The Graph and the Vertex | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Which way the curve opens

a>0

→ opens up → minimum

a<0

→ opens down → maximum

Swipe left or right to see the whole diagram →

The sign of a decides which way the parabola opens. Either way the vertex is the turning point, and the curve is symmetric about the dashed line through it.

Reading the vertex from the completed square

The vertex is the turning point. The line of symmetry is the vertical line through it, which divides the parabola into matching halves.

From $y=a(x-h)^2+k$ : the vertex is at $(h,k)$ and the line of symmetry is $x=h$ . If $a>0$ the vertex is a minimum, and if $a<0$ it is a maximum.

Move the vertex sideways: h1Move the vertex up or down: k2

y = (x − 1)² + 2 · the vertex is at (1, 2)

Common mistake

The bracket sign reverses:

y=(x+5)^2-16

has vertex

(-5,-16)

Sketching one from scratch

A sketch needs the shape, intercepts and vertex. An intercept is a point where the curve meets an axis.

Worked example

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

$a=1>0$ , so the curve opens upward.
Put $x=0$ : the $y$ -intercept is $(0,-4)$ .
Put $y=0$ and factorise.
$(x+1)(x-4)=0$
$x=-1 \quad\text{or}\quad x=4$
The symmetry line is halfway between the roots.
$x=\tfrac{-1+4}{2}=\tfrac{3}{2}$
Substitution gives $y=-\tfrac{25}{4}$ .

Answer

\text{Vertex } \left(\tfrac{3}{2},-\tfrac{25}{4}\right)

y\text{-intercept } (0,-4)

x\text{-intercept } (-1,0)

x\text{-intercept } (4,0)

The completed sketch places every feature found in the working: upward shape, both roots, the y-intercept, the vertex and the line of symmetry.

Completing the square gives the vertex x-coordinate $-\tfrac{b}{2a}$ . You can also read the symmetry line from completed-square form, or average two real roots.

Common mistake

Roots solve

y=0

; the vertex is the highest or lowest point. For

y=x^2-8x+5=(x-4)^2-11

, the minimum is

(4,-11)

(9709/12/O/N/23 Q6(a)).

Form	Read off
$ax^2+bx+c$	shape from $a$ ; $y$ -intercept $c$
$a(x-h)^2+k$	vertex $(h,k)$ ; symmetry line $x=h$
$a(x-p)(x-q)$	roots $p$ and $q$

Form: $ax^2+bx+c$
Read off: shape from $a$ ; $y$ -intercept $c$

Form: $a(x-h)^2+k$
Read off: vertex $(h,k)$ ; symmetry line $x=h$

Form: $a(x-p)(x-q)$
Read off: roots $p$ and $q$

Your turnIndependent sketch [5 marks]

Sketch $y=x^2+2x-3$ , labelling both intercepts and the vertex.

Show worked answer

y=(x+3)(x-1)

y=(x+1)^2-4

Answer

x\text{-intercepts } (-3,0),\ (1,0)

y\text{-intercept } (0,-3)

\text{Vertex } (-1,-4)

Which way the curve opens

a>0

→ opens up → minimum

a<0

→ opens down → maximum

Swipe left or right to see the whole diagram →

The sign of a decides which way the parabola opens. Either way the vertex is the turning point, and the curve is symmetric about the dashed line through it.

Reading the vertex from the completed square

The vertex is the turning point. The line of symmetry is the vertical line through it, which divides the parabola into matching halves.

From $y=a(x-h)^2+k$ : the vertex is at $(h,k)$ and the line of symmetry is $x=h$ . If $a>0$ the vertex is a minimum, and if $a<0$ it is a maximum.

Move the vertex sideways: h1Move the vertex up or down: k2

y = (x − 1)² + 2 · the vertex is at (1, 2)

Common mistake

The bracket sign reverses:

y=(x+5)^2-16

has vertex

(-5,-16)

Sketching one from scratch

A sketch needs the shape, intercepts and vertex. An intercept is a point where the curve meets an axis.

Worked example

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

$a=1>0$ , so the curve opens upward.
Put $x=0$ : the $y$ -intercept is $(0,-4)$ .
Put $y=0$ and factorise.
$(x+1)(x-4)=0$
$x=-1 \quad\text{or}\quad x=4$
The symmetry line is halfway between the roots.
$x=\tfrac{-1+4}{2}=\tfrac{3}{2}$
Substitution gives $y=-\tfrac{25}{4}$ .

Answer

\text{Vertex } \left(\tfrac{3}{2},-\tfrac{25}{4}\right)

y\text{-intercept } (0,-4)

x\text{-intercept } (-1,0)

x\text{-intercept } (4,0)

The completed sketch places every feature found in the working: upward shape, both roots, the y-intercept, the vertex and the line of symmetry.

Completing the square gives the vertex x-coordinate $-\tfrac{b}{2a}$ . You can also read the symmetry line from completed-square form, or average two real roots.

Common mistake

Roots solve

y=0

; the vertex is the highest or lowest point. For

y=x^2-8x+5=(x-4)^2-11

, the minimum is

(4,-11)

(9709/12/O/N/23 Q6(a)).

Form	Read off
$ax^2+bx+c$	shape from $a$ ; $y$ -intercept $c$
$a(x-h)^2+k$	vertex $(h,k)$ ; symmetry line $x=h$
$a(x-p)(x-q)$	roots $p$ and $q$

Form: $ax^2+bx+c$
Read off: shape from $a$ ; $y$ -intercept $c$

Form: $a(x-h)^2+k$
Read off: vertex $(h,k)$ ; symmetry line $x=h$

Form: $a(x-p)(x-q)$
Read off: roots $p$ and $q$

Your turnIndependent sketch [5 marks]

Sketch $y=x^2+2x-3$ , labelling both intercepts and the vertex.

Show worked answer

y=(x+3)(x-1)

y=(x+1)^2-4

Answer

x\text{-intercepts } (-3,0),\ (1,0)

y\text{-intercept } (0,-3)

\text{Vertex } (-1,-4)