Maximum and Minimum Points and the Line of Symmetry

How to find the turning point of a quadratic graph and its line of symmetry: from a drawn graph, from two equal values of $y$ in a table, or halfway between the roots.

Before you start

In lesson 5.4 you drew the graph of $y = ax^2 + bx + c$ from a table of values: it is ∪-shaped when $a > 0$ and ∩-shaped when $a < 0$ .
In lesson 5.5 you found where a quadratic graph meets the $x$ -axis by factorising: $x^2 - 4x + 3 = 0$ gives $(x - 1)(x - 3) = 0$ , so $y = x^2 - 4x + 3$ meets the $x$ -axis at $(1, 0)$ and $(3, 0)$ .

Turning Points and the Line of Symmetry

Turning points and the line of symmetry

A ∪-shaped graph ( $a > 0$ ) has a lowest point, its minimum point. A ∩-shaped graph ( $a < 0$ ) has a highest point, its maximum point. Either is called the turning point.

The $y$ -coordinate of the minimum point is the minimum value of $y$ ; the $y$ -coordinate of the maximum point is the maximum value of $y$ .

The graph is symmetrical about the vertical line through its turning point, its line of symmetry. If the turning point is $(1, -4)$ , the line of symmetry is $x = 1$ .

Example 1: Reading from a drawn graph

The graph of $y = x^2 - 2x - 3$ for $-2 \le x \le 4$ is drawn below. Write down
(a) the minimum point
(b) the minimum value of $y$
(c) the equation of the line of symmetry.

Solution

(a) The lowest point of the curve is on the grid lines $x = 1$ and $y = -4$ : the minimum point is $(1, -4)$ .

(b) The minimum value of $y$ is its $y$ -coordinate, $-4$ .

(c) The vertical line through $(1, -4)$ is $x = 1$ . Check with two points at the same height: $(0, -3)$ and $(2, -3)$ are each 1 unit from it.

(a) $(1, -4)$
(b) $-4$
(c) $x = 1$

The Line of Symmetry from a Table

In the table for $y = x^2 - 2x - 3$ , $x = 0$ and $x = 2$ both give $y = -3$ . Two points at the same height are mirror images of each other, so the line of symmetry lies exactly halfway between them. The number halfway between $p$ and $q$ is $\frac{p + q}{2}$ : halfway between $0$ and $2$ is $\frac{0 + 2}{2} = 1$ .

Finding the turning point from equal values of y

1. Find two values of $x$ , say $p$ and $q$ , that give the same value of $y$ .

2. The line of symmetry is halfway between them: $x = \frac{p + q}{2}$ .

3. The turning point lies on this line. Substitute its $x$ into the equation to find the $y$ -coordinate.

Example 2: Equal values in a table

Use the table of values to find the line of symmetry and the minimum point of $y = x^2 - 2x - 3$ .

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

Solution

$x = 0$ and $x = 2$ both give $y = -3$ . Halfway between them: $x = \frac{0 + 2}{2} = 1$ . (The pairs $-1$ and $3$ , and $-2$ and $4$ , give $1$ too.)

Substitute $x = 1$ : $y = 1 - 2 - 3 = -4$ .

The line of symmetry is $x = 1$ and the minimum point is $(1, -4)$ , as Example 1 read from the graph.

Example 3: A turning point between the whole numbers

Draw up a table for $y = x^2 - 5x + 6$ for $0 \le x \le 5$ , and find the minimum point.

Solution

At $x = 0$ : $6$ . At $x = 1$ : $1 - 5 + 6 = 2$ . At $x = 2$ : $4 - 10 + 6 = 0$ . At $x = 3$ : $9 - 15 + 6 = 0$ . At $x = 4$ : $16 - 20 + 6 = 2$ . At $x = 5$ : $25 - 25 + 6 = 6$ .

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

$x = 2$ and $x = 3$ both give $0$ , so the line of symmetry is $x = \frac{2 + 3}{2} = 2\frac{1}{2}\text{.}$

Substitute $x = 2\frac{1}{2} = \frac{5}{2}$ : $y = \frac{25}{4} - \frac{25}{2} + 6 = \frac{25}{4} - \frac{50}{4} + \frac{24}{4} = -\frac{1}{4}\text{.}$

The lowest value in the table is $0$ , but the curve dips a little below the $x$ -axis between $x = 2$ and $x = 3$ . The minimum point is not always in the table.

