DfE content area B (Algebra and functions) asks you to understand and use the graphs of functions; to sketch curves defined by simple equations, including polynomials, $y = \dfrac{a}{x}$ and $y = \dfrac{a}{x^2}$ with their vertical and horizontal asymptotes; to interpret the intersection points of graphs as solutions of equations; and to use and understand the graphs of functions in direct and inverse proportion.
- Shape: what type of curve is it, and is the leading coefficient positive or negative?
- $y$-intercept: put $x = 0$.
- $x$-intercepts: put $y = 0$ — factorise first if needed (see Polynomials and the Factor Theorem).
- Repeated roots and asymptotes: mark them clearly.
| Leading coefficient | As $x \to -\infty$ | As $x \to +\infty$ | Overall shape |
|---|---|---|---|
| Positive, e.g. $y = x^3$ | $y \to -\infty$ | $y \to +\infty$ | Starts bottom-left, ends top-right |
| Negative, e.g. $y = -x^3$ | $y \to +\infty$ | $y \to -\infty$ | Starts top-left, ends bottom-right |
Squared factor $(x - a)^2$: the curve touches the axis at $x = a$ and turns back
Cubed factor $(x - a)^3$: the curve flattens and passes through (a point of inflection, like $y = x^3$ at $0$)
Sketch $y = (x + 2)(x - 1)(x - 3)$.
Sketch $y = (1 - x)(x + 2)^2$.
Sketch $y = x^3 - x^2 - 6x$.
A quartic (degree $4$) has both ends going the same way, like a quadratic: both up if the leading coefficient is positive, both down if negative. It has up to four real roots and up to three turning points — a "W" or "M" shape.
Sketch $y = (x + 2)(x + 1)(x - 1)(x - 3)$.
| Graph | $a > 0$ | $a < 0$ | Symmetry |
|---|---|---|---|
| $y = \dfrac{a}{x}$ | Branches in 1st and 3rd quadrants | Branches in 2nd and 4th quadrants | Rotational, about the origin |
| $y = \dfrac{a}{x^2}$ | Both branches above the $x$-axis | Both branches below the $x$-axis | Reflection in the $y$-axis |
Describe the sketches of (a) $y = -\dfrac{2}{x}$ (b) $y = \dfrac{3}{x^2}$, and state their asymptotes.
$y = \mathrm{f}(x)$ and $y = \mathrm{g}(x)$ intersect
So the number of real solutions of an awkward equation can be found by sketching two simple graphs and counting crossings — even when you cannot solve the equation algebraically.
On the same axes sketch $y = x(x - 4)$ and $y = \dfrac{3}{x}$. Hence show that $x^3 - 4x^2 - 3 = 0$ has exactly one real root.
| Statement | Symbol | Equation | Graph of $y$ against $x$ |
|---|---|---|---|
| $y$ is directly proportional to $x$ | $y \propto x$ | $y = kx$ | Straight line through the origin |
| $y$ is directly proportional to $x^2$ | $y \propto x^2$ | $y = kx^2$ | Parabola with vertex at the origin |
| $y$ is inversely proportional to $x$ | $y \propto \dfrac1x$ | $y = \dfrac{k}{x}$ | Reciprocal curve, asymptotes the axes |
| $y$ is inversely proportional to $x^2$ | $y \propto \dfrac1{x^2}$ | $y = \dfrac{k}{x^2}$ | Inverse-square curve |
$y$ is inversely proportional to $x^2$, and $y = 5$ when $x = 2$. Find $y$ when $x = 4$, and the positive value of $x$ when $y = 80$.
The time $T$ hours taken to dig a trench is inversely proportional to the number of workers $n$. Six workers take $10$ hours. Find a formula for $T$, sketch the graph for $n > 0$, and find how long $15$ workers take.
Positive cubic
Bottom-left to top-right.
Negative cubic
Top-left to bottom-right.
Quartic
Both ends up ($a>0$) or both down ($a<0$).
