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MakcuM [25]
4 years ago
7

What is the shape graphed by the function r = 1+ sin theta

Mathematics
1 answer:
IRISSAK [1]4 years ago
3 0

Answer: Is known as a "heart" shape.

Step-by-step explanation:

r = 1 + sin(θ)

Let's do it without a graph, let's use only math and logic:

remember that θ is measured from the x-axis

when θ = 0, we have r = 1 (so we have a radius of 1 over the x-axis)

when θ = pi/2, we have r = 1 + sin(pi/2) = 2

when θ = pi, we have r = 1 + sin(pi) = 1

when θ = 3*pi/2, we have r = 1 + sin(3*pi/2) = 0

First, we have symetry around the y-axis,

now, notice that the value of x in θ = 0, θ = pi and θ = 3*pi/2 is the same. so this is not a circle, this is actually a circle where the bottom part is flatted.

But not actually flat, because between θ = pi and θ = 2pi we are in the negative y-axis, so in this region we have two small bumps that connect in the point (0, 0)

This is a kinda "heart" shape.

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A function f is even if the graph of f is symmetric with respect to the y-axis. Algebraically, f is even if and only if f(-x) = f(x) for all x in the domain of f.

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- First function (to the left)

The function:

g(x)=5x^2

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The function:

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We have the following results:

- First function (to the left): even

- Second graph (to the right): odd

- First function (to the left): even

- Second function (to the right): odd

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