How do you draw the graph of $y = 1 + sin x$ $y=1+sinx$ for $0 \leq x < 2 π$ $0<=x<2pi$ ?

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How do you draw the graph of $y = 1 + sin x$ $y=1+sinx$ for $0 \leq x < 2 π$ $0<=x<2pi$ ?

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Answer 1 · 2016-10-28T13:33:05

You need to understand the base sine graph, and how to do basic transformations.

The graph of $sin x$ $sinx$ will have its $y$ $y$ -intercept at $x = 0$ $x = 0$ . It has an amplitude of $1$ $1$ , so it will always have a maximum of $y = 1$ $y= 1$ and a minimum of $y = - 1$ $y = -1$ . It first goes up, and then comes back down to reach a minimum, passing through the line $y = 0$ $y= 0$ , which is in fact the axis of symmetry.

The period of sine is $2 π$ $2pi$ , that's to say it takes $2 π$ $2pi$ units for it to repeat itself.

As for $y = 1 + sin x$ $y = 1 + sinx$ , this is the graph of $y = sin x$ $y = sinx$ , with the axis of symmetry moved up $1$ $1$ unit to $y = 1$ $y = 1$ . The graph of $y = 1 + sin x$ $y = 1 + sinx$ is shown in the following image.

enter image source here

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How do you draw the graph of $y = 1 + sin x$ $y=1+sinx$ for $0 \leq x < 2 π$ $0<=x<2pi$ ?

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How do you draw the graph of y=1+sinxy=1+sinxy=1+sinx for 0<=x<2pi0≤x<2π0<=x<2pi?

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How do you draw the graph of $y = 1 + sin x$ $y=1+sinx$ for $0 \leq x < 2 π$ $0<=x<2pi$ ?