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प्रश्न
For the curve y = x3 given in Figure 1.67, draw
y = −x3 
उत्तर
y = – x3
| x | 0 | 1 | – 1 | 2 | – 2 |
| y | 0 | – 1 | 1 | – 8 | 8 |
The graph y = – x3 is the reflection of the graph y = x3 about x-axis.
The graph of y = – f(x) is the reflection of the graph of y = f(x) about x-axis.
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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
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)^((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 = 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 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
