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प्रश्न
From the curve y = x, draw y = x + 1
उत्तर
| x | 0 | 1 | 2 | 3 | − 1 | − 2 | − 3 |
| y | 0 | 1 | 2 | 3 | − 1 | − 2 | − 3 |
y = x + 1
| x | 0 | 1 | 2 | 3 | − 1 | − 2 | − 3 | − 4 |
| y | 1 | 2 | 3 | 4 | 0 | − 1 | − 2 | − 3 |
The graph of y = x + 1 causes the graph y = x shift to 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 = `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))`
Graph the functions f(x) = x3 and g(x) = `root(3)(x)` on the same coordinate plane. Find f o g and graph it on the plane as well. Explain your results
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 = 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
From the curve y = sin x, draw y = sin |x| (Hint: sin(−x) = − sin x)
