Question

can some help me plot g(X)=abs(x+2)-abs(x-3)+1

Answer 1

Maka a table of x-y points and plot it.

Answer 2

yeah,but what i am wanting is turn it into a piecewise

Answer 3

theres no easy way about it, its intuitive, turning it into a peicewise function

Answer 4

perhaps using a number line?

Answer 5

you see in reality i have already turned it into a piecewise, expecpt i cannot get the third piece correct, so if someone could kind of explaine that.....i know i am really not making much sense lol

Answer 6

technically theres nothing to do cause its not an equation.. is it = 0?

Answer 7

nvm lol sry, i missted PLOT

Answer 8

yes, i need plot it

Answer 9

oh ok then this is prolly simplier than you realize, you just make a table and plug in x values and compute y values, then plot points

Answer 10

but, can anybbody help me graph it by turing it into a piece wise adn graphing the point on the given intervals?

Answer 11

or graphing the functiosn on the intervals

Answer 12

what i would do, is graph and then turn it into a peicewise

Answer 13

how can you do that?

Answer 14

i gotta go my boss is yelling at me :/ good luck? ... y= mx +b ...

Answer 15

yeah yeah, thanks tell you boss hi lol:)

Answer 16

If you wish to turn it into a piecewise, all you need to do is to first figure out where the "pieces" will be.

Once you have done that, just evaluate the abs() function.
Remember that abs(x) = x when x >= 0 and abs(x) = -x when x < 0

So for example, if you have abs(2x+3) and you want to turn it into a piecewise,
first realise that the point where it "flips" is when 2x+3 = 0 i.e. x = -3/2

So, abs(2x+3) = 2x+3 when 2x+3 >= 0 or x >= -3/2
abs(2x+3) = -2x -3 when 2x+ 3 < 0 or x < -3/2

Answer 17

I've been working on this since my first response, and this is what I came up with.
The two points of interest are -2 and 3
g(x) = -(x + 2) + (x - 3) + 1 = -4, for x <= -2
g(x) = (x + 2) + (x - 3) + 1 = 2x, for -2 <= x <= 3
g(x) = (x + 2) - (x - 3) + 1 = 6, for x >= 3

Answer 18

the function will be greater than zero so long as abs(x+2)>abs(x-3)+1 and less than zero while abs(x+2)<abs(x-3)+1 it just so happens that it equals zero at x=0 too

Answer 19

The way to find points of interest in an absolute value function is to find when it's inner function becomes 0. That's when the sign changes and the 'absolute' part kicks in.

Answer 20

So when x+2 = 0 and when x-3 = 0. You'd then solve for x.