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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 | 2 | 0 | 9 | – 7 |
The graph of y = x3 + 1, causes the graph y = x3 a shift to the upward by 1 unit.
The graph of y = f(x) + d, d > 0 causes the graph y = f(x) a shift to the upward by d units.
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संबंधित प्रश्न
For the curve y = x3 given in Figure 1.67, draw
y = −x3
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))`
Write the steps to obtain the graph of the function y = 3(x − 1)2 + 5 from the graph y = x2
From the curve y = sin x, graph the function.
y = sin(− x)
From the curve y = x, draw y = x + 1
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
