Graphs of quadratic functions

AS · P1 · 13 min

The graph of $y = ax^2 + bx + c$ is a parabola, a symmetric $\cup$ or $\cap$ shaped curve. Being able to sketch it quickly, with its intercepts and vertex labelled, turns many algebra questions into pictures: how many roots an equation has, where an expression is positive, what the range of a function is. Cambridge asks for these sketches directly, and expects you to use them silently in inequality, range and intersection questions.

The shape and what controls it

Every quadratic graph has the same basic shape, the graph of $y = x^2$ , moved and stretched.

The sign of $a$ decides the direction. If $a > 0$ the parabola opens upwards ( $\cup$ ) and has a lowest point. If $a < 0$ it opens downwards ( $\cap$ ) and has a highest point.
The size of $a$ decides how narrow it is. $y = 3x^2$ is narrower than $y = x^2$ ; $y = \tfrac{1}{3}x^2$ is wider.
The vertex is the turning point. The parabola is symmetric about the vertical line through it, the line of symmetry.

y = x^2 y = 3x^2 y = (1/3)x^2 y = -x^2

From top to bottom near $x = 2$ : $y = 3x^2$ , $y = x^2$ , $y = \tfrac{1}{3}x^2$ , then $y = -x^2$ opening downwards.

Three forms, three pieces of information

The same quadratic can be written in three ways. Each one shows a different feature of the graph immediately.

Key result

Form	What you can read off
$y = ax^2 + bx + c$	$y$ -intercept $(0, c)$ ; direction from the sign of $a$
$y = a(x - \alpha)(x - \beta)$	$x$ -intercepts $(\alpha, 0)$ and $(\beta, 0)$
$y = a(x - p)^2 + q$	vertex $(p, q)$ ; line of symmetry $x = p$

The line of symmetry is halfway between the roots and is also given by

x = -\frac{b}{2a}

The formula $x = -\dfrac{b}{2a}$ comes from the completed square $a\left(x + \dfrac{b}{2a}\right)^2 + \dots$ , or from the fact that the two roots $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ are placed symmetrically either side of $-\dfrac{b}{2a}$ .

Definition

A sketch is a freehand drawing that shows the correct shape and the key features, with coordinates written on. For a parabola that means: the direction it opens, the vertex, the $y$ -intercept, and any $x$ -intercepts. It does not need to be to scale, but the features must be in the right order and position relative to each other.

Sketching a quadratic graph

Look at the sign of $a$ : $\cup$ if $a > 0$ , $\cap$ if $a < 0$ .
Put $x = 0$ to find the $y$ -intercept $(0, c)$ .
Put $y = 0$ and solve to find the $x$ -intercepts. If $b^2 - 4ac < 0$ there are none.
Find the vertex by completing the square, or by taking the midpoint of the roots and substituting.
Draw a smooth symmetric curve (not a V) through these points, and label every point with its coordinates.

Sketching from factorised form

Sketch the graph of $y = 2(x + 1)(x - 3)$ , showing the coordinates of the vertex and of the points where the graph meets the axes.

Solution

$a = 2 > 0$ , so the graph is $\cup$ -shaped.
$x$ -intercepts: $(-1, 0)$ and $(3, 0)$ .
$y$ -intercept: $y = 2(1)(-3) = -6$ , so $(0, -6)$ .
Line of symmetry: halfway between $-1$ and $3$ , so $x = 1$ . Then $y = 2(2)(-2) = -8$ . Vertex $(1, -8)$ .

y = 2(x + 1)(x - 3) (-1, 0) (3, 0) (0, -6) (1, -8) x = 1

A parabola that does not meet the x-axis

Sketch $y = -x^2 + 4x - 7$ .

Solution

$a = -1 < 0$ , so the graph is $\cap$ -shaped.
$y$ -intercept $(0, -7)$ .
Discriminant: $b^2 - 4ac = 16 - 4(-1)(-7) = 16 - 28 = -12 < 0$ , so there are no $x$ -intercepts.
Completing the square: $-x^2 + 4x - 7 = -\left(x^2 - 4x\right) - 7 = -\left[(x - 2)^2 - 4\right] - 7 = -(x - 2)^2 - 3$ . Vertex $(2, -3)$ , which is a maximum.

The whole curve lies below the $x$ -axis, which is another way of saying $-x^2 + 4x - 7 < 0$ for all $x$ .

y = -x^2 + 4x - 7 (2, -3) (0, -7) (4, -7) x = 2

By symmetry the point $(4, -7)$ is also on the curve, which helps you draw it accurately.

Surd intercepts

Sketch $y = x^2 - 4x + 1$ , giving exact coordinates of the intercepts with the $x$ -axis.

