Advertisements
Advertisements
प्रश्न
From the curve y = x, draw y = − x
Advertisements
उत्तर
| x | 0 | 1 | 2 | 3 | − 1 | − 2 | − 3 |
| y | 0 | 1 | 2 | 3 | − 1 | − 2 | − 3 |
y = − x
| x | 0 | 1 | 2 | 3 | − 1 | − 2 | − 3 |
| y | 0 | − 1 | − 2 | − 3 | 1 | 2 | 3 |
Graph of y = – x is the reflection of the graph of y = x about the x-axis.
The graph of y = – f(x) is the reflection of the graph of y = f(x) about x-axis.
APPEARS IN
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 = 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))`
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 = x, draw y = 2x
From the curve y = x, draw y = x + 1
From the curve y = |x|, draw y = |x − 1| + 1
From the curve y = |x|, draw y = |x + 2| − 3
