I've drawn everything for you, see if you can figure anything out
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 is symmetric
with respect to the line , the line and/or the origin.
We have: .
Replace: .
. . Multiply by -1: . . . . the original equation!
The graph is symmetric to the origin.
Symmetry to
Replace: .
. . This cannot be made equal to the original equation.
The graph is not symmetric to
Symmetry to
Replace: .
. . This cannot be made to equal the orignal equation.
The graph is not symmetric to