Hello,
I am having a difficult time with this so could someone please explain and show me how to exactly do this:
Determine whether the graph of f(x)=x³+6x is symmetric with respect to the line y=x the line y=-x and/or the origin.
Thanks,
Nick
I dont think that quite answers the question!
The point that is "symmetric" to (4, 0) in the line y= x is the point (0, 4) (i.e. a line from (4,0) to (0,4) is perpendicular to y= x and those point are equidistant from the line). In general, the point that is "symmetric" to (x,y) in the line y= x is the point (y, x). Just "swap" x and y coordinates.
To determine if the graph of y= x³+6x is "symmetric with respect to the line y= x", swap x and y to get x= y³+ 6y. Is that the same graph?
To determine if it is "symmetric with respect to the line y= -x", replace y with -x and x with -y: -x= (-y)³+ 6(-y). Is that the same graph?
Finally, to determine "if it is symmetric with respect to the origin", replace x by -x, y by -y: -y= (-x)³+ 6(-x). Is that the same graph?
Hello, Nick!
There are tests for these symmetries . . .
Symmetry to the originDetermine whether the graph of $\displaystyle f(x)\:=\:x^3+6x$ is symmetric
with respect to the line $\displaystyle y=x$, the line $\displaystyle y=-x$ and/or the origin.
$\displaystyle \text{Replace }x\text{ with }\text{-}x\text{ and }y\text{ with }\text{-}y.$
$\displaystyle \text{If the result is the original equation, the graph is symmetric to the origin.}$
We have: .$\displaystyle y \:=\:x^3 + 6x$
Replace: .$\displaystyle \text{-}y \:=\:(\text{-}x)^3 + 6(\text{-}x)\quad\Rightarrow\quad \text{-}y \:=\:\text{-}x^3 - 6x$
. . Multiply by -1: .$\displaystyle y \:=\:x^3 + 6x$ . . . the original equation!
The graph is symmetric to the origin.
Symmetry to $\displaystyle y = x$
$\displaystyle \text{Replace }x\text{ with }y\text{ and }y\text{ with }x.$
$\displaystyle \text{If the result is the original equation, the graph is symmetric to }y = x.$
Replace: .$\displaystyle x \:=\:y^3 + 6y$
. . This cannot be made equal to the original equation.
The graph is not symmetric to $\displaystyle y = x.$
Symmetry to $\displaystyle y = -x$
$\displaystyle \text{Replace }x\text{ with }y\text{ and }y\text{ with }\text{-}x.$
$\displaystyle \text{If the result is the original equation, the graph is symmetric to }y = -x.$
Replace: .$\displaystyle \text{-}x \:=\:y^3 + 6y$
. . This cannot be made to equal the orignal equation.
The graph is not symmetric to $\displaystyle y = -x.$