# Homework Help: Intersection of two graphs

1. Sep 7, 2011

### pb23me

1. The problem statement, all variables and given/known data The graphs y=abs(x) and y=(x^2)-6 intersect at x=3 and x= -3
What is confusing me is when I set them equal to eachother and solve (x^2)-x-6=0 and (x^2)+x-6=0 I get -3,+3,-2,+2
What is the deal with the negative 2 and pos 2?

2. Relevant equations

3. The attempt at a solution

2. Sep 7, 2011

### wukunlin

abs(x) = x only if x is positive, so when you solve x^2-x-6=0, you need to disregard negative answers, and vice versa

3. Sep 7, 2011

### rock.freak667

y= |x| is the graph of y=x with the bottom half reflected in the x-axis (so any value of x gives a positive value of y). It looks like a v basically.

Since both y=|x| and y=x2-6 are symmetrical about the y-axis, you will get 'mirrored' answers, so if you get x=2, you will get x=-2 on the next side of the graphs.

Draw them out and you will see the symmetry I am talking about.

4. Sep 7, 2011

### pb23me

Thanks, I did draw them out and the graphs intersect at x=3 and x=-3
They do not intersect at x=2 and x=-2 even though 2 and -2 solve the equation when I set the two functions equal to eachother.

5. Sep 7, 2011

### rock.freak667

Yeah drawing you would see +3 and -3 but it's like wukunlin said.

When you solved x2-x-6=0 you had the constraint of x>0, you would ignore the x=-2.

6. Sep 7, 2011

### pb23me

ohhhhhhhh ok I get it now. Thanks guys