how do I show my work?

Question

g (x) = |x|-3

Answer 1

From what you've written, it's not entirely clear what the problem is asking you to do, which is why you've got two very different answers here.

 

 

It may be that you're looking to find the zeroes of the function g(x) = |x| - 3. Zeros are values of x for which g(x) = 0. In other words, plug 0 in for g(x) and solve for x:

 

0 = |x| - 3

3 = |x|

 

There are two values of x which have absolute values of 3:

 

|3| = 3

|-3| = 3

 

So, like Brian says, x = 3 and -3

 

 

It may be that, instead, you're looking to graph the function g(x) = |x| - 3. In that case, think about what a basic absolute value function f(x) = |x| looks like. Like Larry says, it's a V-shape. Its point is at (0, 0), and the legs of the V go diagonally up through (1, 1) on one side and (-1, 1) on the other.

 

Your function is almost the basic absolute value function, except that it has a -3 outside the absolute value sign. This means that you should shift the basic function down by 3. It will have the exact same V-shape, but its point will be at (0, -3) instead of (0, 0) and its legs will go diagonally up through (1, -2) and (-1, -2). Larry has listed several other useful points out for you, if you want to check your work.

Answer 2

Tamica:  Just in case you don't know what the other tutor is telling you, I would like to offer something simpler.  Do you know what a t-chart is?  When you plug -3 in the absolute value of x, you end up with +3 - 3 = zero.  When you plug -2, the outcome is -1 because 2 - 3 = -1.  Do this for -1 and you have l -1 l - 3 = +1 - 3 = -2.  Do this for x values of -3, -2, -1, 0, +1, +2 and +3.  When you do this, your graph will consist of the points (-3,0), (-2,-1), (-1,-2), (0,-3), (1,-2), (2,-1) and (3,0).  You have a "v" graph with the lowest point on the y axis at -3.  Is that the kind of explanation you need, or did Brian's explanation suffice?  Good luck  Larry

 

 

Answer 3

The function is not a quadratic, actually. But because it does have 2 zeroes.

 

Whenever you have |x| in a function, you can always express it as a piecewise defined function, broken up where the expression within the absolute values is equal to zero. In general:

|f(x)| = f(x) if f(x) ≥ 0, -f(x) if f(x)<0

 

so |4x+2| = 4x+2 if 4x+2 ≥ 0, -(4x+2) if 4x+2<0

but those conditions are usually expressed in terms of x, so 4x+2 ≥ 0 would be written x≥ -1/2  and the other condition would be x < -1/2

 

Back to your problem. You can write

g(x) = x - 3 if x≥ 0, -x-3 if x<0

 

So for the two pieces of the function, look for zeroes on the respective domains.

x-3=0 if x=3, and x=3 is indeed in the interval x≥ 0.

-x-3 = 0 if x=-3, and -x is in the interval x<0

So the two zeroes of this function are x=3 and x=-3