balanced above and below its principal axis.

Asymptotes

Some graphs approach a straight line more and more closely without settling on it. On a GDC the curve and the line look almost the same far from the origin, or the curve shoots up beside a vertical line where the function is not defined.

Two kinds occur in this course, and both are written as equations of lines.

A vertical asymptote usually comes from division by zero. In $f(x) = \frac{4}{x - 1} + 3$, the denominator is zero at $x = 1$. Inputs just above 1 make $x - 1$ a tiny positive number, so $\frac{4}{x - 1}$ is huge and positive; inputs just below 1 make it huge and negative.

For the same function, as $x$ becomes very large, $\frac{4}{x - 1}$ becomes tiny, so $f(x)$ gets closer and closer to 3. The table shows both behaviours numerically, which is how a GDC's table function identifies an asymptote when a formula is unfamiliar.

The outputs explode as $x$ approaches 1 from either side: vertical asymptote $x = 1$. They settle towards 3 at both ends: horizontal asymptote $y = 3$.

Finding asymptotes with technology follows a routine:

1. Look for inputs where the formula breaks down (a zero denominator). Graph near them and use the table with inputs very close on each side; outputs that grow without limit confirm a vertical asymptote $x = a$.

2. Use the table with very large positive and negative inputs (1000, 10 000, −10 000). If the outputs settle towards a number $b$, the line $y = b$ is a horizontal asymptote.

3. Write each asymptote as an equation and draw it dashed on any sketch, with the curve approaching it but never touching it near the ends.

A graph never crosses a vertical asymptote, because the function is undefined there. A horizontal asymptote describes only the ends of the graph, so a curve may cross it at a finite input and still approach it far out: $y = \frac{x}{x^2 + 1}$ crosses its asymptote $y = 0$ at the origin. Asymptotes of the standard models ($y = c$ for an exponential model, the axes for $y = ax^n$ with $n < 0$) are applied in subtopic 2.5.

Points of intersection

When two graphs are drawn on the same axes, the points they share solve a problem that involves both functions.

At a point of intersection of $y = f(x)$ and $y = g(x)$ the two graphs have the same $y$-value at the same $x$, so $f(x) = g(x)$. The $x$-coordinates of the intersections are therefore exactly the solutions of that equation. This gives a method that works for almost any equation in one unknown, including ones with no algebraic method of solution at this level.

Solving f(x) = g(x) with technology

The method has four steps, and each produces something worth writing down:

1. Write the equation being solved, for example $2^x = x + 3$.

2. Enter each side as its own function: $Y_1 = 2^x$, $Y_2 = x + 3$.

3. Choose a window that shows every intersection: zoom out first, count the crossings, then zoom in on each.

4. Use the intersect tool once for each crossing (on a TI-84 Plus CE: select the first curve, the second curve, then move the cursor near the crossing as a guess), and write the solutions as $x = \ldots$

The same method solves $f(x) = k$ for a constant $k$: graph $Y_2 = k$, a horizontal line, and find where it meets the curve. Solving $f(x) = 0$ is the special case $k = 0$, where the line is the $x$-axis and the intersections are the zeros. An equation can also be rearranged to $f(x) - g(x) = 0$ and solved with the zero tool; both methods give the same $x$-values.

Inequalities and lengths of time

Many questions in context ask for how long, or for which inputs, a quantity is above or below a level. Graphing the level as a horizontal line and finding where it meets the curve answers both: the intersections are the boundaries, and the graph shows which side of each boundary is above the line.

Case study: India overtaking China

For most of recorded history China was the world's most populous country. The United Nations' population estimates show India's population growing by more than 11 million a year in the early 2020s while China's peaked in 2021 and began to fall, so the two population curves had to cross.

Source: United Nations, World Population Prospects 2024, via ourworldindata.org, accessed October 2026.

