Graphs of Polynomials (Edexcel IGCSE Further Maths)

Quadratic Functions & Graphs

Spec point — 4A

What are the key features of quadratic graphs?

• A quadratic graph can be written in the form $y=ax^{2}+bx+c$ where $a\neq0$

• The shape of the graph is known as a parabola

• The value of a affects the shape of the curve

  • If a is positive the shape is 'u-shaped' ∪
  • If a is negative the shape is 'upside down u-shaped' ∩

• The y-intercept is at the point (0, c)

• The roots are the solutions to $ax^{2}+bx+c=0$

  • These are also known as the x-intercepts or zeroes

  • They can be found by

    • Factorising
    • Quadratic formula
    • Completing the square
    • Your calculator may also be able to find these for you

  • There can be 0, 1 or 2 *x-*intercepts

    • This is determined by the value of the discriminant

• There is an axis of symmetry at $x=-\frac{b}{2a}$

  • If there are two x-intercepts then the axis of symmetry goes through the midpoint between them

• The vertex lies on the axis of symmetry

  • It can be found by completing the square
  • The x-coordinate is $x=-\frac{b}{2a}$
  • The y-coordinate can be found by calculating y when $x=-\frac{b}{2a}$
  • If a is positive then the vertex is the minimum point
  • If a is negative then the vertex is the maximum point

What are the equations of a quadratic function?

• $f(x)=ax^{2}+bx+c$

  • This is the general form
  • It clearly shows the y-intercept (0, c)
  • You can find the axis of symmetry by $x=-\frac{b}{2a}$

• $f(x)=a(x-p)(x-q)$

  • This is the factorised form
  • It clearly shows the roots (p, 0) & (q, 0)
  • You can find the axis of symmetry by $x=\frac{p+q}{2}$

• $f(x)=a(x-h)^{2}+k$

  • This is the vertex form (or completed square form)

  • It clearly shows the vertex (h, k)

  • The axis of symmetry is therefore $x=h$

  • It clearly shows how the function can be transformed from the graph of $y=x^{2}$

    • Vertical stretch by scale factor ­a
    • Translation by vector $( \begin{gathered} h \\ k \end{gathered} )$

How do I find an equation of a quadratic?

• If you have the roots x = p and x = q...

  • Write in factorised form $y=a(x-p)(x-q)$
  • You will need a third point to find the value of a

• If you have the vertex (h, k) then...

  • Write in vertex form$y=a(x-h)^{2}+k$
  • You will need a second point to find the value of a

• If you have three random points (x₁, y₁), (x₂, y₂) & (x₃, y₃) then...

  • Write in the general form $y=ax^{2}+bx+c$
  • Substitute the three points into the equation
  • Form and solve a system of three linear equations to find the values of a, b & c

Exam Hint

• Your calculator may be able to find the roots and turning point of a quadratic function

  • Even on a 'show that' question this can be used to check your answers

Worked Example The diagram below shows the graph of $y=f(x)$, where $f(x)$ is a quadratic function.

The vertex and the intercept with the $y$-axis have been labelled.

Find an expression for $y=f(x)$.

Method 1

Since we know the vertex (turning point), it will be easiest to start with the completed square version of the equation This is $f(x)=a(x-h)^{2}+k$, where the vertex is at $( h , k )$

$\begin{gathered} \mathrm{f} ( x ) & = a ( x - ( - 1 ) )^{2} + 8 \\ = a ( x + 1 )^{2} + 8 \end{gathered}$

We also know the curve $y = \mathrm{f} ( x )$ goes through (0, 6) Put those coordinates into the equation and solve for $a$

$\begin{gathered} 6 = a ( 0 + 1 )^{2} + 8 \\ 6 = a + 8 \\ a = - 2 \end{gathered}$

Substitute $a=-2$ into the expression for $\mathrm{f} ( x )$ Expand the brackets and rearrange into the form required

$\begin{gathered} \mathrm{f} ( x ) & = - 2 ( x + 1 )^{2} + 8 \\ = - 2 ( x^{2} + 2 x + 1 ) + 8 \\ = - 2 x^{2} - 4 x - 2 + 8 \end{gathered}$

$f(x)=-2x^{2}-4x+6$

Method 2

It is also possible to start with the $y=ax^{2}+bx+c$ form *Because the y- intercept is (0, 6) we know that *$c=6$

Goes through $( 0 , 6 )$ means $\mathrm{f} ( x ) = a x^{2} + b x + 6$

It also goes through (-1, 8) Substitute those coordinates into the equation of $y = f ( x )$

Goes through $( - 1 , 8 )$ means

$\begin{gathered} 8 = a ( - 1 )^{2} + b ( - 1 ) + 6 \\ 8 = a - b + 6 \\ a - b & = 2 \end{gathered}$

We need one more piece of information You may remember that the turning point lies on the line $x=-\frac{b}{2a}$ *If not, then use the fact that the x- coordinate of the turning point satisfies *$\mathrm{f}^{'} ( x ) = 0$

Turning point at $( - 1 , 8 )$* means** *$\mathrm{f}^{'} ( - 1 ) = 0$

Differentiate to find $\mathrm{f}^{'} ( x )$ Then solve $\mathrm{f}^{'} ( - 1 ) = 0$ to find another equation with $a$and $b$

