This is a brief summary of quadratics that should be known before commencing maths methods 3&4

## Basic shape

The basic shape of a quadratic or parabola is a U, shown in the blue function in the graph below.

The basic equation is the power function of : $y=x^{2}$ The red function is when it is reflected in the x axis and the equation becomes $y=-x^{2}$ $y=ax^{2}+bx+c$

The easiest way to learn how to draw a quadratic is to do an example such as: $y=2x^{2}-13x+11$

Step 1: Find the y intercept.

Let x=0

When the graph is in the form of  the y intercept will always be the c, which in this case is $y-intercept=11$

Step 2: Find the x intercept.

To find the x intercept we make y = 0 $y=2x^{2}-13x+11=0$

There are two methods that can be used to solve this equation, factorising and using the quadratic formula.

### Factorisation

To factorise we’ll use the cross method. To do this method we just guess and check our factors, and then cross multiply the terms and add them together to ensure we get our b term.  $a=2x\times x=2x^{2}$ $b=-11x+-2x= -13x$ $c=-11\times-1=11$

Therefore our a, b and c terms match up and we get the function $y=(2x -11)(x-1) = 0$ $x-intercepts =\frac{11}{2} or x = 1$ $x-intercepts = \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

So we substitute in our values of a,b and c from the original equation. $x-intercepts = \frac{13\pm\sqrt{(-13)^{2}-4\times2\times11}}{2\times2}\newline =\frac{13\pm\sqrt{81}}{4}\newline =\frac{4}{4} or \frac{22}{4}\newline=1or\frac{11}{2}$

The same as previously.

Remember that some graphs will not have x intercepts. This happens when the graph is entirely above the x axis or entirely below it. In this case the quadratic formula will have no solution and we will not be able to factorise the equation with real numbers.

Step 3: Find the turning point.

we have both our x intercepts we know that the x coordinate of the turning point is exactly halfway between these. $x-TP=\frac{1+\frac{11}{2}}{2}=\frac{13}{4}$

Once we have the x-coordinate, we can substitute this into our equation to find the y –coordinate. $y-TP=2(\frac{13}{4})^{2}-13(\frac{13}{4})+11=-\frac{81}{8}$

TP = (13/4, -81/8)

If you do not have x intercepts we can find the x-coordinate of the TP by using by using the formula: $x=-\frac{b}{2a}\newline =\frac{13}{4}$

Step 4 Draw the shape

Now that we have the intercepts and the turning point we can plot these on our graph and join the dots to draw a curve. Before you move onto the next question make sure your graph contains the following:

• all intercepts
• x and y axis are labelled
• TP
• Correct shape (in the exam you will get marked on shape so it might be a good idea to spend some time practising your graphing technique)

## Turning point form

Sometimes the equation may appear in what we call turning point form: $y=a(x-h)+k$

This is just a form which lets us know where the turning point is straight away. $TP=(h,k)$

To draw the rest of the graph we go through the same process above of setting x to zero to calculate the y intercept and then y to zero to calculate the x intercept.

If in doubt you can always get the same form of the equation as above by expanding out all the terms of the turning point form equation.