Quadratic Graphs

A quadratic is a function of the form $y = a x^{2} + b x + c$ where $a$ is not zero
They are a very common type of function in mathematics, so it is important to know their key features

What does a quadratic graph look like?

The shape made by a quadratic graph is known as a parabola
The parabola shape of a quadratic graph can either look like a “u-shape” or an “n-shape”
- A quadratic with a positive coefficient of $x^{2}$ will be a u-shape
- A quadratic with a negative coefficient of $x^{2}$ will be an n-shape
A quadratic will always cross the $y$ -axis
A quadratic may cross the $x$ -axis twice, once, or not at all
- The points where the graph crosses the $x$ -axis are called the roots
If the quadratic is a u-shape, it has a minimum point (the bottom of the u)
If the quadratic is an n-shape, it has a maximum point (the top of the n)
Minimum and maximum points are both examples of turning points

Quadratic Graphs Notes Diagram 1

How do I sketch a quadratic graph?

We could create a table of values for the function and then plot it accurately, however we often only require a sketch to be drawn, showing just the key features
The most important features of a quadratic are

Its overall shape; a u-shape or an n-shape
Its $y$ -intercept
Its $x$ -intercept(s), these are also known as the roots
Its minimum or maximum point (turning point)

If it is a positive quadratic ( $a$ in $a x^{2} + b x + c$ is positive) it will be a u-shape
If it is a negative quadratic ( $a$ in $a x^{2} + b x + c$ is negative) it will be an n-shape
The $y$ -intercept of $y = a x^{2} + b x + c$ will be $(0, c)$
The roots, or the $x$ -intercepts will be the solutions to $y = 0$ ; $a x^{2} + b x + c = 0$

You can solve a quadratic by factorising, completing the square, or using the quadratic formula
There may be 2, 1, or 0 solutions and therefore 2, 1, or 0 roots

The minimum or maximum point of a quadratic can be found by;

Completing the square

Once the quadratic has been written in the form $y = p {(x - q)}^{2} + r$ , the minimum or maximum point is given by $(q, r)$
Be careful with the sign of the x-coordinate. E.g. if the equation is $y = {(x - 3)}^{2} + 2$ then the minimum point is $(3, 2)$ but if the equation is $y = {(x + 3)}^{2} + 2$ then the minimum point is $(- 3, 2)$

Using differentiation

Solving $\frac{d y}{d x} = 0$ will find the $x$ -coordinate of the minimum or maximum point
You can then substitute this into the equation of the quadratic to find the $y$ -coordinate

Worked example

Sketch the graph of $y = x^{2} - 5 x + 6$ showing the $x$ and $y$ intercepts

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

The $+ c$ at the end is the $y$ -intercept, so this graph crosses the $y$ -axis at (0,6)

Factorise

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

Solve $y = 0$

$(x - 2) (x - 3) = 0$

$x = 2 or x = 3$

So the roots of the graph are

(2,0) and (3,0)

cie-igcse-quadratic-graphs-we-1

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

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

The $+ c$ at the end is the $y$ -intercept, so this graph crosses the y-axis at

(0,13)

We can find the minimum point (it will be a minimum as it is a positive quadratic) by completing the square:

$x^{2} - 6 x + 13 = {(x - 3)}^{2} - 9 + 13 = {(x - 3)}^{2} + 4$

This shows that the minimum point will be

(3,4)

As the minimum point is above the $x$ -axis, this means the graph will not cross the $x$ -axis i.e. it has no roots

We could also show that there are no roots by trying to solve $x^{2} - 6 x + 13 = 0$

If we use the quadratic formula, we will find that $x$ is the square root of a negative number, which is not a real number, which means there are no real solutions, and hence no roots

cie-igcse-quadratic-graphs-we-2

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

It is a negative quadratic, so will be an n-shape

The $+ c$ at the end is the $y$ -intercept, so this graph crosses the $y$ -axis at (0, -4)

We can find the maximum point (it will be a maximum as it is a negative quadratic) by completing the square:

$- x^{2} - 4 x - 4 = - 1 (x^{2} + 4 x + 4) = - 1 ({(x + 2)}^{2} - 4 + 4) = - {(x + 2)}^{2}$

This shows that the maximum point will be

(-2, 0)

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

We could also show that there is only one root by solving $- x^{2} - 4 x - 4 = 0$

If you use the quadratic formula, you will find that the two solutions for $x$ are the same number; in this case -2

cie-igcse-quadratic-graphs-we-3

Quadratic Graphs (CIE IGCSE Maths: Extended)

Revision Note