Quadratic Graphs (Cambridge International IGCSE Further Maths)

Quadratic graphs

A quadratic is a function of the form $y=ax^{2}+bx+c$ 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 “∪-shape” or a “∩-shape”

  • A quadratic with a positive coefficient of $x^{2}$ will be a ∪-shape
  • A quadratic with a negative coefficient of $x^{2}$ will be a ∩-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 ∪-shape, it has a minimum point (the bottom of the ∪)

• If the quadratic is a ∩-shape, it has a maximum point (the top of the ∩)

• Minimum and maximum points are both examples of turning points

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 key features needed to be able to sketch a quadratic graph are

  • the overall shape

    • ∪-shape graphs occur when $a>0$ (positive quadratic)
    • ∩-shape graphs occur when $a<0$ (negative quadratic)

  • the $x$-intercept(s), these are also known as the roots (there may be none!)

    • roots are found by setting the quadratic function (or $y$) equal to zero

    • i.e. solve $ax^{2}+bx+c=0$

    • if there are no (real) solutions (i.e. no roots), the graph does not intersect the $x$-axis

      • the *discriminant can be used to determine whether a quadratic function has 0, 1 or 2 roots

  • the $y$-intercept

    • this is found by setting $x=0$ in the quadratic function
    • so for $ax^{2}+bx+c$ the coordinates of the $y$-intercept will be $( 0 , c )$

  • the minimum or maximum point (turning point)

    • sometimes a rough idea of where this should lie is enough

    • sometimes the specific coordinates of the turning point will be needed

    • when required the coordinates of the turning point can be found by either completing the square or differentiation

      • in cases where the quadratic has just one root, the graph will touch (rather than cross) the $x$-axis and so this will be the turning point

Worked Example a) Sketch the graph of $y=x^{2}-5x+6$, labelling any intercepts with the coordinate axes.

It is a positive quadratic, so will be a $\cup$-shape

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

Factorise

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

Solve $y=0$

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

$x=2\mathrm{or}x=3$

So the roots of the graph are

(2,0) and (3,0)

b) Sketch the graph of $y=x^{2}-6x+13$, labelling any intercepts with the coordinate axes.

It is a positive quadratic, so will be a $\cup$-shape

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

(0,13)

The discriminant of the quadratic is '$b^{2}-4ac$'

$( - 6 )^{2} - 4 ( 1 ) ( 13 ) = 36 - 52 = - 16$

As the discriminant is negative, there are no (real) roots and the graph does not intersect the $x$-axis

(Note we have included the coordinates of the turning point, $( 3 , 4 )$ to help you visualise the graph, but there was no requirement from the question to do this - on a sketch like this, the turning point should be in the correct quadrant)

c) Sketch the graph of $y=-x^{2}-4x-4$, labelling any intercepts with the coordinate axes and the turning point.

It is a negative quadratic, so will be an $\cap$-shape

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

Factorising

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

This shows that there is only one root and the graph will touch the $x$-axis at the point (-2, 0) This point will also be the turning point - and as this is a negative quadratic - will be a maximum point

Sketching Graphs by Completing the Square

How does completing the square help me sketch graphs?

• Completing the square can quickly tell us the coordinates of the turning point on a quadratic graph

• This is based on the fact that a squared term (e.g. $( x + 1 )^{2}$) cannot be negative

• STEP 1 Complete the square - rewrite $ax^{2}+bx+c=0$ in the form $a ( x + p )^{2} + q$

• STEP 2 Deduce the $x$-coordinate of the turning point

• $( x + p ) ^{2} \ge 0$ for all values of $x$

  • Therefore it's minimum value is 0, and this occurs when $x=-p$

The $x$-coordinate is $-p$

• STEP 3 Deduce the $y$-coordinate of the turning point

• $a ( x + p )^{2} = 0$

  • Therefore $y=q$

The $y$-coordinate is $q$

• STEP 4 The turning point has coordinates $( - p , q )$ This can be considered when sketching the graph of the quadratic function

• Note that the turning point could be a maximum or minimum point - this will depend on the value of $a$

  • $a$ is the coefficient of the $x^{2}$ term
  • If $a$ is positive, the graph is $\cup$- shaped and will have a minimum point
  • If $a$ is negative, the graph is $\cap$- shaped and will have a maximum point

How do I use the graph of a quadratic function to find its range?

• The range of a quadratic function will be shown on its graph by the values $y$ takes

  • i.e. the turning point from a quadratic graph will determine its range

• For the quadratic function $\mathrm{f} ( x )$ whose graph has a minimum point $( x_{min} , y_{min} )$

  • the range of the $\mathrm{f} ( x )$ will be $f\geq y_{min}$

• For the quadratic function $\mathrm{f} ( x )$ whose graph has a maximum point $( x_{max} , y_{max} )$

  • the range of $\mathrm{f} ( x )$ will be $f\leq y_{max}$

• If there any restrictions on the domain of $\mathrm{f} ( x )$ then they could affect the range of $\mathrm{f} ( x )$

Worked Example Sketch the graph of $y = \mathrm{f} ( x )$ where $\mathrm{f} ( x ) = 2 x^{2} - 4 x - 6$, giving the coordinates of the turning point, and any points where the graph intercepts the coordinate axes. Use your graph to write down the range of $\mathrm{f} ( x )$.

STEP 1 - Complete the square.

$\begin{gathered} \mathrm{f} ( x ) & = 2 ( x^{2} - 2 x ) - 6 \\ = 2 [ ( x - 1 )^{2} - 1 ] - 6 \\ = 2 ( x - 1 )^{2} - 8 \end{gathered}$

STEP 2 - Deduce the $x$-coordinate.

$x = - ( - 1 ) = 1$

STEP 3 - Deduce the $y$-coordinate.

$y=-8$

STEP 4 - Label the turning point when sketching the graph of the quadratic function.

The graph has a minimum point so the range will be greater than or equal to the $y$-coordinate of this point.

Final answer: The range of$f(x)$is $f>-8$