Quadratic Graphs (Edexcel IGCSE Maths A (Modular))

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Naomi C

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Quadratic Graphs

What is a quadratic graph?

A quadratic graph has the form $y = a x^{2} + b x + c$
- where $a$ is not zero

What does a quadratic graph look like?

A quadratic graph is a smooth curve with a vertical line of symmetry
- A positive number in front of $x^{2}$ gives a u-shaped curve
- A negative number in front of $x^{2}$ gives an n-shaped curve
The shape made by a quadratic graph is known as a parabola
A quadratic graph will always cross the $y$ -axis
A quadratic graph crosses the $x$ -axis twice, once, or not at all
- The points where the graph crosses the $x$ -axis are called the roots
If the graph is a u-shape, it has a minimum point
If the graph is an n-shape, it has a maximum point
Minimum and maximum points are both examples of turning points
- A turning point can also be called a vertex

Diagram showing a positive quadratic curve with a minimum point and a negative quadratic curve with a maximum point.

How do I sketch a quadratic graph?

It is important to know how to sketch a quadratic curve
- A simple drawing showing the key features is often sufficient
- (For a more accurate graph, create a table of values and plot the points)
To sketch a quadratic graph:
- First sketch the $x$ and $y$ -axes
- Identify the $y$ -intercept and mark it on the $y$ -axis
  - The $y$ -intercept of $y = a x^{2} + b x + c$ will be $(0, c)$
  - It can also be found by substituting in $x = 0$
- Find all root(s) (0, 1 or 2) of the equation and mark them on the $x$ -axis
  - The roots will be the solutions to $y = 0$ ; $a x^{2} + b x + c = 0$
  - You can find the solutions by factorising, completing the square or using the quadratic formula
- Identify if the number $a$ in $a x^{2} + b x + c$ is positive or negative
  - A positive value will result in a u-shape
  - A negative value will result in an n-shape
- Sketch a smooth curve through the $x$ and $y$ -intercepts
  - Mark on any axes intercepts
  - Mark on the coordinates of the maximum/minimum point if you know it

How do I find the coordinates of the turning point by completing the square?

The coordinates of the turning point (vertex) of a quadratic graph can be found by completing the square
For a quadratic graph written in the form $y = a {(x - p)}^{2} + q$
- the minimum or maximum point has coordinates $(p, q)$
Beware: there is a sign change for the $x$ -coordinate
- A curve with equation $y = {(x - 3)}^{2} + 2$ , has a minimum point at $(3, 2)$
- A curve with equation $y = {(x + 3)}^{2} + 2$ , has a minimum point at $(- 3, 2)$
The value of $a$ does not affect the coordinates of the turning point but it will change the shape of the graph
- If it is positive, the graph will be a u-shape
  - The curve $y = 5 {(x - 3)}^{2} + 2$ has a minimum point at $(3, 2)$
- Of it is negative, the graph will be an n-shape
  - The curve $y = - 8 {(x - 3)}^{2} + 2$ has a maximum point at $(3, 2)$

How do I find the coordinates of the turning point using differentiation?

The coordinates of the turning point (maximum/minimum) of a quadratic can be found through differentiation
To find the coordinates of the turning point
- Differentiate the quadratic equation $y = a x^{2} + b x + c$
  - This will give you $\frac{d y}{d x}$
- Set $\frac{d y}{d x} = 0$ and solve for $x$
  - The solution will be the $x$ -coordinate of the turning point
- Substitute the $x$ value into $y = a x^{2} + b x + c$
  - This will give you the $y$ -coordinate of the turning point

Worked Example

(a) Sketch the graph of $y = x^{2} - 5 x + 6$ showing the $x$ and $y$ intercepts clearly.

The $+ c$ at the end is the $y$ -intercept

$y$ -intercept: (0, 6)

Factorise the quadratic expression

$y = (x - 2) (x - 3)$

Solve $y = 0$

$(x - 2) (x - 3) = 0, so x = 2 or x = 3$

So the x-intercepts are given by the coordinates

(2, 0) and (3, 0)

It is a positive quadratic graph, so will be a u-shape

Graph of y=x²-5x+6 with y-intercept (06) and roots (2, 0), (3, 0) marked on.

(b) Sketch the graph of $y = x^{2} - 6 x + 13$ showing the $y$ -intercept and the coordinates of the turning point.

