Home / Graphical Methods / Graph sketching and quadratic and cubic models
3.11b · Graphical Methods · Sub-skill
Sketch quadratic and cubic graphs from their key features, and use them to model real situations.
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Build it up, step by step
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For a sketch, find: where the curve crosses the $y$-axis (put $x = 0$), where it crosses the $x$-axis (the roots: put $y = 0$), the turning points, and what happens at the ends. A sketch needs the correct shape and labelled key points, not exact accuracy.
A quadratic $y = ax^2 + bx + c$ is a parabola: $\cup$ if $a > 0$ (a minimum) and $\cap$ if $a < 0$ (a maximum). If it factorises, $y = (x - p)(x - q)$ has roots $p$ and $q$. The turning point lies halfway between the roots; the $y$-intercept is $c$.
A cubic usually has an S-shape, going from bottom-left to top-right if the $x^3$ term is positive (the reverse if negative). It can cross the $x$-axis up to three times and has up to two turning points.
Quadratics model the height of a thrown object, the area of a rectangle with a fixed perimeter, or profit against price. The turning point gives the greatest or least value; the roots show when the quantity is zero (for a ball: when it is launched and when it lands). Solve inequalities by looking at where the curve is above or below a line.
Sketch $y = x^2 - 2x - 3$, marking the roots, the $y$-intercept and the turning point.
Factorise: $y = (x - 3)(x + 1)$, so the roots are $x = 3$ and $x = -1$.
$y$-intercept: $x = 0 \Rightarrow y = -3$.
Turning point: halfway between the roots at $x = 1$, where $y = 1 - 2 - 3 = -4$. The curve is a $\cup$ shape with minimum $(1, -4)$.
Test yourself
The height $h$ metres of a ball $t$ seconds after it is thrown is $h = 20t - 5t^2$.
(a) Sketch the graph for $0 \le t \le 4$, showing the greatest height and when the ball lands.
(b) For how long is the ball higher than 15 m? [5 marks]
(a) $h = 5t(4 - t)$: roots $t = 0$ and $t = 4$, so the ball lands after 4 seconds. The turning point is at $t = 2$: $h = 40 - 20 = $ 20 m. ∩-shaped curve through (0, 0), (2, 20), (4, 0). (3 marks)
(b) $20t - 5t^2 = 15 \Rightarrow t^2 - 4t + 3 = 0 \Rightarrow (t - 1)(t - 3) = 0$, so $t = 1$ and $t = 3$. The ball is above 15 m between these, for 2 seconds. (2 marks)
Practice
Five short questions. Work through them, then reveal the mark scheme to check.