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प्रश्न
From the curve y = |x|, draw y = |x + 2| − 3
उत्तर
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 the curve y = |x + 2|
y = `{{:((x + 2), "if" x + 2 ≥ 0),(-(x + 2), "if" x + 2 < 0):}`
y = `{{:(x + 2, "if" x ≥ - 2),(- x - 2, "if" x < - 2):}`
x = 0 ⇒ y = x + 2 ⇒ y = 2
x = 1 ⇒ y = x + 2 ⇒ y = 3
x = 2 ⇒ y = x + 2 ⇒ y = 4
x = 3 ⇒ y = x + 2 ⇒ y = 5
x = – 1 ⇒ y = x + 2 ⇒ y = 1
x = – 2 ⇒ y = x + 2 ⇒ y = 0
x = – 3 ⇒ y = – x – 2 ⇒ y = 1
| x | 0 | 1 | 2 | 3 | – 1 | – 2 | – 3 |
| y | 2 | 3 | 4 | 5 | 1 | 0 | 1 |
The graph of y = |x + 2| shifts the graph y = |x| to the left by 2 units.
The graph of y = f(x + c), c > 0 causes the graph y = f(x) a shift to the left by c units.
(b) Consider the curve y = |x + 2| – 3
y = `{{:((x + 2) - 3, "if" x + 2 ≥ 0),(-(x + 2) - 3, "if" x + 2 < 0):}`
y = `{{:(x - 1, "if" x ≥ - 2),(- x - 5, "if" x < - 2):}`
x = 0 ⇒ y = x – 1 ⇒ y = – 1
x = 1 ⇒ y = x – 1 ⇒ y = 0
x = 2 ⇒ y = x – 1 ⇒ y = 1
x = 3 ⇒ y = x – 1 ⇒ y = 2
x = – 1 ⇒ y = x – 1 ⇒ y = – 2
x = – 2 ⇒ y = x – 1 ⇒ y = – 3
x = – 3 ⇒ y = – x – 5 ⇒ y = – 2
| x | 0 | 1 | 2 | 3 | – 1 | – 2 | – 3 |
| y | – 1 | 0 | 1 | 2 | – 2 | – 3 | – 2 |
The graph of y = |x + 2| – 3 shifts the graph y = |x| to the left by 2 units and causes a shift downward by 3 units.
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 down ward by d units.
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