Symetry

**calc_help123** · February 2nd 2009, 06:54 PM

Hi, i hadd a problem y= ((8)/(x^3))-((6)/(x))
I need to know hoe the graph is symetric with respect to the
1)x axis
2)y axis
3)origin
I just needed to know the analytical way to find it without looking at the graph

**earboth** · February 2nd 2009, 10:35 PM

Originally Posted by calc_help123

Hi, i hadd a problem y= ((8)/(x^3))-((6)/(x))
I need to know hoe the graph is symetric with respect to the
1)x axis
2)y axis
3)origin
I just needed to know the analytical way to find it without looking at the graph

The graph of a function is symmetric with respect to:

- the y-axis if f(x) = f(-x) for all $x \in domain$

- the origin if f(x) = -f(-x) for all $x \in domain$

If a graph is symmetric to the x-axis it is not the graph of a function:

For every x-value you have pairs of y-values such that $y_1 = -y_2$

To your function: $f(x)=\dfrac8{x^3} - \dfrac6x = \dfrac{8-6x^2}{x^3}$

a) Symmetry to the y-axis:

$\dfrac{8-6x^2}{x^3} = \dfrac{8-6(-x)^2}{(-x)^3}$

$\dfrac{8-6x^2}{x^3} \neq -\dfrac{8-6(x)^2}{(x)^3}$

That means: No symmetry to the y-axis.

b) Symmetry to the origin:

$\dfrac{8-6x^2}{x^3} = -\dfrac{8-6(-x)^2}{(-x)^3}$

$\dfrac{8-6x^2}{x^3} = \dfrac{8-6(x)^2}{(x)^3}$

That means: Symmetry to the origin.

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