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प्रश्न
For the curve y = x3 given in Figure 1.67, draw
y = x3 − 1 
उत्तर
y = x3 – 1
| x | 0 | 1 | – 1 | 2 | – 2 |
| y | – 1 | 0 | – 2 | 7 | – 9 |
The graph of y = x3 – 1, causes the graph y = x3 a shift to the downward by 1 unit.
The graph of y = f(x) – d, d > 0 causes the graph y = f(x) a shift to the downward by d units.
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संबंधित प्रश्न
For the curve y = x3 given in Figure 1.67, draw
y = x3 + 1
For the curve y = x3 given in Figure 1.67, draw
y = (x + 1)3 with the same scale
For the curve y = `x^((1/3))` given in Figure 1.68, draw
y = `- x^((1/3))`
For the curve y = `x^((1/3))` given in Figure 1.68, draw
y = `x^((1/3)) + 1`
For the curve y = `x^((1/3))` given in Figure 1.68, draw
y = `x^((1/3)) - 1`
For the curve y = `x^((1/3))` given in Figure 1.68, draw
y = `(x + 1)^((1/3))`
From the curve y = sin x, graph the function.
y = sin(− x)
From the curve y = sin x, graph the function
y = − sin(−x)
From the curve y = sin x, graph the function
y = `sin(pi/2 + x)` which is cos x
From the curve y = sin x, graph the function
y = `sin (pi/2 - x)` which is also cos x (refer trigonometry)
From the curve y = x, draw y = − x
From the curve y = x, draw y = 2x
From the curve y = x, draw y = `1/2 x + 1`
From the curve y = x, draw 2x + y + 3 = 0
From the curve y = |x|, draw y = |x − 1| + 1
From the curve y = |x|, draw y = |x + 1| − 1
From the curve y = |x|, draw y = |x + 2| − 3
