 
        Dan F.
asked  10/16/12y=x+ abs x+1 abs
I dont know what to do. I graph it on calculator but what do i do next
2 Answers By Expert Tutors
 
FREDERICK S. answered  10/19/12
Doc Fred, Super Tutor Kitsap - my specialties are: math, SAT, ACT...
Look Dan, Robert in CA gave you an answer; I hope that helps.
But the stump for you was that your calculator graphed it and you did not understand the steps, right?
the given function was y = x + |x+1|
you should know that if x+1 is >= o, then the function is y = x + (x+1) = 2x+1, but if x+1 < 0 [ x < -1],
then the function is y = x + -(x+i) = -1 or a flat line down there...
That is wat you device did, in milliseconds = graph a flat line to x=-1
and then a rising line with slope 2 after that. Get it? Got it? Good!
 
Robert C. answered  10/16/12
Dr. Robert can help you with Math and Science
I am interpreting yoru question as:
y = x + | x + 1|
The absolute value of a number is either the number or its opposite, whichever is positive.
For x < 0, |x| = -x. For x ≥ 0, |x| = x
Examples:
|-2| = -(-2) = 2
|7| = 7
For your equation, you have |x+1|. When x + 1 < 0, |x+1| = -(x+1) = -x-1. When x+1 ≥ 0, |x+1| = x+1.
x+1 ≥ 0 is the same as saying x ≥ -1.
x + 1 < 0 is the same as saying x < -1
So your equation could be described as two separate equations depending on what the vlaue of x is:
y = x + | x + 1|
For x < -1:
y = x + -(x+1)
or y = -1
For x ≥ -1:
y = x + x +1
or y = 2x + 1
To summarize:
y = -1 {x < -1} and y = 2x + 1 {x ≥ -1}
This is called a piecewise function since it is defined in pieces. All functions with absolute values can be turned into piecewise functions.
This graph should look like a steep upward climbing line to the right of x = -1 and a horizontal line below the y axis to the left of x = -1. At x = -1, the y value is -1 for both halves of the equation, so it is a continuous function.
I'm not sure where the question wants you to go with this since you are not given any y or x values to plug in and use to solve. Perhaps there is more to the problem you can add with the discussion feature.
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Tamara J.
What is the problem asking you to solve for?
10/16/12