Solution

$a > 0$ : $\cup$ -shaped. $y$ -intercept $(0, 1)$ .
Completing the square: $x^2 - 4x + 1 = (x - 2)^2 - 3$ . Vertex $(2, -3)$ .
$x$ -intercepts: $(x - 2)^2 = 3$ , so $x = 2 \pm \sqrt{3}$ . The points are $\left(2 - \sqrt{3}, 0\right)$ and $\left(2 + \sqrt{3}, 0\right)$ , roughly $(0.27, 0)$ and $(3.73, 0)$ .

y = x^2 - 4x + 1 (2 - sqrt(3), 0) (2 + sqrt(3), 0) (0, 1) (2, -3)

Label the exact values on the sketch; decimals are only to help you place them.

Finding the equation of a parabola

Choose the form that uses the information you are given.

Given both $x$ -intercepts: start from $y = a(x - \alpha)(x - \beta)$ , then use one more point to find $a$ .
Given the vertex: start from $y = a(x - p)^2 + q$ , then use one more point to find $a$ .
Given three general points: substitute into $y = ax^2 + bx + c$ and solve three simultaneous equations. (Rare in P1, but quick if two points are intercepts.)

Equation from intercepts

A parabola crosses the $x$ -axis at $(-1, 0)$ and $(5, 0)$ and crosses the $y$ -axis at $(0, 10)$ . Find its equation in the form $y = ax^2 + bx + c$ and state the coordinates of its vertex.

Solution

$y = a(x + 1)(x - 5)$ . At $(0, 10)$ : $10 = a(1)(-5)$ , so $a = -2$ .

y = -2(x + 1)(x - 5) = -2\left(x^2 - 4x - 5\right) = -2x^2 + 8x + 10

The line of symmetry is $x = \dfrac{-1 + 5}{2} = 2$ , and $y = -2(3)(-3) = 18$ . The vertex is $(2, 18)$ , a maximum since $a < 0$ .

Equation from roots and a greatest value

The curve $y = ax^2 + bx + c$ meets the $x$ -axis at $x = 1$ and $x = 3$ , and the greatest value of $y$ is $2$ . Find $a$ , $b$ and $c$ .

Solution

$y = a(x - 1)(x - 3)$ . The greatest value occurs on the line of symmetry $x = 2$ :

a(2 - 1)(2 - 3) = 2 \quad\Rightarrow\quad -a = 2 \quad\Rightarrow\quad a = -2

So $y = -2(x - 1)(x - 3) = -2x^2 + 8x - 6$ , giving $a = -2$ , $b = 8$ , $c = -6$ . A negative $a$ is consistent with the curve having a greatest value.

The graph and the equation

The roots of $ax^2 + bx + c = 0$ are the $x$ -coordinates where the graph $y = ax^2 + bx + c$ meets the line $y = 0$ . More generally:

Key result

The solutions of $f(x) = k$ are the $x$ -coordinates of the points where $y = f(x)$ meets the horizontal line $y = k$ .

For a $\cup$ -shaped parabola with vertex $(p, q)$ , the equation $f(x) = k$ has

two distinct roots if $k > q$ ,
one repeated root if $k = q$ ,
no real roots if $k < q$ .

This is the picture behind the discriminant, and it is often quicker. Slide a horizontal line up and down the graph and count the crossings.

Counting solutions on a restricted domain

The function $f$ is defined by $f(x) = x^2 - 6x + 5$ for $x \ge 0$ . Find the set of values of $k$ for which the equation $f(x) = k$ has two distinct solutions.

Solution

Complete the square: $f(x) = (x - 3)^2 - 4$ . The vertex is $(3, -4)$ . The graph starts at $x = 0$ , where $f(0) = 5$ , falls to the vertex and then rises for ever.

y = x^2 - 6x + 5 + 0 sqrt(x) y = 5 y = -4 (0, 5) (3, -4)

A horizontal line $y = k$ :

for $k < -4$ misses the curve;
for $k = -4$ touches it once, at the vertex;
for $-4 < k \le 5$ crosses it twice (one crossing on each side of $x = 3$ , and at $k = 5$ one of those is the end point $(0, 5)$ , which is included because $x \ge 0$ );
for $k > 5$ crosses it only once, on the right-hand branch.

So $f(x) = k$ has two distinct solutions for $-4 < k \le 5$ .

Watch out

Drawing a V or a U with vertical sides. A parabola is smooth at the vertex and keeps getting steeper; it never becomes vertical.

Wrong vertex sign. $y = (x + 2)^2 - 1$ has vertex $(-2, -1)$ .

Unlabelled sketches. A sketch without coordinates usually earns only the shape mark. Label the intercepts and the vertex.

Ignoring a domain restriction. If $x \ge 0$ , the graph has an end point at $x = 0$ . Draw only the allowed part, and include or exclude the end point correctly.

Exam tip

"Sketch" means shape plus key points; you will not need graph paper. Draw a large, clear diagram with axes labelled $x$ and $y$ .
If the question says "showing the coordinates of the vertex", the vertex coordinates must be written on the sketch or clearly stated next to it.
A quick sketch is the safest way to finish a quadratic inequality or a "set of values of $k$ " question, even when it is not asked for. It costs ten seconds and prevents the most common sign error.
When a range of values of $k$ comes from a graph, think carefully about the end points: is $k = q$ included? Is the value at a domain end point included?

