How many intercepts does #y = x^2 + x − 6 # have?

1 Answer
Aug 16, 2015

This function has a total of three intercepts.

Explanation:

There are two types of intercepts a function can have, #y#-intercepts, which are calculated for #x = 0#, and #x#-intercepts, which are calculated for #y = 0#.

So, starting with the #y#-intercept, you have #x = 0#, which results in

#y = (0)^2 + 0 - 6#

#y = -6#

This means that the #y#-intercept, which is the point where the graph of the function intercepts the #y#-axis, will be #(0,-6)#.

Now for the #x#-intercept. In order to find the #x#-intercept, you need to have #y=0#, which means that

#y = x^2 + x - 6 = 0#

The discriminant of this quadratic equation will actually tell you how many #x#-intercepts you have.

For ageneral form quadratic equation

#color(blue)(ax^2 + bx + c = 0)#

the discriminant is defined as

#color(blue)(Delta = b^2 - 4ac)#

For your quadratic, #a = 1#, #b=1#, and #c = -6#. The discriminant will be equal to

#Delta = 1^2 - 4 * 1 * (-6)#

#Delta = 1 + 24 = 25#

When #Delta>0#, the quadratic equation will have two disctinct real roots, which is another way of saying that the graph of the function will intercept the #x#-axis in two points.

These roots will be equal to

#color(blue)(x_(1,2) = (-b +- sqrt(Delta))/(2a)#

In your case, you have

#x_(1,2) = (-1 +- sqrt(25))/(2 * 1)#

#x_(1,2) = (-1 +- 5)/2 = {(x_1 = (-1 - 5)/2 = -3), (x_2 = (-1 + 5)/2 = 2):}#

The two #x#-intercepts will thus be #(-3,0)# and #(2, 0)#.

Therefore, the function will have a total of three intercepts, one #y#-intercept and two #x#-intercepts.

graph{x^2 + x - 6 [-20.27, 20.28, -10.14, 10.13]}