# The graph of y = (x+2) * (x-3) intersects the x-axis at points A and B. Find the length of AB.

*print*Print*list*Cite

y= (x+2) (x-3)

First we need to determine the points where the function y intersects with the x-axis

Since it intersects with x-axis ==> y = 0

Let us substitute:

0 = (x+2)(x-3)

==> x = -2 and x= 3

Then the points of intersection are:

A(-2,0) and B(3,0)

Now let us calculate the length of AB using the formula:

AB= sqrt(3+2)^2 + (0)^2

= sqrt(5)^2 = 5

**==> the length of AB = 5 units.**

The given function is y = (x+2)(x-3).

To find the length of AB , where A and B are the intercepts of y= (x+2)*(x-3) with x and y axis.

To get the the point A , we put y = 0 in the given equation y = (x+2)(x-3). Then Then (x+2)(x-3) = 0. So x+2 = 0, Or x-3 = 0. This gives x = -2 , or x = -3 . So there are two points A1(-2,0) and A2 (3, 0) on x axis where the graph intercepts.

To get the point B, we put x = 0 in y = (x+2)(x-3). So y = (0+2)(0-3) = -6. Therefore the coordintes of B are (0,-6)

Therefore by distance formula, the distance between the two points (x1,y1) and (x2,y2) is given by d = sqrt{(x2-x1)^2+(y2-y1)^2}.

Therefore the distance between A1and B is = sqrt {(0-(-2))^2+(-6-0)^2} = sqrt {4+36} = sqrt40 = 2sqrt10.

The distance between A2 and B = sqrt{(0-3)^2+(-6-0)^2} = sqrt45 = 3sqrt 5.