Sec 3 · Additional Mathematics

Quadratic functions

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1. Maximum and minimum values

Form	Turning point	Value
$y = a(x - h)^{2} + k$ , $a > 0$	minimum point $(h, k)$	minimum value $k$ at $x = h$
$y = a(x - h)^{2} + k$ , $a < 0$	maximum point $(h, k)$	maximum value $k$ at $x = h$

Left: U-shaped curve y = (x − 1)² + 2 with minimum point (1, 2). Right: ∩-shaped curve y = −(x − 1)² + 3 with maximum point (1, 3).

Minimum: complete the square

\displaystyle y = x^{2} - 4x + 7

e.g.

\displaystyle x^{2} - 4x + 7

\displaystyle = (x - 2)^{2} - 2^{2} + 7

\displaystyle = (x - 2)^{2} + 3

minimum value $3$ , when $x = 2$

Maximum: take out −1 first

\displaystyle y = -5 + 4x - x^{2}

e.g.

\displaystyle -5 + 4x - x^{2}

\displaystyle = -(x^{2} - 4x) - 5

\displaystyle = -((x - 2)^{2} - 4) - 5

\displaystyle = -1 - (x - 2)^{2}

maximum value $-1$ , when $x = 2$

📖 Worked example

By completing the square, find the maximum value of $y = -5 + 4x - x^2$ .

Step 1 of 2

Take out −1, then complete the square inside

-5 + 4x - x^2 = -(x^2 - 4x) - 5

= -[(x - 2)^2 - 4] - 5

= -(x - 2)^2 + 4 - 5

= -(x - 2)^2 - 1

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2. Always positive or always negative

$y = ax^{2} + bx + c$	Shown by
always positive	$a > 0$ and $b^{2} - 4ac < 0$ , or $a(x - h)^{2} + k$ with $a > 0$ , $k > 0$
always negative	$a < 0$ and $b^{2} - 4ac < 0$ , or $a(x - h)^{2} + k$ with $a < 0$ , $k < 0$

Always positive

\displaystyle x^{2} - 6x + 11

e.g.

\displaystyle x^{2} - 6x + 11

\displaystyle = (x - 3)^{2} + 2

$(x - 3)^{2} \geq 0$ , so $x^{2} - 6x + 11 \geq 2$ : always positive

📖 Worked example

Determine whether $2 - 2x - x^2$ is always positive, always negative or neither for all real values of $x$ .

Step 1 of 2

Take out −1, then complete the square

2 - 2x - x^2 = -(x^2 + 2x) + 2

= -[(x + 1)^2 - 1] + 2

= -(x + 1)^2 + 1 + 2

= -(x + 1)^2 + 3

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3. Maximum and minimum problems

80 m of fencing, three sides, the wall is the fourth

\displaystyle A = x(80 - 2x)

e.g.

\displaystyle x(80 - 2x)

\displaystyle = -2(x^{2} - 40x)

\displaystyle = -2((x - 20)^{2} - 400)

\displaystyle = 800 - 2(x - 20)^{2}

greatest area $800$ m², when $x = 20$

📖 Worked example

A ball is thrown upwards. Its height, $h$ m, $t$ seconds after it is thrown is given by $h = 15 + 20t - 5t^2$ . By completing the square, find the greatest height of the ball.

Step 1 of 3

Take out −5 from the t terms

h = -5(t^2 - 4t) + 15

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Practice questions by skill

1. Maximum and minimum values

Form	Turning point	Value
$y = a(x - h)^{2} + k$ , $a > 0$	minimum point $(h, k)$	minimum value $k$ at $x = h$
$y = a(x - h)^{2} + k$ , $a < 0$	maximum point $(h, k)$	maximum value $k$ at $x = h$

Minimum: complete the square

\displaystyle y = x^{2} - 4x + 7

e.g.

\displaystyle x^{2} - 4x + 7

\displaystyle = (x - 2)^{2} - 2^{2} + 7

\displaystyle = (x - 2)^{2} + 3

minimum value $3$ , when $x = 2$

Maximum: take out −1 first

\displaystyle y = -5 + 4x - x^{2}

e.g.

\displaystyle -5 + 4x - x^{2}

\displaystyle = -(x^{2} - 4x) - 5

\displaystyle = -((x - 2)^{2} - 4) - 5

\displaystyle = -1 - (x - 2)^{2}

maximum value $-1$ , when $x = 2$

📖 Worked example

By completing the square, find the maximum value of $y = -5 + 4x - x^2$ .

Step 1 of 2

Take out −1, then complete the square inside

-5 + 4x - x^2 = -(x^2 - 4x) - 5

= -[(x - 2)^2 - 4] - 5

= -(x - 2)^2 + 4 - 5

= -(x - 2)^2 - 1

Practise this →

2. Always positive or always negative

$y = ax^{2} + bx + c$	Shown by
always positive	$a > 0$ and $b^{2} - 4ac < 0$ , or $a(x - h)^{2} + k$ with $a > 0$ , $k > 0$
always negative	$a < 0$ and $b^{2} - 4ac < 0$ , or $a(x - h)^{2} + k$ with $a < 0$ , $k < 0$

Always positive

\displaystyle x^{2} - 6x + 11

e.g.

\displaystyle x^{2} - 6x + 11

\displaystyle = (x - 3)^{2} + 2

$(x - 3)^{2} \geq 0$ , so $x^{2} - 6x + 11 \geq 2$ : always positive

📖 Worked example

Determine whether $2 - 2x - x^2$ is always positive, always negative or neither for all real values of $x$ .

Step 1 of 2

Take out −1, then complete the square

2 - 2x - x^2 = -(x^2 + 2x) + 2

= -[(x + 1)^2 - 1] + 2

= -(x + 1)^2 + 1 + 2

= -(x + 1)^2 + 3

Practise this →

3. Maximum and minimum problems

80 m of fencing, three sides, the wall is the fourth

\displaystyle A = x(80 - 2x)

e.g.

\displaystyle x(80 - 2x)

\displaystyle = -2(x^{2} - 40x)

\displaystyle = -2((x - 20)^{2} - 400)

\displaystyle = 800 - 2(x - 20)^{2}

greatest area $800$ m², when $x = 20$

📖 Worked example

A ball is thrown upwards. Its height, $h$ m, $t$ seconds after it is thrown is given by $h = 15 + 20t - 5t^2$ . By completing the square, find the greatest height of the ball.

Step 1 of 3

Take out −5 from the t terms

h = -5(t^2 - 4t) + 15

Practise this →