Between the estimates for 1 July 2021 and 1 July 2022, each population can be modelled by a straight line, with $t$ the time in years after 1 July 2021:

Graphing $Y_1 = 1414.20 + 11.22x$ and $Y_2 = 1426.44 - 1.26x$ for $0 \le x \le 1$ and using the intersect tool gives $t = 0.980\ldots$ and a population of about 1425.2 million. The crossing is about 0.98 of a year after 1 July 2021, in the middle of 2022, when each country had about 1.425 billion people.

The result depends on both the data and the model. Joining two yearly estimates with a straight line assumes steady change within the year, and population estimates themselves are revised as new census and survey data arrive. The UN's population dashboard in 2023, based on the previous revision of the estimates, showed India passing China by 1 July 2023; the 2024 revision used in the table moves the crossing to 2022. Intersecting the two curves is the right method in both cases, but a revision of the data moves the point of intersection.

Break-even and market equilibrium

Two applications of intersections from business and economics appear again and again, because they answer the most basic questions a business asks: how much must be sold to stop losing money, and at what price will the market settle.

Break-even points

A business makes a loss when its costs are greater than its revenue and a profit when revenue is greater. In subtopic 2.3 the profit was found as the difference $R(x) - C(x)$; the same boundary appears here as the crossing of two graphs.

Market equilibrium

In a market, the demand for a product falls as its price rises, while the supply that producers offer rises with the price. Graphed with quantity $Q$ on the horizontal axis and price $P$ on the vertical axis (the economists' convention), the demand curve slopes down and the supply curve slopes up.

At a higher price, sellers offer more than buyers want and prices tend to fall; at a lower price, buyers want more than is offered and prices tend to rise. Only at the equilibrium price are the two quantities equal, so there is no pressure on the price to change.

Exploring parameters with sliders

A formula such as $y = ax^2 + bx + c$ describes a whole family of graphs, one for each choice of the parameters $a$, $b$ and $c$. Graphing software such as Desmos and GeoGebra can turn a parameter into a slider, so that its effect on the key features can be watched as it changes.

Sliders show how each key feature depends on the parameters. With $y = ax^2 + bx + c$, dragging $c$ moves the whole graph up and down and changes only the $y$-intercept $(0, c)$; dragging $a$ makes the parabola narrower or wider and turns it upside down when $a$ becomes negative; dragging $b$ moves the vertex along a curved path while the $y$-intercept stays fixed. For $y = \frac{k}{x - h} + v$, sliders for $h$ and $v$ move the asymptotes $x = h$ and $y = v$, and the graph moves with them.

A slider does not replace finding a feature accurately. It suggests a relationship, such as "the horizontal asymptote is always $y = v$", which is then confirmed with the table or the analysis tools for particular values. Subtopic 2.5 uses this approach to interpret the parameters of each model.

Exam focus

Key features are examined on both papers. On Paper 1, a short question of 4 to 6 marks may ask you to write down a zero, the coordinates of a local minimum or the equation of an asymptote, then solve $f(x) = g(x)$. On Paper 2, key features are parts of longer modelling questions: the maximum height of a projectile, the break-even point of a business, the time a quantity stays above a level, often after a sketch. Questions that ask you to determine whether something fits or is possible need a reason as well as a conclusion.

What must be written down:

• for a "write down" feature, the answer in the right form: coordinates in brackets, asymptotes and axes of symmetry as equations, zeros and solutions as $x = \ldots$;

• for "find" or "solve", the equation you entered (for example "$f(x) = 3 - x$") or a sketch showing the intersection, then the answer to 3 s.f.;

• for "determine whether" or "justify", the values compared and a conclusion in context.

Part (d) shows the most frequent structure in context questions: the boundaries come from intersections and the answer is their difference. Giving one intersection value, or substituting 2 for $t$, loses most of the marks.

The vertical asymptote of a function given on a domain such as $x > 0$ may lie at the edge of the domain (the $y$-axis), so it still counts as a key feature on a sketch and as an answer to "write down the equation of the vertical asymptote".