Summary

$a > 0$ gives $\cup$ (minimum); $a < 0$ gives $\cap$ (maximum).
$ax^2 + bx + c$ shows the $y$ -intercept, $a(x - \alpha)(x - \beta)$ the $x$ -intercepts, $a(x - p)^2 + q$ the vertex $(p, q)$ .
The line of symmetry is $x = -\dfrac{b}{2a}$ , halfway between the roots.
A sketch needs the correct shape and labelled vertex and intercepts.
To find an equation, pick the form matching the given information, then use one more point to find $a$ .
Solutions of $f(x) = k$ are where $y = f(x)$ meets $y = k$ ; count crossings by sliding the line.

Practice questions

Question

Sketch $y = x^2 - 6x + 5$ , showing the vertex and the intercepts with both axes.
Sketch $y = 9 - x^2$ .
Sketch $y = 2x^2 + 4x + 5$ , showing that it does not meet the $x$ -axis.
A parabola passes through $(-2, 0)$ , $(4, 0)$ and $(0, 8)$ . Find its equation in the form $y = ax^2 + bx + c$ , and its vertex.
A parabola has vertex $(3, 4)$ and passes through $(1, -4)$ . Find its equation in the form $y = ax^2 + bx + c$ .
The curve $y = x^2 + bx + c$ has line of symmetry $x = -1.5$ and passes through $(1, 2)$ . Find $b$ and $c$ .
Find the set of values of $k$ for which $x^2 - 4x + 7 = k$ has (a) two distinct real roots, (b) no real roots.
The function $f$ is defined by $f(x) = -x^2 + 4x + 1$ for $0 \le x \le 5$ . Find the set of values of $k$ for which $f(x) = k$ has exactly two solutions.
The curve $C$ has equation $y = x^2 - 2px + 3p$ , where $p$ is a constant. (a) Find, in terms of $p$ , the coordinates of the vertex of $C$ . (b) Find the values of $p$ for which the vertex lies on the line $y = 2x$ . (c) Find the set of values of $p$ for which $C$ lies entirely above the $x$ -axis.

Answers

$(x - 1)(x - 5)$ : $x$ -intercepts $(1, 0)$ , $(5, 0)$ ; $y$ -intercept $(0, 5)$ ; $(x - 3)^2 - 4$ gives vertex $(3, -4)$ ; $\cup$ -shaped.
$9 - x^2 = (3 - x)(3 + x)$ : $x$ -intercepts $(\pm 3, 0)$ ; vertex and $y$ -intercept $(0, 9)$ ; $\cap$ -shaped.
$2x^2 + 4x + 5 = 2(x + 1)^2 + 3$ , vertex $(-1, 3)$ , $y$ -intercept $(0, 5)$ , $\cup$ -shaped. Its least value is $3 > 0$ (equivalently $b^2 - 4ac = 16 - 40 < 0$ ), so it never meets the $x$ -axis.
$y = a(x + 2)(x - 4)$ ; $8 = a(2)(-4)$ gives $a = -1$ . $y = -x^2 + 2x + 8$ . Line of symmetry $x = 1$ , $y = 9$ : vertex $(1, 9)$ .
$y = a(x - 3)^2 + 4$ ; $-4 = 4a + 4$ gives $a = -2$ . $y = -2(x - 3)^2 + 4 = -2x^2 + 12x - 14$ .
$-\dfrac{b}{2} = -1.5$ gives $b = 3$ . Then $1 + 3 + c = 2$ gives $c = -2$ .
$x^2 - 4x + 7 = (x - 2)^2 + 3$ , vertex $(2, 3)$ . (a) $k > 3$ . (b) $k < 3$ .
$f(x) = 5 - (x - 2)^2$ : maximum $5$ at $x = 2$ , $f(0) = 1$ , $f(5) = -4$ . On the left of $x = 2$ the curve rises from $(0, 1)$ to $(2, 5)$ ; on the right it falls from $(2, 5)$ to $(5, -4)$ . A line $y = k$ meets both parts when $1 \le k < 5$ . So exactly two solutions for $1 \le k < 5$ . (At $k = 5$ there is one; for $-4 \le k < 1$ there is one.)
(a) $y = (x - p)^2 - p^2 + 3p$ , so the vertex is $\left(p,\ 3p - p^2\right)$ . (b) $3p - p^2 = 2p$ gives $p^2 - p = 0$ , so $p = 0$ or $p = 1$ . (c) The minimum value $3p - p^2$ must be positive: $p(3 - p) > 0$ , so $0 < p < 3$ .

The shape and what controls it

Three forms, three pieces of information

Finding the equation of a parabola

The graph and the equation

Practice questions

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