$\mathrm{f}^{'} ( x ) = 2 a x + b$

$\begin{gathered} 2 a ( - 1 ) + b & = 0 \\ - 2 a + b & = 0 \end{gathered}$

We now have two simultaneous equations that we can solve to find $a$ and $b$

$\begin{gathered} a - b & = 2 [ 1 ] \\ - 2 a + b & = 0 [ 2 ] \end{gathered}$

Add [1] and [2] together to eliminate $b$

$\begin{gathered} - a & = 2 \\ a = - 2 \end{gathered}$

Substitute into [2] and solve to find $b$

$\begin{gathered} - 2 ( - 2 ) + b & = 0 \\ 4 + b & = 0 \\ b = - 4 \end{gathered}$

Write final answer in form requested

$f(x)=-2x^{2}-4x+6$

Cubic Functions & Graphs

Spec point — 4A

What are the key features of cubic graphs?

• A cubic graph can be written in the form $y=ax^{3}+bx^{2}+cx+d$ where $a\neq0$

• When asked to consider the equation of a cubic and its graph, think about the following

  • y-*axis intercept

    • This occurs at $( 0 , d )$

  • x-*axis intercepts (roots)

    • *x-*coordinates are the solutions to $ax^{3}+bx^{2}+cx+d=0$

  • turning points (maximum and/or minimum)

    • A cubic can have 0 or 2 turning points
    • *x-*coordinates are the solutions to $3ax^{2}+2bx+c=0$
    • these are when the derivative of $ax^{3}+bx^{2}+cx+d$ equals 0

  • The value of $a$ affects the shape of the curve

    • If $a$ is positive the ends of the curve will go 'down on the left and up on the right'
    • If $a$ is negative the ends of the curve will go 'up on the left and down on the right'

How do I sketch the graph of a cubic?

• STEP 1
  Find the y-*axis intercept by setting x = 0 **

• STEP 2 Find the x-axis intercepts (roots) by setting y = 0** *

  • This may require factorising the cubic

• STEP 3 Consider the shape and “start”/”end” of the graph

  • a positive cubic graph starts in third quadrant (“bottom left”) and ends in first quadrant (“top right”)
  • a negative cubic graph starts in second quadrant (“top left”) and ends in fourth quadrant (“bottom right”)

• STEP 4 Consider where any turning points should go

  • Differentiate the equation of the curve and set equal to zero

• STEP 5 Draw with a smooth curve (this takes practice!)

Worked Example Consider the function $\mathrm{f} ( x ) = - x^{3} + b x^{2} + c x$, where $b$ and $c$ are constants.

The graph of $y = \mathrm{f} ( x )$ has two turning points, the $x$-coordinates of which are $1$ and $3$.

(a) Find the values of $b$ and $c$.

The turning points occur when $\mathrm{f}^{'} ( x ) = 0$

Start by differentiating to find $\mathrm{f}^{'} ( x )$

$\mathrm{f}^{'} ( x ) = - 3 x^{2} + 2 b x + c$

That is equal to 0 when $x=1$ and when $x=3$

Substitute those values in to find two equations in $b$ and $c$

$\begin{gathered} - 3 ( 1 )^{2} + 2 b ( 1 ) + c & = 0 \\ - 3 + 2 b + c & = 0 \\ 2 b + c & = 3 \end{gathered}$

$\begin{gathered} - 3 ( 3 )^{2} + 2 b ( 3 ) + c & = 0 \\ - 27 + 6 b + c & = 0 \\ 6 b + c & = 27 \end{gathered}$

Those are simultaneous equations in $b$ and $c$ Subtract the first from the second to eliminate $c$ and find $b$ Then substitute that value into either equation to find $c$

$b=6c=-9$

(b) Sketch the graph of $y = \mathrm{f} ( x )$.

There is no constant term in $\mathrm{f} ( x )$, so the y-intercept will be 0

Substitute the x-coordinates of the turning points into $\mathrm{f} ( x )$ to find the corresponding y-coordinates

$- ( 1 )^{3} + 6 ( 1 )^{2} - 9 ( 1 ) = - 1 + 6 - 9 = - 4

• ( 3 )^{3} + 6 ( 3 )^{2} - 9 ( 3 ) = - 27 + 54 - 27 = 0$

So the turning points are at $( 1 , - 4 )$ and $( 3 , 0 )$

*Find the x- intercepts by solving *$\mathrm{f} ( x ) = 0$

$\begin{gathered} - x^{3} + 6 x^{2} - 9 x & = 0 \\ - x^{3} ( x^{2} - 6 x + 9 ) & = 0 \\ - x^{3} ( x - 3 )^{2} & = 0 \end{gathered}$

$x=0,x=3$

*So the x- intercepts are *$( 0 , 0 )$ and $( 3 , 0 )$ Note that the first is also the y- intercept. and the second is also a turning point

Finally consider the shape of the curve It's a 'negative cubic', so it's going to be 'up on the left and down on the right' Draw a smooth curve incorporating all these features Label the turning points and intercepts