Massey logo
Home > College of Sciences > Institute of Fundamental Sciences >
Maths First > Online Maths Help > Algebra > Quadratic Polynomials > Graphs
SEARCH
MASSEY
MathsFirst logo College of Science Brandstrip
  Home  |  Study  |  Research  |  Extramural  |  Campuses  |  Colleges  |  About Massey  |  Library  |  Fees  |  Enrolment

 

Quadratic Polynomials

Graphs

The graph of a quadratic function is a parabola.

Here are some examples:

y=2*x^2 is an upward facing parabola with vertex at the origin. y=x^2 is an upward facing parabola with vertex at the origin. y=(-1)*x^2 is a downward facing parabola with vertex at the origin. It is the reflection of the parabola y=x^2. y=(-2)*x^2 is a downward facing parabola with vertex at the origin. It is the reflection of the parabola y=2*x^2.

y = 2x2
y = x2
y = -x2
y = -2x2

Experiment with the applet below, varying the value of a in the quadratic equation y = ax2 to see how changing a changes the graph. The initial graph shown is y = x2 and remains stationary for comparison.

Can't see the above java applet? Click here to see how to enable Java on your web browser. (This applet is based on free Java applets from JavaMath )

 

Note that the parabola given by y = ax2 opens up (concave upward) if a > 0 and it opens down (concave downward) if a < 0.


Each parabola is symmetric about a vertical line called the axis which passes through a point called the vertex. In the parabola y = ax2, the vertex is the origin.


In general the shape of the parabola given by y = ax2 + bx+ c is the same as that given by y = ax2 except the vertex has been shifted. Launch the applet below to see how the graph changes as you vary each of a, b and c in y = ax2 + bx+ c. The black function is y = ax2 and remains stationary for comparison.


Can't see the above java applet? Click here to see how to enable Java on your web browser. (This applet is based on free Java applets from JavaMath )


Example 1.

Now we look at examples of how to draw the graph of a quadratic function. We do this by first finding the vertex through which the axis can be drawn.

Draw the parabola given by the equation y = x2 + x +

Set x = 0 in the above equation. When x = 0, y = .

We cut the parabola by the horizontal line y = . Substitute y = into the original equation to find the points of intersection of the horizontal line and the parabola:

x2 + x + =

x2 + x = 0

x( x + ) = 0

x = ,


The axis of symmetry passes through the mid-point of
x = and x = .


That is, x =
, and y = = .
The vertex is at (
, )

You can now use the Quadratic Polynomial applet to see this graph.

When using the applet, enter the function you wish to be drawn on the function line and click on new function. The applet starts with f(x) = x^2 + 2x + 1. To look at a particular point on the graph (the vertex for example) enter the value of x in the box provided and the point will be highlighted by a grey '+'.

Can't see the above java applet? Click here to see how to enable Java on your web browser. (This applet is based on free Java applets from JavaMath )


Exercise 1.

Now try a few on your own.

Sketch the graph y = x2 + x +

Check if your sketch is correct by using the Quadratic Polynomial Applet above.

<< Quadratic Polynomials and Functions (Definition) | Quadratic Polynomials Index | y-intercepts of Quadratics

 

   Contact Us | About Massey University | Sitemap | Disclaimer | Last updated: November 21, 2012     © Massey University 2003