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Quadratic Polynomials

x-intercepts

The points on the graph cut by the x-axis (that is, y = 0) are called the x-intercepts.

For a general parabola given by y = ax² + bx+ c , these might not exist, as the parabola may lie wholly above or below the x-axis.
The x-coordinate of an x-intercept is given by a solution of the equation ax² + bx+ c = 0.

We consider a few easy cases:

Case 1:    c = 0, ax² + bx = 0 

Note: x is a common factor. Thus x(ax + b) = 0.

Hence either x = 0 or x = -b/a.
Thus the x-intercepts in this case are (0, 0) and (-b/a, 0). These coincide if b = 0.

Example 3.

Study a few more examples.

Exercise 3.

Now try a few on your own.

Case 2:     b = 0, ax² + c = 0 

We rewrite the equation as

Or

This has no real solution if both a and c are the same sign (that is, both are positive or both are negative).
It has two solutions otherwise, one for each of the positive and negative cases of the square root.

Example 4.

Here are some worked examples. The notation sqrt stands for and is common notation used in programs such as Excel and MATLAB.

Exercise 4.

Feeling confident? Good. Try a few of then on your own.

Case 3:     y = (ax - b)(cx - d)

In this situation, the x-intercepts are (b/a, 0) and (d/c, 0).

Exercise 5.

Case 4:     y = ax² + bx + c

Regardless of the values of a, b and c, we can use the quadratic formula to find the x-intercepts. We will look at this in a later section titled "The Quadratic Formula".

Note: If the x-intercepts exist and are distinct then the x-coordinate of the vertex is the mid-point of the intercepts.

Example 5.

Draw the parabola given by the equation y = x² − 6x.

The x-intercepts are (0, 0) and (6, 0).

Hence the vertex is situated at x = 6/2 = 3.

And y = 3² – 6(3) = -9. The vertex is (3, -9).

Exercise 6.

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. 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 )

Example 6.

Draw the parabola given by the equation y = x² − 4.

The x-intercepts are (2, 0) and (-2, 0), hence the vertex is (0, -4).

The graph is