The minimum point is $\left(2\frac{1}{2}, -\frac{1}{4}\right)\text{.}$

Example 4: A maximum point

Find the line of symmetry and the maximum point of $y = -2x^2 + 8x - 5$ , using its values at $x = 0$ and $x = 4$ .

Solution

At $x = 0$ : $y = -5$ . At $x = 4$ : $y = -2(16) + 32 - 5 = -32 + 32 - 5 = -5\text{.}$

The two values are equal, so the line of symmetry is $x = \frac{0 + 4}{2} = 2$ .

Substitute $x = 2$ : $y = -2(4) + 16 - 5 = -8 + 16 - 5 = 3\text{.}$

$a = -2 < 0$ , so the graph is ∩-shaped and this turning point is a maximum.

The line of symmetry is $x = 2$ , and the maximum point is $(2, 3)$ : the maximum value of $y$ is $3$ .

The Line of Symmetry from the Roots

In lesson 5.5 you found the points where a curve meets the $x$ -axis: their $x$ -coordinates are the roots of $ax^2 + bx + c = 0$ . For $y = x^2 - 4x + 3$ they are $(1, 0)$ and $(3, 0)$ . Both points have $y = 0$ , so they are two points at the same height, and the line of symmetry is halfway between them.

This needs two different roots. When the equation has two equal roots, the curve touches the $x$ -axis at one point, and that point is the turning point. A curve that does not meet the $x$ -axis is dealt with in the next section.

Finding the turning point from the roots

1. Find where the curve meets the $x$ -axis, as in lesson 5.5: solve $ax^2 + bx + c = 0$ by factorising. Call the roots $p$ and $q$ .

2. The line of symmetry is $x = \frac{p + q}{2}$ .

3. Substitute this $x$ into $y = ax^2 + bx + c$ to find the $y$ -coordinate of the turning point.

Example 5: Halfway between the roots

Find the line of symmetry and the minimum point of $y = x^2 - 4x + 3$ .

Solution

Set $y = 0$ : $x^2 - 4x + 3 = 0$ , so $(x - 1)(x - 3) = 0$ , and $x = 1$ or $x = 3$ .

Line of symmetry: $x = \frac{1 + 3}{2} = 2$ .

Substitute $x = 2$ : $y = 4 - 8 + 3 = -1$ .

The line of symmetry is $x = 2$ and the minimum point is $(2, -1)$ .

Example 6: A minimum and a maximum

Find the turning point of
(a) $y = x^2 + 6x + 5$
(b) $y = -x^2 + 2x + 3$ , and say whether it is a maximum or a minimum point.

Solution

(a) $x^2 + 6x + 5 = 0$ gives $(x + 1)(x + 5) = 0$ , so the roots are $-1$ and $-5$ .

Line of symmetry: $x = \frac{-1 + (-5)}{2} = -3\text{.}$ Then $y = 9 - 18 + 5 = -4$ . Since $a = 1 > 0$ , $(-3, -4)$ is a minimum point.

(b) $-x^2 + 2x + 3 = 0$ . Multiply by $-1$ : $x^2 - 2x - 3 = 0$ , so $(x - 3)(x + 1) = 0$ , and the roots are $3$ and $-1$ .

Line of symmetry: $x = \frac{3 + (-1)}{2} = 1\text{.}$ Then $y = -1 + 2 + 3 = 4$ . Since $a = -1 < 0$ , $(1, 4)$ is a maximum point.

(a) minimum point $(-3, -4)$
(b) maximum point $(1, 4)$

Curves That Do Not Meet the x-Axis

Some curves do not meet the $x$ -axis (lesson 5.5): then $ax^2 + bx + c = 0$ has no solution, and there are no roots to use. Use two other values of $x$ that give the same $y$ instead. The easiest pair often starts at $x = 0$ , where $y = c$ .

To find the partner of $x = 0$ , solve $y = c$ by factorising. For $y = x^2 - 2x + 5$ , $c = 5$ : $x^2 - 2x + 5 = 5$ gives $x^2 - 2x = 0$ , so $x(x - 2) = 0$ and $x = 0$ or $x = 2$ . The line of symmetry is halfway between, $x = 1$ . The number $c$ is on both sides and cancels, so the equation always has the common factor $x$ .

