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प्रश्न
From the curve y = x, draw 2x + y + 3 = 0
उत्तर
| x | 0 | 1 | 2 | 3 | − 1 | − 2 | − 3 |
| y | 0 | 1 | 2 | 3 | − 1 | − 2 | − 3 |
2x + y + 3 = 0
y = – 2x – 3
When x = 0 ⇒ y = – 2 × 0 – 3 = – 3
x = 1 ⇒ y = – 2 × 1 – 3 = – 5
x = `1/2` ⇒ y = – 2 × `1/2` – 3 = – 1 – 3 = – 4
x = 2 ⇒ y = – 2 × 2 – 3 = – 4 – 3 = – 7
x = – 1 ⇒ y = – 2 × – 1 – 3 = 2 – 3 = – 1
x = – 2 ⇒ y = 2 × – 2 – 3 = 4 – 3 = 1
x = – 3 ⇒ y = – 2 × – 3 – 3 = 6 – 3 = 3
| x | 0 | 1 | `1/2` | 2 | − 1 | − 2 | − 3 |
| y | − 3 | − 5 | − 4 | − 7 | − 1 | 1 | − 3 |
The graph of y = – 2x – 3 stretches the graph y = x towards the x-axis since the multiplying factor is – 2 which is less than 1 and causes the shift to the downward by 3 units 5.
The graph of y = kf(x), k > 0 moves towards the x-axis if k is less than 1.
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 = x3 − 1
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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)) + 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
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y = − sin(−x)
From the curve y = sin x, graph the function
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From the curve y = |x|, draw y = |x − 1| + 1
From the curve y = |x|, draw y = |x + 1| − 1
