Quadratic functions have graphs that are parabolas.
The general equation of a quadratic function can be given in two forms:
In both forms the highest exponent of x is 2.
The vertex, stationary point or turning point gives either the maximum or minimum value of the function.
Sketching Parabolas
Method 1: Factorisation.
If the equation is in the form y = ax^{2} + bx + c the following method should be used:
Step (a) Factorise the function.
Step (b) Find the x and yintercepts by putting y = 0 and x = 0.
Step (c) Find the axis of symmetry − always midway between the two xintercepts.
Step (d) Find the coordinates of the vertex or turning point.
Answer


Sketch the graph of the function y = x^{2} − 6x + 8 
Step (a) y = x^{2} − 6x + 8 = (x − 4)(x − 2) Step (b) Put x = 0 into the equation. y = 0^{2} − 6 × 0 + 8 The yintercept is 8 Put y = 0 0 = (x − 4)(x − 2) The xintercepts are 4 and 2 Step (c) Axis of symmetry is always midwaybetween the xintercepts. Axis of symmetry is the line x = 3 Step (d) The vertex occurs at x = 3 When x = 3 The vertex is at the point (3, 1) 
Now sketch the graph 
:
Method 2: Transformation.
This method finds the coordinates of the vertex.
The yintercept is found by putting x = 0.
The graph of the basic parabola y = x^{2} is shown in the diagram.
This basic parabola is moved or transformed as follows.
(a) y = ax^{ 2} The a has the effect of changing theparabola in the y direction. It affects the steepness of the graph.


(b) y = x^{ 2} + k The k has the effect of moving the parabola along the yaxis by k units.


(c) y = (x − h)^{2} The h has the effect of moving the basic parabola along the xaxis by h units.