Example 7: No roots

For $y = x^2 - 6x + 11$ , find the value of $x$ , other than $0$ , at which $y$ takes the same value as at $x = 0$ . Hence find the minimum point, and explain why the curve does not meet the $x$ -axis.

Solution

At $x = 0$ : $y = 11$ . Solve $y = 11$ : $x^2 - 6x + 11 = 11$ , so $x^2 - 6x = 0$ and $x(x - 6) = 0$ : $x = 0$ or $x = 6$ . Check: $36 - 36 + 11 = 11$ ✓.

The values are equal, so the line of symmetry is $x = \frac{0 + 6}{2} = 3$ . Then $y = 9 - 18 + 11 = 2$ .

The minimum value of $y$ is $2$ . Every point of the curve has $y \ge 2$ , so the curve never reaches $y = 0$ : it does not meet the $x$ -axis.

The minimum point is $(3, 2)$ ; the curve stays above the $x$ -axis because its lowest value of $y$ is $2$ .

When $b = 0$ , as in $y = x^2 + 2$ , the values at $x$ and at $-x$ are always equal, because $(-x)^2 = x^2$ . So the line of symmetry is the $y$ -axis, $x = 0$ , and the turning point is on the $y$ -axis at $(0, c)$ .

Example 8: The y-axis as the line of symmetry

Find the turning point and the line of symmetry of
(a) $y = x^2 + 2$
(b) $y = 9 - x^2$ .

Solution

(a) At $x = 1$ and $x = -1$ , $y = 3$ ; at $x = 2$ and $x = -2$ , $y = 6$ . The line of symmetry is $x = 0$ , and at $x = 0$ , $y = 2$ . Since $a = 1 > 0$ , $(0, 2)$ is a minimum point.

(b) At $x = \pm 3$ , $y = 0$ , so the line of symmetry is $x = \frac{-3 + 3}{2} = 0$ . At $x = 0$ , $y = 9$ . Since $a = -1 < 0$ , $(0, 9)$ is a maximum point.

(a) minimum point $(0, 2)$
(b) maximum point $(0, 9)$ ; in both, the line of symmetry is the $y$ -axis, $x = 0$ .

Using the Symmetry

Once you know the line of symmetry, each point of the curve has a partner at the same height, the same distance away on the other side.

Example 9: A partner point

A quadratic curve has the line of symmetry $x = 3$ and passes through $(1, 7)$ . Find another point on the curve with $y = 7$ .

Solution

$x = 1$ is $3 - 1 = 2$ units to the left of the line $x = 3$ .

Its partner is $2$ units to the right: $x = 3 + 2 = 5$ , at the same height.

$(5, 7)$

Greatest and Least Values in a Setting

When a quadratic function describes something real, its maximum or minimum point answers questions such as “how high?” or “how much at most?”.

Example 10: The height of a ball

A ball is thrown upwards. Its height, $h$ metres, after $t$ seconds is $h = 30t - 5t^2$ . The graph is drawn below. Find the greatest height of the ball and the time at which it is reached.

Solution

From the graph: the highest point is at $t = 3$ , on the grid line for $45$ .

Check by symmetry: $h = 0$ at $t = 0$ and at $t = 6$ (since $30 \times 6 - 5 \times 36 = 180 - 180 = 0\text{),}$ so the line of symmetry is $t = 3$ . Then $h = 90 - 45 = 45$ .

The greatest height is 45 m, reached 3 seconds after the ball is thrown.

Common Mistakes

Taking the lowest value in the table as the minimum point. The turning point may lie between two values of $x$ , as in Example 3.
Giving the minimum value as a point, or the minimum point as one number. The minimum point is $(1, -4)$ ; the minimum value of $y$ is $-4$ .
Writing the line of symmetry as $y = 1$ or as the number $1$ . It is a vertical line, $x = 1$ .
Finding the midpoint of two $x$ values with different $y$ values. The two points must be at the same height.
Forgetting the sign of $a$ : a ∩-shaped graph has a maximum point, not a minimum.

Practice Questions

Show how you find each line of symmetry, and check each turning point by substituting. The questions get harder as you go.

