- #1

Bashyboy

- 1,421

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Hello everyone,

I have the functions [itex]y_1 = \frac{c}{b} + d e^{-bx}[/itex] and[itex]y_2 = \frac{c}{b} - d e^{-bx} [/itex], where [itex]c \in \mathbb{R}[/itex], and [itex]b,d \in \mathbb{R}^+[/itex].

What I would like to know is how to show that these two functions are symmetric about the line [itex]y = \frac{c}{d}[/itex].

What I thought was that if y_1 and y_2 are symmetric about the line [itex]y = \frac{c}{d}[/itex], then the distance between y_1 and y, and the distance between y_2 and y, will be the same. That is, [itex]d_1 = \sqrt{(y_1 - y)^2 + (x - x_0)^2}[/itex] and [itex]d_2 = \sqrt{(y_2 - y)^2 + (x - x_0)^2}[/itex], where [itex]d_1 = d_2[/itex].

Is this a correct way of determining symmetry? Is it true in general? Are there any other ways in which I could prove symmetry?

I have the functions [itex]y_1 = \frac{c}{b} + d e^{-bx}[/itex] and[itex]y_2 = \frac{c}{b} - d e^{-bx} [/itex], where [itex]c \in \mathbb{R}[/itex], and [itex]b,d \in \mathbb{R}^+[/itex].

What I would like to know is how to show that these two functions are symmetric about the line [itex]y = \frac{c}{d}[/itex].

What I thought was that if y_1 and y_2 are symmetric about the line [itex]y = \frac{c}{d}[/itex], then the distance between y_1 and y, and the distance between y_2 and y, will be the same. That is, [itex]d_1 = \sqrt{(y_1 - y)^2 + (x - x_0)^2}[/itex] and [itex]d_2 = \sqrt{(y_2 - y)^2 + (x - x_0)^2}[/itex], where [itex]d_1 = d_2[/itex].

Is this a correct way of determining symmetry? Is it true in general? Are there any other ways in which I could prove symmetry?

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