Roots and the x-Intercepts of a Quadratic Graph

How the points where the graph of $y = ax^2 + bx + c$ meets the $x$ -axis match the roots of $ax^2 + bx + c = 0$ , how to find them by factorising without drawing, and why a graph may meet the $x$ -axis twice, once or not at all.

Before you start

In lesson 5.4 you drew the graph of $y = ax^2 + bx + c$ from a table of values, and saw that it meets the $y$ -axis at $(0, c)$ .
In lessons 5.1 to 5.3 you solved $ax^2 + bx + c = 0$ by factorising: $x^2 - 2x - 3 = 0$ gives $(x + 1)(x - 3) = 0$ , so $x = -1$ or $x = 3$ .

Where the Curve Meets the x-Axis

Every point on the $x$ -axis has $y = 0$ . So the points where the graph of $y = ax^2 + bx + c$ meets the $x$ -axis are the points where $ax^2 + bx + c = 0$ : their $x$ -coordinates are the roots of the equation.

Roots and the x-intercepts

The roots of $ax^2 + bx + c = 0$ are the $x$ -coordinates of the points where the graph of $y = ax^2 + bx + c$ meets the $x$ -axis. These numbers are also called the $x$ -intercepts of the graph.

If the equation factorises as $(x - p)(x - q) = 0$ , the graph meets the $x$ -axis at $(p, 0)$ and $(q, 0)$ .

A root is a number, such as $3$ ; the point where the graph meets the $x$ -axis is $(3, 0)$ . (In lesson 5.4 the $y$ -intercept was the number $c$ ; the point where the graph meets the $y$ -axis is $(0, c)$ .)

Example 1: Reading the roots, then checking

The graph of $y = x^2 - 2x - 8$ for $-3 \le x \le 5$ is drawn below. Use it to write down the roots of $x^2 - 2x - 8 = 0$ , and check them by factorising. Where does the graph meet the $y$ -axis?

Solution

The curve meets the $x$ -axis at $x = -2$ and $x = 4$ , so the roots are $-2$ and $4$ .

Check by factorising: product $-8$ , sum $-2$ : $-4$ and $2$ . $x^2 - 2x - 8 = (x - 4)(x + 2)\text{,}$ and $(x - 4)(x + 2) = 0$ gives $x = 4$ or $x = -2$ ✓.

Check by substituting: $16 - 8 - 8 = 0$ ✓ and $4 + 4 - 8 = 0$ ✓.

The $y$ -axis: at $x = 0$ , $y = c = -8$ .

The roots are $x = -2$ and $x = 4$ ; the graph meets the $y$ -axis at $(0, -8)$ .

You do not need a graph to find the intercepts. Put $y = 0$ and solve the equation by factorising for the $x$ -intercepts; put $x = 0$ for the $y$ -intercept.

Example 2: The intercepts without drawing

Without drawing, find where the graph of $y = (x + 1)(x - 3)$ meets the axes.

Solution

The $x$ -axis: $y = 0$ when $(x + 1)(x - 3) = 0$ , that is, when $x + 1 = 0$ or $x - 3 = 0$ : $x = -1$ or $x = 3$ .

The $y$ -axis: at $x = 0$ , $y = (0 + 1)(0 - 3) = -3\text{.}$

Expanded, $y = x^2 - 2x - 3$ : the curve you drew in lesson 5.4, which does meet the axes at these three points.

It meets the $x$ -axis at $(-1, 0)$ and $(3, 0)$ , and the $y$ -axis at $(0, -3)$ .

Example 3: Factorise first

Without drawing, find where the graph of $y = 2 - x - x^2$ meets the axes. Say whether it is ∪-shaped or ∩-shaped.

Solution

The $x$ -axis: $2 - x - x^2 = 0$ . Multiply both sides by $-1$ , so that the $x^2$ term is positive: $x^2 + x - 2 = 0$ .

Product $-2$ , sum $1$ : $2$ and $-1$ . $(x + 2)(x - 1) = 0$ , so $x = -2$ or $x = 1$ .

Check in the equation as given: $x = -2$ : $2 + 2 - 4 = 0$ ✓; $x = 1$ : $2 - 1 - 1 = 0$ ✓.

The $y$ -axis: at $x = 0$ , $y = 2$ .

The coefficient of $x^2$ is $-1$ , so the graph is ∩-shaped. Drawn, it looks like this:

It meets the $x$ -axis at $(-2, 0)$ and $(1, 0)$ , and the $y$ -axis at $(0, 2)$ ; it is ∩-shaped.

Two, One or No x-Intercepts

A quadratic graph can cross the $x$ -axis at two points, touch it at one point, or miss it altogether. Which it does depends on the equation $ax^2 + bx + c = 0$ .

