Mathematics 9709 · Pure Mathematics 1

The quadratic graph

The graph puts the roots (§1.1), the vertex (§1.2), and the axis crossings into one sketch.
You only need to mark which way it opens, where it crosses, and the turning point.

Which way it opens

The sign of $a$ (the number in front of $x^2$ ) decides it: $a>0$ opens up (vertex = lowest point), $a<0$ opens down (vertex = highest).

y = x²: a = 1 (positive), so it opens up; the vertex is the lowest point.

y = −x²: a = −1 (negative), so it opens down; the vertex is the highest point.

The size of $a$ only changes how steep it is. The sign is what turns it up or down. (Why squaring is never negative: §1.2.)

What the exam asks

The form a quadratic is written in tells you what you can read at once.

Written as…	Read at once	Still must work out
$(x-p)(x-q)$	the roots p, q	y-int, vertex
$a(x-h)^2+k$	the vertex (h, k)	roots, y-int
$ax^2+bx+c$	the y-intercept c	roots, vertex

Sketch it — mark the crossings and the vertex.
How many roots (§1.5).
Find the max or min — the vertex.

The turning point

From completed-square form $a(x-h)^2+k$ : vertex $(h,k)$ , line of symmetry $x=h$ . (How to get there, and why: §1.2.)

y = (x − 2)² + 1: vertex (2, 1), symmetric about the dashed line x = 2.

y = (x − 1)² + 2·vertex (1, 2)

Drag h and k, and the vertex follows the numbers in the bracket.

Where it crosses the axes

$y$ -axis: put $x=0$ → you get $c$ . $x$ -axis: put $y=0$ and solve — those crossings are the roots (calculator, §1.1).

y = x² − 4: crosses the x-axis at the roots ±2; the y-intercept is c = −4.

A parabola can meet the $x$ -axis twice, once, or not at all — that count is the number of real roots (§1.5):

2 roots

1 (repeated)

0 real

Putting a sketch together

Four marks, four steps: direction (sign of $a$ ) · y-intercept ( $x=0$ ) · roots ( $y=0$ , §1.1) · vertex (complete the square, §1.2).

Worked example

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

Up, since $a=1>0$ .
y-intercept $(0,-3)$ .
Roots: $(x+3)(x-1)=0\Rightarrow x=-3,\,1$ .
Vertex: $(x+1)^2-4\Rightarrow(-1,-4)$ .

Plot the five facts and join with a U: roots −3 & 1, y-int −3, vertex (−1, −4).

Worked example

Sketch $y=x^2-3x-4$

Up; y-intercept $(0,-4)$ .
Roots: $(x-4)(x+1)=0\Rightarrow x=-1,\,4$ .
Vertex $\left(\tfrac32,-\tfrac{25}{4}\right)$ — its $x$ sits halfway between the roots.

Roots −1 & 4, y-int −4, vertex (1.5, −6.25), halfway between the roots.

Finding a max or min fast

If you only want the turning point: use $x=\dfrac{-b}{2a}$ , then put that $x$ back in to get $y$ . No need to complete the square. And $a>0$ gives a minimum, $a<0$ gives a maximum.

Worked example

Maximum of $y=16x-7-4x^2$

Order it, read $a,b$ : $y=-4x^2+16x-7,\ \ a=-4,\ b=16$ .
$x=\dfrac{-b}{2a}=\dfrac{-16}{2(-4)}=2$ .
Sub back: $y=-4(2)^2+16(2)-7=9$ .
$a<0$ , so it is a maximum.

Answer

\text{maximum }(2,\,9)

That $x$ is halfway between the roots. If you need the form $a(x-h)^2+k$ , complete the square (§1.2).

y = −4x² + 16x − 7 opens down; the vertex (2, 9) is the maximum.

Marks, and one to finish

Common mistake

Write a range in

f(x)

, never in

x

(§1.2). And when

a<0

, take the negative out before you complete the square, or the maximum ends up in the wrong place.

Your turn— tap to reveal the worked answer (sketch + a downward max)

Sketch $y=x^2-2x-8$ : up; y-int $(0,-8)$ ; roots $(x-4)(x+2)=0\Rightarrow -2,\,4$ ; vertex $(x-1)^2-9\Rightarrow(1,-9)$ .

Max of $y=5-6x-3x^2$ : order $\to-3x^2-6x+5$ , complete the square $\to-3(x+1)^2+8$ .