Question 1

The graph of $y = 5 + 4x - x^2$ for $-1 \le x \le 5$ is drawn below. Write down the maximum point, the maximum value of $y$ and the line of symmetry.

ShowHide solution

The highest point is on the grid lines $x = 2$ and $y = 9$ . Check: $5 + 8 - 4 = 9$ ✓.

Maximum point $(2, 9)$ ; maximum value $9$ ; line of symmetry $x = 2$ .

Question 2

Use the table to find the line of symmetry and the minimum point of $y = x^2 + 2x - 8$ .

$x$	$-5$	$-4$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$y$	$7$	$0$	$-5$	$-8$	$-9$	$-8$	$-5$	$0$	$7$

ShowHide solution

$x = -2$ and $x = 0$ both give $-8$ : the line of symmetry is $x = \frac{-2 + 0}{2} = -1\text{.}$

At $x = -1$ : $y = 1 - 2 - 8 = -9$ .

Line of symmetry $x = -1$ ; minimum point $(-1, -9)$ .

Question 3

Find the roots of $x^2 - 6x + 8 = 0$ . Hence find the line of symmetry and the minimum point of $y = x^2 - 6x + 8$ .

ShowHide solution

$(x - 2)(x - 4) = 0$ , so $x = 2$ or $x = 4$ .

Line of symmetry: $x = \frac{2 + 4}{2} = 3$ . At $x = 3$ : $y = 9 - 18 + 8 = -1$ .

Roots $2$ and $4$ ; line of symmetry $x = 3$ ; minimum point $(3, -1)$ .

Question 4

Find the minimum point of $y = x^2 - 3x - 10$ .

ShowHide solution

$x^2 - 3x - 10 = 0$ gives $(x - 5)(x + 2) = 0$ , so the roots are $5$ and $-2$ .

Line of symmetry: $x = \frac{5 + (-2)}{2} = \frac{3}{2} = 1\frac{1}{2}\text{.}$

At $x = \frac{3}{2}$ : $y = \frac{9}{4} - \frac{9}{2} - 10 = \frac{9}{4} - \frac{18}{4} - \frac{40}{4} = -\frac{49}{4} = -12\frac{1}{4}\text{.}$

The minimum point is $\left(1\frac{1}{2}, -12\frac{1}{4}\right)\text{.}$

Question 5

Find the maximum point of $y = 12 - 4x - x^2$ , and state the maximum value of $y$ .

ShowHide solution

Set $y = 0$ : $12 - 4x - x^2 = 0$ . Multiply by $-1$ : $x^2 + 4x - 12 = 0$ , so $(x + 6)(x - 2) = 0$ , and the roots are $-6$ and $2$ .

Line of symmetry: $x = \frac{-6 + 2}{2} = -2\text{.}$ At $x = -2$ : $y = 12 + 8 - 4 = 16$ .

The coefficient of $x^2$ is $-1$ , so the graph is ∩-shaped and this is a maximum.

Maximum point $(-2, 16)$ ; maximum value $16$ .

Question 6

For $y = x^2 + 4x + 7$ , find the value of $x$ , other than $0$ , at which $y$ takes the same value as at $x = 0$ . Hence find the minimum point, and explain why the curve does not meet the $x$ -axis.

ShowHide solution

At $x = 0$ : $y = 7$ . Solve $y = 7$ : $x^2 + 4x + 7 = 7$ , so $x^2 + 4x = 0$ and $x(x + 4) = 0$ : $x = 0$ or $x = -4$ . Check: $16 - 16 + 7 = 7$ ✓.

Line of symmetry: $x = \frac{0 + (-4)}{2} = -2\text{.}$ At $x = -2$ : $y = 4 - 8 + 7 = 3$ .

The graph is ∪-shaped with lowest value $3$ , so $y$ is never $0$ : the curve does not meet the $x$ -axis.

Minimum point $(-2, 3)$ ; the least value of $y$ is $3$ , which is above $0$ .

Question 7

Write down the turning point of each graph and say whether it is a maximum or a minimum.
(a) $y = 2x^2 - 8$
(b) $y = 4 - x^2$

ShowHide solution

(a) $b = 0$ , so the line of symmetry is the $y$ -axis. At $x = 0$ , $y = -8$ . $a = 2 > 0$ : minimum.