How many x-intercepts

Two different roots: the graph crosses the $x$ -axis at two points.

Two equal roots, as in $(x - 2)^2 = 0$ : the graph touches the $x$ -axis at one point, its lowest (or highest) point.

No solution, as in $x^2 = -2$ : the graph does not meet the $x$ -axis.

Example 4: Three curves

The graphs are drawn below. Use factorising to explain why (a) crosses the $x$ -axis twice, (b) touches it once and (c) does not meet it.
(a) $y = x^2 - 5x + 4$
(b) $y = x^2 - 4x + 4$
(c) $y = x^2 + 2$

Solution

(a) $x^2 - 5x + 4 = (x - 1)(x - 4)\text{,}$ and $(x - 1)(x - 4) = 0$ has two different roots, $1$ and $4$ . So the curve crosses the $x$ -axis twice, at $(1, 0)$ and $(4, 0)$ . Check: $1 - 5 + 4 = 0$ ✓; $16 - 20 + 4 = 0$ ✓.

(b) $x^2 - 4x + 4 = (x - 2)^2\text{,}$ and $(x - 2)^2 = 0$ has two equal roots, $x = 2$ . So the curve touches the $x$ -axis at one point, $(2, 0)$ , its lowest point.

(c) $x^2 + 2 = 0$ means $x^2 = -2$ , and no number squared is negative, so there is no solution: the curve does not meet the $x$ -axis. Since $x^2$ is never negative, $y = x^2 + 2$ is never less than $2$ .

(a) two different roots, $1$ and $4$ : two crossings
(b) two equal roots, $2$ : it touches at $(2, 0)$
(c) no solution: it does not meet the $x$ -axis

Roots in a Setting

When a quadratic function describes something real, the roots answer questions such as “when does it land?” or “when is the profit zero?”. Keep only the roots that make sense: a time after the start, a length that is positive.

Example 5: When does the ball land?

A ball is thrown upwards from the top of a building 60 m high. Its height above the ground, $h$ metres, after $t$ seconds is $h = 60 + 20t - 5t^2$ . After how many seconds does it hit the ground?

Solution

The ball hits the ground when $h = 0$ : $60 + 20t - 5t^2 = 0$ .

Every term is a multiple of $5$ . Divide both sides by $-5$ , which also makes the $t^2$ term positive: $t^2 - 4t - 12 = 0$ .

Product $-12$ , sum $-4$ : $-6$ and $2$ . $(t - 6)(t + 2) = 0$ , so $t = 6$ or $t = -2$ .

The time $t = -2$ is before the ball was thrown, so it does not answer the question.

Check $t = 6$ : $60 + 120 - 180 = 0$ ✓.

The graph is drawn only for $0 \le t \le 6$ , the times the ball is in the air; the root $t = -2$ is not on it.

The ball hits the ground after 6 seconds.

Challenge: The Equation from the Intercepts

Example 6: Working backwards

A curve $y = x^2 + bx + c$ meets the $x$ -axis at $(-1, 0)$ and $(3, 0)$ . Find its equation, and check where it meets the $y$ -axis.

Solution

The roots $-1$ and $3$ come from the factors $(x + 1)$ and $(x - 3)$ . Since the coefficient of $x^2$ is $1$ , $y = (x + 1)(x - 3)$ .

Expand: $y = x^2 - 3x + x - 3 = x^2 - 2x - 3\text{.}$

At $x = 0$ , $y = -3$ : the curve meets the $y$ -axis at $(0, -3)$ , as Example 2 found.

$y = x^2 - 2x - 3$

Common Mistakes

Changing the signs wrongly from the factors: $y = (x + 1)(x - 3)$ meets the $x$ -axis at $x = -1$ and $x = 3$ , not at $1$ and $-3$ .
Giving a single number when a point is asked for: the curve meets the $x$ -axis at the point $(3, 0)$ ; $3$ is the root (the $x$ -intercept).
Mixing up the intercepts: the $y$ -intercept is where $x = 0$ , and the $x$ -intercepts are where $y = 0$ .
Losing the root $0$ : $y = x^2 + 5x = x(x + 5)$ meets the $x$ -axis at $(0, 0)$ as well as at $(-5, 0)$ .
Keeping a root that makes no sense in the setting, such as a negative time.

Practice Questions

Give the roots as numbers and the places where a graph meets an axis as points, and check each root by substituting. The questions get harder as you go.

Question 1

A table of values for $y = x^2 - 7x + 6$ is given below. Use it to write down the points where the graph meets the $x$ -axis and the $y$ -axis. Then check the roots by factorising.

