How do you graph $y = 1 + | x - 1 |$ $y = 1 + |x-1|$ ?

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Answer 1 · 2016-04-05T14:56:15

See explanation

The two vertical lines mean that for this equation the content is always positive and is of generic shape $\lor$ $vv$

You produce a table setting x against y.

Technically you only need 3 points and these are

The lowest value the equation can achieve and any other two pints so that you can form the V shape

Table to demonstrate what happens to the numbers:
Tony B Tony B

Tony B Tony B

Answer 2 · 2016-04-05T15:09:29

At (1, 1) make a right angle by drawing the straight lines $y = x, x > 1 and y = 2 - x, x < 1$ $y = x, x > 1 and y =2-x, x < 1$ ..

$| x - 1 | = x - 1, x > 1 . S o, y = x, x > 1$ $|x-1| = x-1, x > 1. So, y = x, x > 1$ .

$| x - 1 | = - (x - 1), x < 1 . S o, y = 2 - x, x < 1$ $|x-1|=-(x-1), x < 1. So, y = 2-x, x < 1$

The lines are half lines above the common point (1, 1 )..

How do you graph y = 1 + |x-1|y=1+|x−1|y = 1 + |x-1|?