Question

Can someone help to graph g(X)=abs(x+2)-abs(x-3)+1 , by turing it into a piecewise

Answer 1

by looking at some pdfs, i was able to see that we could first work with the term abs(x+2),

we woudl get :

g(x)={ - (x+2)-|x-3|+1,x+2<0
{(x+2)-|x-3|+1,x+2>=0

Answer 2

then that would simplify to:

g(x)={-x-1-|x-3|, if x<-2
{x+3-|x-3|, if x>=-2

Answer 3

then we work with the term abs(x-3), which goes like this:

|x-3|={-(x-3), if x-3<0
{(x-3), if x-3>=0



which simplifies too:


|x-3|={-x+3, if x<3
{x-3, if x>=3

Answer 4

now here the author, i feel kinda of just went straigt to the solution, without really expaling. So what i did was try and figure it out by my own terms,

i said, well we have a number line broken up by 2 points:

Answer 5

Start with f(x)=abs(x), which is a V-shaped graph at 45 degrees with the axes.
Using the rules of transformations, f(x+2) means shifting f(x) to the left by 2 units, and similarly f(x-3) means shifting f(x) to the right by 3 units.
Also, -f(x) = inverting the graph, i.e. flipping the graph about the x-axis.
Combining all these, we have to combine the three together.