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$y$	$6$	$0$	$-4$	$-6$	$-6$	$-4$	$0$	$6$

ShowHide solution

In the table, $y = 0$ at $x = 1$ and $x = 6$ : the graph meets the $x$ -axis at $(1, 0)$ and $(6, 0)$ .

At $x = 0$ , $y = 6$ : the graph meets the $y$ -axis at $(0, 6)$ .

Check: $x^2 - 7x + 6 = (x - 1)(x - 6)\text{,}$ and $(x - 1)(x - 6) = 0$ gives $x = 1$ or $x = 6$ ✓.

$(1, 0)$ and $(6, 0)$ on the $x$ -axis; $(0, 6)$ on the $y$ -axis.

Question 2

The graph of $y = x^2 - x - 6$ is drawn below. Write down the roots of $x^2 - x - 6 = 0$ , and check them by factorising.

ShowHide solution

The curve meets the $x$ -axis at $x = -2$ and $x = 3$ .

Check: $x^2 - x - 6 = (x + 2)(x - 3)\text{,}$ and $(x + 2)(x - 3) = 0$ gives $x = -2$ or $x = 3$ ✓.

$x = -2$ or $x = 3$

Question 3

The graph of $y = x^2 - 4$ is drawn below. Use it to write down the roots of $x^2 - 4 = 0$ , and check them by factorising.

ShowHide solution

The curve meets the $x$ -axis at $x = -2$ and $x = 2$ .

Check: $x^2 - 4 = (x + 2)(x - 2)\text{,}$ and $(x + 2)(x - 2) = 0$ gives $x = \pm 2$ ✓.

$x = \pm 2$

Question 4

Without drawing, find the points where the graph of $y = (x - 2)(x + 4)$ meets the $x$ -axis and the $y$ -axis.

ShowHide solution

$y = 0$ when $x - 2 = 0$ or $x + 4 = 0$ : $x = 2$ or $x = -4$ .

At $x = 0$ : $y = (-2)(4) = -8$ .

$(2, 0)$ and $(-4, 0)$ on the $x$ -axis; $(0, -8)$ on the $y$ -axis.

Question 5

Without drawing, find the points where each graph meets the axes.
(a) $y = x^2 + 8x + 12$
(b) $y = 2x^2 - x - 3$

ShowHide solution

(a) Product $12$ , sum $8$ : $2$ and $6$ . $(x + 2)(x + 6) = 0$ gives $x = -2$ or $x = -6$ . At $x = 0$ , $y = 12$ . Check: $4 - 16 + 12 = 0$ ✓ and $36 - 48 + 12 = 0$ ✓.

(b) $ac = -6$ ; two numbers with product $-6$ and sum $-1$ : $-3$ and $2$ . Split the middle term: $2x^2 - 3x + 2x - 3 = x(2x - 3) + (2x - 3) = (2x - 3)(x + 1)\text{.}$

$(2x - 3)(x + 1) = 0$ gives $x = \frac{3}{2} = 1\frac{1}{2}$ or $x = -1$ . At $x = 0$ , $y = -3$ .

Check $x = \frac{3}{2}$ : $2 \times \frac{9}{4} - \frac{3}{2} - 3 = \frac{9}{2} - \frac{3}{2} - 3 = 0$ ✓. Check $x = -1$ : $2 + 1 - 3 = 0$ ✓.

(a) $(-2, 0)$ and $(-6, 0)$ ; $(0, 12)$
(b) $\left(1\frac{1}{2}, 0\right)$ and $(-1, 0)$ ; $(0, -3)$

Question 6

Without drawing, say how many times each graph meets the $x$ -axis, and give the points.
(a) $y = x^2 - 4x$
(b) $y = x^2 - 6x + 9$
(c) $y = x^2 + 4$

ShowHide solution

(a) $x^2 - 4x = x(x - 4) = 0$ gives $x = 0$ or $x = 4$ : twice, at $(0, 0)$ and $(4, 0)$ .

(b) $x^2 - 6x + 9 = (x - 3)^2 = 0$ gives two equal roots, $x = 3$ : the curve touches the $x$ -axis once, at $(3, 0)$ .

(c) $x^2 + 4 = 0$ means $x^2 = -4$ , which has no solution: the curve does not meet the $x$ -axis ( $y$ is never less than $4$ ).

(a) twice, at $(0, 0)$ and $(4, 0)$
(b) once, touching at $(3, 0)$
(c) not at all

Question 7

Without drawing, find where the graph of $y = 7 - 6x - x^2$ meets the axes, and say whether it is ∪-shaped or ∩-shaped.

ShowHide solution

$7 - 6x - x^2 = 0$ . Multiply both sides by $-1$ : $x^2 + 6x - 7 = 0$ .

Product $-7$ , sum $6$ : $7$ and $-1$ . $(x + 7)(x - 1) = 0$ , so $x = -7$ or $x = 1$ .