Practice questions

1. Let $f(x) = x + \frac{12}{x^2}$, $x \ne 0$. (a) Write down the zero of $f$. (b) Write down the coordinates of the local minimum point. [4]

2. Let $g(x) = \frac{6}{x + 2} - 1$. Write down the equations of the vertical and horizontal asymptotes of the graph of $g$, and the $y$

3. Solve the equation $x^3 - 3x = 1$. [3]

4. A parabola passes through the points $(-1, 6)$ and $(7, 6)$ and has a maximum point. Write down the equation of its axis of symmetry. [2]

5. A café's monthly revenue from selling $x$ cakes is $R(x) = 4.5x$ and its monthly cost is $C(x) = 900 + 1.5x$, both in dollars. (a) Find the break-even point. (b) Find the profit when 500 cakes are sold. [4]

6. The demand for a product is $P = 60 - 2Q$ and the supply is $P = 0.5Q + 10$. Find the equilibrium price and quantity. [3]

7. The height of a ball, $h$ m, $t$ seconds after it is kicked is $h(t) = 15t - 4.9t^2$, $t \ge 0$

8. Two internet plans charge $A(x) = 25 + 0.8x$ and $B(x) = 5 \times 1.12^{x}$ dollars for $x$

9. An archway is modelled by $y = -0.9x^2 + 4.5x$, where $x$ and $y$ are in metres and $y$

Summary

• Analysis tools on the GDC (value, zero, maximum, minimum, intersect) find key features accurately; the trace cursor does not.

• The $y$-intercept is $(0, f(0))$; the zeros of $f$ are the roots of $f(x) = 0$ and the $x$-coordinates of the $x$-intercepts.

• A local maximum or minimum is a turning point, highest or lowest compared with nearby points; the greatest and least values on a domain may instead be at an end point.

• A vertical line of symmetry $x = a$ means $f(a - d) = f(a + d)$; a parabola's axis of symmetry passes through its vertex.

• A vertical asymptote $x = a$ occurs where outputs grow without limit (often a zero denominator); a horizontal asymptote $y = b$ is the value the outputs approach for very large $|x|$. Both are written as equations.

• The solutions of $f(x) = g(x)$ are the $x$-coordinates of the points of intersection of the two graphs; $f(x) = k$ uses the horizontal line $y = k$

• Lengths of time or intervals above a level come from two intersections and their difference.

• Break-even: revenue graph meets cost graph. Market equilibrium: demand curve meets supply curve.

• Sliders in graphing software show how parameters change key features; the features are then confirmed with the analysis tools.

Self-check

• To do:

  I can use the value, zero, maximum, minimum and intersect tools on my GDC, choosing bounds correctly.

• To do:

  I can write down a $y$-intercept as $(0, f(0))$.

• To do:

  I can find the zeros of a function and give them as $x = \ldots$

• To do:

  I can find the coordinates of local maximum and minimum points to 3 significant figures.

• To do:

  I can find the greatest and least values of a function on a domain by checking turning points and end points.

• To do:

  I can write the equation of a line of symmetry and use symmetry to find a vertex or a missing point.

• To do:

  I can identify vertical and horizontal asymptotes with the GDC's graph and table, and write them as equations.

• To do:

  I can solve $f(x) = g(x)$ and $f(x) = k$ graphically, finding every solution.

• To do:

  I can find the values of $x$ for which one graph is above another.

• To do:

  I can find the length of time for which a quantity is above or below a given level.

• To do:

  I can find a break-even point and interpret it in context.

• To do:

  I can find the market equilibrium from linear supply and demand functions.

• To do:

  I can check whether an object fits under a curve by testing the height at its edges.

• To do:

  I can describe how a slider for a parameter changes the key features of a graph.

1. (a)

   Calculate the discriminant of $f(x)$.

   [1]1 mark

   Markscheme (a)

   $\Delta = 36 - 4(1)(34) = -100$ A1

   [1 mark]

2. (b)

   Hence explain why the graph of $y = f(x)$ does not intersect the $x$-axis.

   [1]1 mark

   Markscheme (b)

   $\Delta < 0$ so the equation $f(x) = 0$ has no real roots (so the graph has no $x$-intercepts) R1

   [1 mark]

3. (c)

   Solve the equation $f(x) = 0$, giving your answers in the form $a + b\mathrm{i}$.

