Question

From the curve y = |x|, draw y = |x + 1| − 1

Answer 1

y = |x|

y = `{{:(x,  "if"  x ≥ 0),(- x,  "if"  x < 0):}`

x	0	1	2	3	− 1	− 2	− 3
y	0	1	2	3	1	2	3

(a) Consider y = |x + 1|

y = `{{:((x + 1),  "if"  x + 1 ≥ 0),(-(x + 1),  "if"  x + 1 < 0):}`

y = `{{:(x + 1,  "if"  x ≥ - 1),(-(x + 1),  "if"  x < - 1):}`

x = 0 ⇒ y = x + 1 ⇒ y = 1

x = 1 ⇒ y = x + 1 ⇒ y = 2

x = 2 ⇒ y = x + 1 ⇒ y = 3

x = 3 ⇒ y = x + 1 ⇒ y = 4

x = – 1 ⇒ y = x + 1 ⇒ y = 0

x = – 2 ⇒ y = – (x + 1) ⇒ y = 1

x = – 3 ⇒ y = – (x + 1) ⇒ y = 2

x	0	1	2	3	1	2	3
y	1	2	3	4	0	1	2

The graph of y = |x + 1| shifts the graph y = |x| to the left by 1 unit.

The graph of y = f( x + c), c > 0 causes the graph y = f(x) a shift to the left by e units.

(b) Consider y = |x + 1| – 1

y = `{{:((x + 1) - 1,  "if"  x + 1 ≥ 0),(-(x + 1) - 1,  "if"  x + 1 < 0):}`

y  = `{{:(x,  "if"  x ≥ - 1),(- 2 - 2,  "if"  x < - 1):}`

x = 0 ⇒ y = x ⇒ y = 0

x = 1 ⇒ y = x ⇒ y = 1

x = 2 ⇒ y = x ⇒ y = 2

x = 3 ⇒ y = x ⇒ y = 3

x = – 1 ⇒ y = x ⇒ y = – 1

x = – 2 ⇒ y = – x – 2 ⇒ y = 0

x = – 3 ⇒ y = – x – 2 ⇒ y = 1

x = – 4 ⇒ y = – x – 5 ⇒ y = – 1

x	0	1	2	3	– 1	– 2	– 3	– 4
y	0	1	2	3	– 1	0	1	– 1

The Graph of y = |x + 1| – 1 shift the graph y = |x| to the left by 1 unit and causes a shift downward by 1 unit.

The graph of y = f(x + c), c > 0 causes the graph y = f(x) a shift to the left by c units.

The graph of y = f(x) – d, d > 0 causes the graph y = f(x) a shift to the downward by d units.