| Command | What is expected |
|---|---|
| Plot | Mark points accurately on the given grid, then join them. |
| Sketch | A diagram, not necessarily to scale, showing the main features. |
| Draw | To an accuracy appropriate to the problem — a judgement call. |
Intersections with the $y$-axis · intersections with the $x$-axis
Behaviour for large $x$ (positive and negative)
$y$-intercept at $(0, c)$ · roots from factorising or the formula
Vertex from completing the square, or midway between the roots
Sketch $y = x^2 - 2x - 8$.
The leading coefficient — which way the ends point
| Polynomial | Ends | Max turning points |
|---|---|---|
| Positive cubic | Down on the left, up on the right | $2$ |
| Negative cubic | Up on the left, down on the right | $2$ |
| Positive quartic | Up at both ends | $3$ |
| Negative quartic | Down at both ends | $3$ |
Squared factor $(x-a)^2$ → the curve touches and turns back
Sketch $y = (x+1)(x-2)(x-3)$.
Sketch $y = x(x-3)^2$.
| Curve | Period | Range | Key features |
|---|---|---|---|
| $y = \sin x$ | $360°$ | $-1$ to $1$ | Through the origin; zeros at $0°, 180°, 360°$ |
| $y = \cos x$ | $360°$ | $-1$ to $1$ | Starts at $(0,1)$; zeros at $90°, 270°$ |
| $y = \tan x$ | $180°$ | All values | Asymptotes at $90°, 270°, \ldots$; zeros at $0°, 180°$ |
$y = \sin x + c$ — shifted up $c$ · $y = \sin(x - d)$ — shifted right $d$
Sketch $y = 3\sin(2x)$ for $0° \leqslant x \leqslant 360°$.
$a > 1$: growth · $0 < a < 1$: decay
The $x$-axis is an asymptote — the curve never reaches it
Sketch $y = 5 \times 2^x$, marking the intercept and describing the behaviour at each end.
Sketch vs plot
Sketch shows features; plot needs accuracy.
Label everything
Intercepts, turning points, asymptotes.
$y$-intercept
Set $x = 0$.
$x$-intercepts
Set $y = 0$ and solve.
Cubic ends
Positive: low-left to high-right.
Squared factor
Touches the axis, does not cross.
Turning points
At most degree $-\,1$ of them.
$\sin$, $\cos$
Period $360°$, range $-1$ to $1$.
$\tan$
Period $180°$, asymptotes at $90°$, $270°$.
Exponential
Through $(0,k)$, asymptote $y=0$, never negative.
State the gradient and $y$-intercept of $y = 5 - 2x$, and sketch it.
▶ Show solution
Gradient $-2$, $y$-intercept $(0, 5)$.
It slopes downwards. The $x$-intercept is where $5 - 2x = 0$, i.e. $(2.5, 0)$.
Find the intercepts of $y = x^2 - 5x + 6$ and state the shape.
▶ Show solution
$y$-intercept $(0, 6)$.
$x^2 - 5x + 6 = (x-2)(x-3)$, so roots at $(2,0)$ and $(3,0)$.
$a = 1 > 0$, so a U shape with a minimum, at $x = 2.5$ where $y = -0.25$.
Sketch $y = -x^2 + 4$, marking all intercepts.
▶ Show solution
$a = -1 < 0$, so an upside-down parabola with a maximum.
$y$-intercept $(0, 4)$ — this is also the vertex, since there is no $x$ term.
$x$-intercepts: $4 - x^2 = 0$ gives $x = \pm 2$, so $(-2, 0)$ and $(2, 0)$.
Describe the shape of $y = (x-1)(x+2)(x-4)$ and give its $y$-intercept.
▶ Show solution
A positive cubic (leading term $x^3$), crossing the axis at $x = -2, 1, 4$.
$y$-intercept: $(-1)(2)(-4) = 8$, so $(0, 8)$.
It rises from the bottom left, turns twice, and rises to the top right.
Where does $y = (x+2)^2(x-1)$ touch the $x$-axis, and where does it cross?
▶ Show solution
Touches at $x = -2$ — the factor is squared.
Crosses at $x = 1$ — a single factor.
State the amplitude and period of $y = 4\cos(3x)$.
▶ Show solution
Amplitude $4$, so the range is $-4$ to $4$.
Period $= \dfrac{360°}{3} = 120°$.
Write down the $y$-intercept of $y = 3 \times 4^x$ and describe its behaviour as $x \to -\infty$.
▶ Show solution
$y$-intercept: $3 \times 4^0 = 3$, so $(0, 3)$.
As $x \to -\infty$, $4^x \to 0$, so $y \to 0$ from above. The $x$-axis is an asymptote.
Sketch $y = 2^{-x}$ and explain its relationship to $y = 2^x$.
▶ Show solution
$2^{-x} = \left(\tfrac12\right)^x$, so this is exponential decay through $(0, 1)$.
It is the reflection of $y = 2^x$ in the $y$-axis: replacing $x$ by $-x$ always reflects a graph in the $y$-axis.
The $x$-axis remains an asymptote, approached as $x \to +\infty$.
State the number of times the line $y = 2$ crosses the curve $y = 3\sin x$ for $0° \leqslant x \leqslant 360°$, and explain why.
▶ Show solution
$y = 3\sin x$ has amplitude $3$, so it rises to $3$ at $x = 90°$ and falls to $-3$ at $x = 270°$.
The horizontal line $y = 2$ lies between $0$ and $3$, so it cuts the curve on the way up and again on the way down within the first hump.
In the second half ($180°$ to $360°$) the curve is negative, so it never reaches $2$.
The line crosses twice.
(Solving: $\sin x = \tfrac23$ gives $x = 41.8°$ and $x = 138.2°$ ✓)
Consider $\mathrm{f}(x) = x^3 - 4x$.
(a) Factorise $\mathrm{f}(x)$ fully and state its roots. (b) Describe the end behaviour. (c) Sketch the curve, and use the sketch to solve $\mathrm{f}(x) > 0$. (d) Explain how the sketch of $y = \mathrm{f}(x) + 5$ differs, and how many roots it has.
▶ Show solution
(a) $x^3 - 4x = x\left(x^2 - 4\right) = x(x-2)(x+2)$
Roots at $x = -2$, $0$ and $2$, all single factors so the curve crosses at each.
(b) A positive cubic: as $x \to -\infty$, $y \to -\infty$; as $x \to +\infty$, $y \to +\infty$.
(c) The curve rises through $(-2,0)$, reaches a maximum, falls through the origin, reaches a minimum, then rises through $(2,0)$.
So it is above the axis between $-2$ and $0$, and again beyond $2$:
$-2 < x < 0$ or $x > 2$
Check at $x = -1$: $-1 + 4 = 3 > 0$ ✓ and at $x = 1$: $1 - 4 = -3 < 0$ ✓
(d) Adding $5$ translates the whole curve up by $5$; the shape is unchanged.
The local maximum of $\mathrm{f}$ occurs at $x = -\tfrac{2}{\sqrt3} \approx -1.155$, where $\mathrm{f}(x) \approx 3.08$, and the local minimum at $x \approx 1.155$ where $\mathrm{f}(x) \approx -3.08$.
After translating up $5$, the maximum sits at about $8.08$ and the minimum at about $1.92$ — both above the axis.
So the translated curve only crosses the $x$-axis once, on its way up from the bottom left. $y = \mathrm{f}(x) + 5$ has exactly one root.