Single root
Curve crosses the axis.
Squared root
Curve touches the axis and turns back.
$y = a/x$
Asymptotes $x=0$, $y=0$; quadrants 1 & 3 if $a>0$.
$y = a/x^2$
Asymptotes $x=0$, $y=0$; both branches on the same side.
Intersections
Solutions of $\mathrm{f}(x)=\mathrm{g}(x)$ are where the graphs meet.
Proportion
$y \propto x^n \Rightarrow y = kx^n$; find $k$ from one pair.
Sketch $y = (x - 1)(x + 2)(x - 4)$, showing the coordinates of the points where the curve meets the axes.
▶ Show solution
Positive cubic (bottom-left to top-right).
Crosses the $x$-axis at $(-2, 0)$, $(1, 0)$ and $(4, 0)$.
$y$-intercept: $(-1)(2)(-4) = 8$, so $(0, 8)$.
Shape: up through $-2$ to a maximum, down through $(0, 8)$ and $(1, 0)$ to a minimum, up through $(4, 0)$.
Sketch $y = x^2(3 - x)$.
▶ Show solution
The $x^3$ term is $-x^3$: negative cubic, top-left to bottom-right.
$x^2$ is squared, so the curve touches the axis at the origin (a minimum there).
Crosses at $(3, 0)$. For $0 < x < 3$, $y > 0$, so the curve rises from the origin to a maximum and falls through $(3, 0)$.
Factorise $x^3 - 6x^2 + 9x$ and hence sketch $y = x^3 - 6x^2 + 9x$.
▶ Show solution
$x\left(x^2 - 6x + 9\right) = x(x - 3)^2$.
Positive cubic. Crosses at $(0, 0)$; touches at $(3, 0)$.
From bottom-left, up through the origin to a maximum, down to touch the axis at $x = 3$ (a minimum), then up.
Sketch $y = (x + 1)^2(x - 2)^2$, giving the coordinates of any points where it meets the axes.
▶ Show solution
Positive quartic, both ends up. Both factors are squared, so $y \geqslant 0$ everywhere.
Touches the $x$-axis at $(-1, 0)$ and $(2, 0)$.
$y$-intercept: $(1)^2(-2)^2 = 4$, so $(0, 4)$.
A "W" shape sitting on the axis at $-1$ and $2$ with a maximum in between (the curve passes through $(0, 4)$).
Sketch $y = -\dfrac{3}{x^2}$ and state the equations of its asymptotes. Explain why the graph has no point in the first quadrant.
▶ Show solution
Asymptotes: $x = 0$ and $y = 0$.
$x^2 > 0$ for every $x \neq 0$, so $-\dfrac{3}{x^2} < 0$: the whole graph lies below the $x$-axis — in the 3rd and 4th quadrants — and none of it is in the first.
The two branches are reflections of each other in the $y$-axis, pass through $(\pm1, -3)$, plunge towards $-\infty$ near $x = 0$ and rise towards $y = 0$ as $x \to \pm\infty$.
$y$ is directly proportional to $x^2$, and $y = 18$ when $x = 3$. (a) Find $y$ when $x = 5$. (b) Find the positive value of $x$ when $y = 8$.
▶ Show solution
$y = kx^2$: $18 = 9k$, so $k = 2$ and $y = 2x^2$.
(a) $y = 2(25) = 50$.
(b) $8 = 2x^2 \Rightarrow x^2 = 4 \Rightarrow x = 2$.
At constant temperature the pressure $P$ kPa of a gas is inversely proportional to its volume $V$ litres. When $V = 2$, $P = 150$. (a) Find $P$ when $V = 5$. (b) Sketch the graph of $P$ against $V$. (c) What graph would give a straight line through the origin?
▶ Show solution
(a) $P = \dfrac{k}{V}$; $150 = \dfrac{k}{2}$ so $k = 300$. $P = \dfrac{300}{5} = 60$ kPa.
(b) The first-quadrant branch of a reciprocal curve, through $(2, 150)$ and $(5, 60)$, approaching both axes.
