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## 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.

 The x-intercepts of the parabola given by     y = x2 + x    are: ( , ) and ( , ).

### Exercise 3.

Now try a few on your own.

 The x-intercepts of the parabola given by     y = x2 + x    are: ( , ) and ( , ).

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.

 The x-intercepts of the parabola given by     y = x2 +    are: ( ± √( ) , )

### Exercise 4.

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

 Find the x-intercepts of the parabola given by     y = x2 + (± √( ) , )      OR

Case 3:     y = (ax - b)(cx - d)

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

### Exercise 5.

 Find the x-intercepts of the parabola given by      y = ( x + )( x + ) Write your answers as unrounded decimals.  ( , )     and     ( , )

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.

 Sketch the graph of    y = x2 + x Check your answer using the Quadratic Polynomial applet below.

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 '+'.

### 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

### Exercise 7.

 Sketch the graph of    y = x2 + Check your answer using the Quadratic Polynomial applet below.

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 + 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 '+'.