Check in the equation as given: $x = -7$ : $7 + 42 - 49 = 0$ ✓; $x = 1$ : $7 - 6 - 1 = 0$ ✓.

At $x = 0$ , $y = 7$ . The coefficient of $x^2$ is $-1$ , so the graph is ∩-shaped.

$(-7, 0)$ and $(1, 0)$ on the $x$ -axis; $(0, 7)$ on the $y$ -axis; ∩-shaped.

Question 8

A ball is kicked up from the ground. Its height, $h$ metres, after $t$ seconds is $h = 15t - 5t^2$ . When is the ball at ground level?

ShowHide solution

At ground level $h = 0$ : $15t - 5t^2 = 0$ .

Take out the common factor $5t$ : $5t(3 - t) = 0$ , so $t = 0$ or $t = 3$ .

$t = 0$ is the moment it is kicked; $t = 3$ is when it lands. Check: $45 - 45 = 0$ ✓.

When it is kicked ( $t = 0$ ) and when it lands, after 3 seconds.

Question 9

A class runs a stall at a school fair selling bookmarks. When the stall sells each bookmark for $x$ dollars, its profit is $P = 12x - x^2 - 20$ dollars.
(a) At which prices does the stall make no profit?
(b) What is the profit when a bookmark is sold for $6?

ShowHide solution

(a) No profit means $P = 0$ : $12x - x^2 - 20 = 0$ . Multiply both sides by $-1$ : $x^2 - 12x + 20 = 0$ .

Product $20$ , sum $-12$ : $-2$ and $-10$ . $(x - 2)(x - 10) = 0$ , so $x = 2$ or $x = 10$ .

Check: $24 - 4 - 20 = 0$ ✓ and $120 - 100 - 20 = 0$ ✓.

(b) $P = 12 \times 6 - 36 - 20 = 72 - 36 - 20 = 16\text{.}$

(a) at $2 and at $10
(b) $16

Question 10

Sam says that the graph of $y = (x - 3)(x + 5)$ meets the $x$ -axis at $(-3, 0)$ and $(5, 0)$ . Explain his mistake and give the correct points.

ShowHide solution

He has copied the numbers from the brackets with the wrong signs. $y = 0$ when $x - 3 = 0$ , so $x = 3$ , or when $x + 5 = 0$ , so $x = -5$ .

His points fail the check: at $x = -3$ , $y = (-6)(2) = -12\text{,}$ not $0$ .

The graph meets the $x$ -axis at $(3, 0)$ and $(-5, 0)$ .

Question 11 (challenge)

A curve $y = x^2 + bx + c$ meets the $x$ -axis at $(2, 0)$ and $(5, 0)$ . Find $b$ and $c$ , and the point where the curve meets the $y$ -axis.

ShowHide solution

The roots $2$ and $5$ come from the factors $(x - 2)$ and $(x - 5)$ : $y = (x - 2)(x - 5) = x^2 - 7x + 10\text{.}$

So $b = -7$ and $c = 10$ . At $x = 0$ , $y = 10$ .

Check: at $x = 5$ , $25 - 35 + 10 = 0$ ✓.

$b = -7$ , $c = 10$ ; the curve meets the $y$ -axis at $(0, 10)$ .

Question 12 (challenge)

A curve $y = x^2 + bx + c$ touches the $x$ -axis at $(-4, 0)$ and nowhere else. Find its equation.

ShowHide solution

Touching the $x$ -axis at one point means two equal roots, $x = -4$ , so the factors are $(x + 4)(x + 4)$ .

$y = (x + 4)^2 = x^2 + 8x + 16\text{.}$

Check: at $x = -4$ , $16 - 32 + 16 = 0$ ✓, and $(x + 4)^2$ is never negative, so the curve stays on or above the $x$ -axis.

$y = x^2 + 8x + 16$

Key Points

On the $x$ -axis, $y = 0$ . The roots of $ax^2 + bx + c = 0$ are the $x$ -coordinates of the points where $y = ax^2 + bx + c$ meets the $x$ -axis.
To find the $x$ -intercepts without drawing, put $y = 0$ and solve by factorising; for the $y$ -intercept, put $x = 0$ .
The graph of $y = (x - p)(x - q)$ meets the $x$ -axis at $(p, 0)$ and $(q, 0)$ , and the $y$ -axis at $(0, pq)$ .
Two different roots: the curve crosses the $x$ -axis twice. Two equal roots: it touches the axis once, at its lowest (or highest) point. No solution: it does not meet the axis.
A root (an $x$ -intercept) is a number, such as $3$ ; the point where the curve meets the $x$ -axis is $(3, 0)$ .
In a setting, keep only the roots that make sense.