Question

How do you graph `y = - abs(x-3) + 5`?

Answer 1

graph{-abs(x-3)+5 [-7.27, 10.93, -2.696, 6.404]}

`y = x+2 , if x<=3`
`y = -x+8, if x>3`

Explanation:

Recall : The "absolute value" function is defined as :

`abs(x)= x, if x>=0`
`abs(x) =-x, if x<0`
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Graph of `abs(x)` :
graph{abs(x) [-, 10, -5, 5]}

Here, we have `abs(x-3)`. Using the definition of `abs(x)` we can write :
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`abs(x-3) = x-3, if x>=3`
`abs(x-3) = -(x-3) = -x+3, if x<3`
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Your function doesn't contain `abs(x-3)` but `-abs(x-3)`, we have to adapt what I wrote in the last paragraph.

`color(red)-abs(x-3) = -x+3, if x>=3`
`color(red)-abs(x-3) = x-3, if x<3`
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Graph of `-abs(x-3)` :
graph{-abs(x-3) [-10, 10, -5, 5]}
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Now we have to add 5 to `-abs(x-3)` without regard of the value of `x`
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Then :

`-abs(x-3)+5 = -x+3+5 = -x+8, if x>=3`
`-abs(x-3) = x-3+5 = x+2, if x<3`
``
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Note : I don't know how to use multi-line equations with socratic, then sorry for heavy notations.