   [2]2 marks

   Markscheme (c)

   $x = \dfrac{-6 \pm \sqrt{-100}}{2}$

4. (d)

   Write down the product of the two solutions.

   [1]1 mark

   Markscheme (d)

   $34$ A1

   [1 mark]

(b) GDC working. Graph $Y_1 = x^3 - 6x^2 + 9x + 1$ in the window $-2 \le x \le 5$, $-5 \le y \le 8$. The curve crosses the $x$-axis once, just to the left of the origin. The zero tool, with bounds $x = -1$ and $x = 0$, gives $x = -0.103803\ldots$

The zero is $x = -0.104$ (3 s.f.).

Zooming out confirms there are no other crossings: to the right the curve dips only to a lowest point of $y = 1$ at $x = 3$ before rising for ever, so it never returns to the axis. A polynomial root finder (subtopic 1.8) gives the same real root together with two non-real ones, which are not zeros on a graph.

The maximum concentration is 19.8 mg per litre, about 4.48 hours after the tablet is taken.

(b) $C(12) = 12 \times 12 \times 0.8^{12} = 9.8956\ldots$, so about 9.90 mg per litre after 12 hours. The least value on the domain is $C(0) = 0$, at the start, and the greatest is the maximum. The range is $0 \le C \le 19.8$.

The local maximum is the greatest value here because the curve has only one turning point and both end values are lower.

(b) Centred on the axis, the banner stretches from $x = -30$ to $x = 30$. The arch is lowest over the banner at its edges, because the parabola falls away from the centre. By symmetry both edges give the same height:

Across the whole width of the banner the arch is at least 173.25 m high, which is more than 150 m, so the banner fits, with at least 23.25 m to spare.

(c) GDC working. Graph $Y_1 = 192 - x^2/48$ and $Y_2 = 150$; the intersect tool gives $x = -44.899\ldots$ and $x = 44.899\ldots$ The width is $2 \times 44.899\ldots = 89.8$ m.

The real arch is a weighted catenary, slightly flatter at the top and steeper at the sides than this parabola, so the model's heights near the base are only approximate.

In context: the soup cools towards the temperature of the room, 21 °C, but according to the model never quite reaches it.

• $x = -0.23606\ldots$, $y = -1.4721\ldots$, so $(-0.236, -1.47)$;

• $x = 2$, $y = 3$, so $(2, 3)$ (exactly: $f(2) = 8 - 24 + 18 + 1 = 3$ and $g(2) = 3$);

• $x = 4.2360\ldots$, $y = 7.4721\ldots$, so $(4.24, 7.47)$.

(b) The cubic is above the line between the first two crossings and after the third. So $f(x) > g(x)$ for $-0.236 < x < 2$ and $x > 4.24$.

(b) GDC working (degree mode). Graph $Y_1 = 14 - 8\cos(30x)$ and $Y_2 = 18$ with window $0 \le x \le 12$, $0 \le y \le 25$. The intersect tool gives $x = 4$ and $x = 8$. Between these the curve is above the line, so the temperature is above 18 °C for

Equating the model to 18 (rather than substituting 18 for $t$) is the step that earns the method mark.

The equilibrium quantity is $Q = 20$, that is 2000 kg a week, at an equilibrium price of $10 per kg.

in the long run the concentration of the drug in the blood approaches zero (the drug leaves the body) R1

(d) $\frac{12t}{t^2 + 4} = 2$ or a sketch of $C(t)$ and $y = 2$ with intersections shown (M1)

$t = 0.763932\ldots$ and $t = 5.23606\ldots$ (A1)

length of time $= 5.23606\ldots - 0.763932\ldots = 4.47$ hours $(4.47213\ldots)$ A1

(e) the maximum concentration is 3 mg per litre, and $3 < 3.5$ R1

so the doctor is not correct A1

Note: do not award R0A1.

Total [10]

$x = -3 \pm 5\mathrm{i}$ A1

[2 marks]

[2]2 marks

[2]2 marks