(b) $b = 0$ again. At $x = 0$ , $y = 4$ . $a = -1 < 0$ : maximum.

(a) minimum point $(0, -8)$
(b) maximum point $(0, 4)$

Question 8

A quadratic curve has the line of symmetry $x = -1$ and passes through $(2, 5)$ . Find another point on the curve with $y = 5$ .

ShowHide solution

$x = 2$ is $2 - (-1) = 3$ units to the right of the line $x = -1$ .

Its partner is $3$ units to the left: $x = -1 - 3 = -4$ .

$(-4, 5)$

Question 9

The graph of a quadratic function has the line of symmetry $x = 2$ . Copy and complete its table of values.

$x$	$-1$	$0$	$1$	$2$	$3$	$4$	$5$
$y$	$10$	$5$	$2$	$1$

ShowHide solution

Each $x$ on the right of $x = 2$ has a partner on the left the same distance away: $3$ pairs with $1$ , $4$ with $0$ , and $5$ with $-1$ .

So $y = 2$ at $x = 3$ , $y = 5$ at $x = 4$ , and $y = 10$ at $x = 5$ .

The missing values are $2, 5, 10$ .

Question 10

Wei made a table for $y = x^2 - x - 2$ for $-2 \le x \le 3$ and found the values $4, 0, -2, -2, 0, 4$ . He says the minimum point is $(0, -2)$ . Explain why he is wrong, and find the minimum point.

ShowHide solution

$x = 0$ and $x = 1$ both give $-2$ , so these two points are at the same height and the turning point is halfway between them, at $x = \frac{0 + 1}{2} = \frac{1}{2}$ , not at $x = 0$ .

At $x = \frac{1}{2}$ : $y = \frac{1}{4} - \frac{1}{2} - 2 = -2\frac{1}{4}\text{,}$ which is lower than $-2$ .

The minimum point is $\left(\frac{1}{2}, -2\frac{1}{4}\right)\text{.}$

Question 11

A ball is kicked into the air. Its height, $h$ metres, after $t$ seconds is $h = 10t - 5t^2$ . The graph is drawn below. Find
(a) the greatest height of the ball and when it is reached
(b) when the ball lands.

ShowHide solution

(a) The highest point of the curve is at $t = 1$ , $h = 5$ . Check: $10 - 5 = 5$ ✓, and by symmetry: $h = 0$ at $t = 0$ and $t = 2$ , and halfway between them is $t = 1$ .

(b) The ball lands when $h = 0$ again: $t = 2$ . Check: $20 - 20 = 0$ ✓

(a) 5 m, after 1 second
(b) after 2 seconds

Question 12

A rectangle has a perimeter of 20 cm. When its width is $x$ cm, its area is $A = 10x - x^2$ square centimetres. By finding the values of $x$ for which $A = 0$ , find the greatest possible area and the width that gives it.

ShowHide solution

$10x - x^2 = 0$ gives $x(10 - x) = 0$ , so $x = 0$ or $x = 10$ .

The line of symmetry is $x = \frac{0 + 10}{2} = 5$ . At $x = 5$ : $A = 50 - 25 = 25$ .

The coefficient of $x^2$ is $-1$ , so this is a maximum. The rectangle is then $5$ cm by $5$ cm, a square.

The greatest area is 25 cm², when the width is 5 cm.

Key Points

A ∪-shaped graph ( $a > 0$ ) has a minimum point; a ∩-shaped graph ( $a < 0$ ) has a maximum point. Either is the turning point.
The minimum (or maximum) value of $y$ is the $y$ -coordinate of the turning point.
The line of symmetry is the vertical line through the turning point, $x =$ its $x$ -coordinate.
Two points at the same height have the line of symmetry halfway between them: $x = \frac{p + q}{2}$ .
When $ax^2 + bx + c = 0$ has two different roots, the curve meets the $x$ -axis at two points at the same height, so the line of symmetry is halfway between the roots. If it has no roots, use two other values of $x$ that give the same $y$ .
Substitute the $x$ of the line of symmetry into the equation to find the turning point.
When $b = 0$ , the line of symmetry is the $y$ -axis and the turning point is $(0, c)$ .