(c) $P = 300 \times \dfrac{1}{V}$, so plotting $P$ against $\dfrac1V$ gives a straight line through the origin with gradient $300$.
The curve $y = \dfrac{k}{x}$ passes through $(2, 6)$. Find $k$, and find the coordinates of the points where the curve meets the line $y = x + 1$.
▶ Show solution
$6 = \dfrac{k}{2}$, so $k = 12$.
At an intersection $\dfrac{12}{x} = x + 1$, so $12 = x^2 + x$ and $x^2 + x - 12 = 0$.
$(x + 4)(x - 3) = 0$, so $x = 3$ or $x = -4$.
Points: $(3, 4)$ and $(-4, -3)$ — one on each branch of the curve.
On the same axes sketch $y = (x - 1)(x - 3)$ and $y = \dfrac{1}{x^2}$. Hence state the number of real solutions of $x^2(x - 1)(x - 3) = 1$.
▶ Show solution
The parabola is $\cup$-shaped through $(1, 0)$, $(3, 0)$ and $(0, 3)$. $y = \dfrac{1}{x^2}$ has both branches above the $x$-axis, asymptotes $x = 0$ and $y = 0$.
For $x < 0$: as $x \to -\infty$ the parabola is huge and $\dfrac1{x^2}$ is tiny; as $x \to 0^-$ the parabola tends to $3$ and $\dfrac1{x^2} \to \infty$. So they cross once.
For $0 < x < 1$: $\dfrac1{x^2} > 1$ while the parabola is less than $3$ and falls to $0$; the reciprocal curve stays above (e.g. at $x = 0.5$: $4$ against $1.25$). No crossing. For $1 < x < 3$ the parabola is negative: no crossing.
For $x > 3$ the parabola rises from $0$ and $\dfrac1{x^2}$ falls: one crossing.
Intersections satisfy $(x-1)(x-3) = \dfrac{1}{x^2}$, i.e. $x^2(x - 1)(x - 3) = 1$. So there are 2 real solutions (approximately $x \approx -0.45$ and $x \approx 3.05$).
$\mathrm{f}(x) = x^3 - 3x^2 - x + 3$.
(a) Factorise $\mathrm{f}(x)$ completely. (b) Sketch $y = \mathrm{f}(x)$, showing where it meets the axes. (c) Add the line $y = 3$ to your sketch and explain how many solutions $\mathrm{f}(x) = 3$ has. (d) Find these solutions exactly.
▶ Show solution
(a) Grouping: $x^2(x - 3) - (x - 3) = (x - 3)\left(x^2 - 1\right) = (x - 3)(x - 1)(x + 1)$.
(b) Positive cubic crossing at $(-1, 0)$, $(1, 0)$, $(3, 0)$, with $y$-intercept $(0, 3)$. It has a maximum between $-1$ and $1$ and a minimum between $1$ and $3$.
(c) The horizontal line $y = 3$ passes through the $y$-intercept $(0, 3)$, so that is one solution. Just to the left, $\mathrm{f}(-0.2) = 3.072 > 3$, while $\mathrm{f}(-1) = 0 < 3$, so the curve crosses $y = 3$ again between $x = -1$ and $x = -0.2$. On the right, the curve falls to its minimum below the axis and then rises without limit, so it crosses $y = 3$ a third time beyond $x = 3$. Three intersections, so three solutions.
(d) $x^3 - 3x^2 - x + 3 = 3 \Rightarrow x^3 - 3x^2 - x = 0 \Rightarrow x\left(x^2 - 3x - 1\right) = 0$.
$x = 0$ or $x = \dfrac{3 \pm \sqrt{9 + 4}}{2} = \dfrac{3 \pm \sqrt{13}}{2}$.
So $x = 0$, $x = \dfrac{3 - \sqrt{13}}{2} \approx -0.303$, $x = \dfrac{3 + \sqrt{13}}{2} \approx 3.303$.