It is a positive quadratic, so will be a u-shape
The turning point will therefore be a minimum

The $+ c$ at the end is the $y$ -intercept

$y$ -intercept: (0, 13)

Find the minimum point by completing the square (or through differentiation)

For example, complete the square by writing the equation in the form $a {(x - p)}^{2} + q$ (you may need to look this method up)

$\begin{array}{rcl} x^{2} - 6 x + 13 & = & {(x - 3)}^{2} - 9 + 13 \\ = & {(x - 3)}^{2} + 4 \end{array}$

The turning point of $y = a {(x - p)}^{2} + q$ has coordinates $(p, q)$
The minimum point is therefore

(3, 4)

As the minimum point is above the $x$ -axis, and the curve is a u-shape, this means the graph will not cross the $x$ -axis (it has no roots)

Graph of y=x²-6x+13, with y-intercept (0,13) and minimum point (3, 4) marked on.

(c) Sketch the graph of $y = - x^{2} - 4 x - 4$ showing the root(s), $y$ -intercept, and the coordinates of the turning point.

It is a negative quadratic, so will be an n-shape
The turning point will therefore be a maximum

The $+ c$ at the end is the $y$ -intercept

$y$ -intercept: (0, -4)

Find the minimum point by completing the square or through differentiation

For example, differentiate the equation

$\frac{d y}{d x} = - 2 x - 4$

Set the derivative equal to zero and solve for $x$

$\begin{array}{rcl} 0 & = & - 2 x - 4 \\ 2 x & = & - 4 \\ x & = & - 2 \end{array}$

Substitute this value of $x$ back into the original equation for $y$

$\begin{array}{rcl} y & = & - {(- 2)}^{2} - 4 (- 2) - 4 \\ y & = & - 4 + 8 - 4 \\ y & = & 0 \end{array}$

This shows that the maximum point has coordinates

(-2, 0)

As the maximum is on the $x$ -axis, there is only one root

Graph of y=-x²-4x-4 with root (-2, 0) and y-intercept (0, -4) marked on.

How do I find the equation of a quadratic from its graph?

If the vertex and one other point are known
- Use the form $y = a {(x - p)}^{2} + q$ to fill in $p$ and $q$
  - The vertex is at $(p, q)$
- Then substitute in the other known point $(x, y)$ to find $a$
If the roots ( $x$ -intercepts) and one other point are known
- Use the form $y = a (x - x_{1}) (x - x_{2})$ to fill in $x_{1}$ and $x_{2}$
  - The roots are at $(x_{1}, 0)$ and $(x_{2}, 0)$
- Then substitute in the other known point $(x, y)$ to find $a$
If $a = 1$ then you only need either the vertex or the roots

Worked Example

(a) Find the equation of the graph below.

Positive u-shaped curve with roots at (2,0) and (3,0) and y-intercept of (0,24)

The graph shows the roots and a point on the curve (in this case the $y$ -intercept)

Use the form $y = a (x - x_{1}) (x - x_{2})$ to fill in $x_{1}$ and $x_{2}$ by inspection

The roots are at $(x_{1}, 0)$ and $(x_{2}, 0)$

$y = a (x - 2) (x - 3)$

Substitute in the other known point (0, 24) to find $a$

$\begin{array}{rcl} 24 & = & a (0 - 2) (0 - 3) \\ 24 & = & a (- 2) (- 3) \\ 24 & = & 6 a \\ 4 & = & a \end{array}$

Write the full equation

$y = 4 (x - 2) (x - 3)$

You could also write this in expanded form: $y = 4 x^{2} - 20 x + 24$

(b) Find the equation of the graph below.

Positive u-shaped curve with vertex at (9,-16) and a point on the curve at (2,82)

The graph shows the vertex and a point on the curve

Use the form $y = a {(x - p)}^{2} + q$ to fill in $p$ and $q$ by inspection

The vertex is at $(p, q)$

$y = a {(x - 9)}^{2} - 16$

Substitute in the other known point (2, 82) to find $a$

$\begin{array}{rcl} 82 & = & a {(2 - 9)}^{2} - 16 \\ 82 & = & a {(- 7)}^{2} - 16 \\ 82 & = & 49 a - 16 \\ 98 & = & 49 a \\ 2 & = & a \end{array}$

Write the full equation

$y = 2 {(x - 9)}^{2} - 16$

You could also write this in expanded form: $y = 2 x^{2} - 36 x + 146$

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Quadratic Graphs (Edexcel IGCSE Maths A (Modular))

Revision Note

Quadratic Graphs

What is a quadratic graph?

What does a quadratic graph look like?

How do I sketch a quadratic graph?

How do I find the coordinates of the turning point by completing the square?

How do I find the coordinates of the turning point using differentiation?

Worked Example

How do I find the equation of a quadratic from its graph?

Worked Example

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Author: Naomi C