Answer

\text{(i) roots }-2,4,\ \text{vertex }(1,-9)\quad\text{(ii) max }(-1,\,8)

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The quadratic graph

The graph puts the roots (§1.1), the vertex (§1.2), and the axis crossings into one sketch.
You only need to mark which way it opens, where it crosses, and the turning point.

Which way it opens

The sign of $a$ (the number in front of $x^2$ ) decides it: $a>0$ opens up (vertex = lowest point), $a<0$ opens down (vertex = highest).

y = x²: a = 1 (positive), so it opens up; the vertex is the lowest point.

y = −x²: a = −1 (negative), so it opens down; the vertex is the highest point.

The size of $a$ only changes how steep it is. The sign is what turns it up or down. (Why squaring is never negative: §1.2.)

What the exam asks

The form a quadratic is written in tells you what you can read at once.

Written as…	Read at once	Still must work out
$(x-p)(x-q)$	the roots p, q	y-int, vertex
$a(x-h)^2+k$	the vertex (h, k)	roots, y-int
$ax^2+bx+c$	the y-intercept c	roots, vertex

Sketch it — mark the crossings and the vertex.
How many roots (§1.5).
Find the max or min — the vertex.

The turning point

From completed-square form $a(x-h)^2+k$ : vertex $(h,k)$ , line of symmetry $x=h$ . (How to get there, and why: §1.2.)

y = (x − 2)² + 1: vertex (2, 1), symmetric about the dashed line x = 2.

y = (x − 1)² + 2·vertex (1, 2)

Drag h and k, and the vertex follows the numbers in the bracket.

Where it crosses the axes

$y$ -axis: put $x=0$ → you get $c$ . $x$ -axis: put $y=0$ and solve — those crossings are the roots (calculator, §1.1).

y = x² − 4: crosses the x-axis at the roots ±2; the y-intercept is c = −4.

A parabola can meet the $x$ -axis twice, once, or not at all — that count is the number of real roots (§1.5):

2 roots

1 (repeated)

0 real

Putting a sketch together

Four marks, four steps: direction (sign of $a$ ) · y-intercept ( $x=0$ ) · roots ( $y=0$ , §1.1) · vertex (complete the square, §1.2).

Worked example

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

Up, since $a=1>0$ .
y-intercept $(0,-3)$ .
Roots: $(x+3)(x-1)=0\Rightarrow x=-3,\,1$ .
Vertex: $(x+1)^2-4\Rightarrow(-1,-4)$ .

Plot the five facts and join with a U: roots −3 & 1, y-int −3, vertex (−1, −4).

Worked example

Sketch $y=x^2-3x-4$

Up; y-intercept $(0,-4)$ .
Roots: $(x-4)(x+1)=0\Rightarrow x=-1,\,4$ .
Vertex $\left(\tfrac32,-\tfrac{25}{4}\right)$ — its $x$ sits halfway between the roots.

Roots −1 & 4, y-int −4, vertex (1.5, −6.25), halfway between the roots.

Finding a max or min fast

If you only want the turning point: use $x=\dfrac{-b}{2a}$ , then put that $x$ back in to get $y$ . No need to complete the square. And $a>0$ gives a minimum, $a<0$ gives a maximum.

Worked example

Maximum of $y=16x-7-4x^2$

Order it, read $a,b$ : $y=-4x^2+16x-7,\ \ a=-4,\ b=16$ .
$x=\dfrac{-b}{2a}=\dfrac{-16}{2(-4)}=2$ .
Sub back: $y=-4(2)^2+16(2)-7=9$ .
$a<0$ , so it is a maximum.

Answer

\text{maximum }(2,\,9)

That $x$ is halfway between the roots. If you need the form $a(x-h)^2+k$ , complete the square (§1.2).

y = −4x² + 16x − 7 opens down; the vertex (2, 9) is the maximum.

Marks, and one to finish

Common mistake

Write a range in

f(x)

, never in

x

(§1.2). And when

a<0

, take the negative out before you complete the square, or the maximum ends up in the wrong place.

Your turn— tap to reveal the worked answer (sketch + a downward max)

Sketch $y=x^2-2x-8$ : up; y-int $(0,-8)$ ; roots $(x-4)(x+2)=0\Rightarrow -2,\,4$ ; vertex $(x-1)^2-9\Rightarrow(1,-9)$ .

Max of $y=5-6x-3x^2$ : order $\to-3x^2-6x+5$ , complete the square $\to-3(x+1)^2+8$ .

Answer

\text{(i) roots }-2,4,\ \text{vertex }(1,-9)\quad\text{(ii) max }(-1,\,8)