SOLUTION: Given the following equation, determine if the following symmetry exists. f(x)=-2x^3+4x^2-2x+1 1. x- axis 2. y=x

Algebra ->  Equations -> SOLUTION: Given the following equation, determine if the following symmetry exists. f(x)=-2x^3+4x^2-2x+1 1. x- axis 2. y=x       Log On


   



Question 1073333: Given the following equation, determine if the following symmetry exists.
f(x)=-2x^3+4x^2-2x+1

1. x- axis
2. y=x

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Neither
1. For the graph of a relation to be symmetrical with respect to the x-axis,for every point (x,y), its reflection over the x-axis, point (X,-Y) must also be in the graph.
That means that changing y to -y, you get the same equation.
That cannot happen with a function,
because by definition a function assigns just one y value to every x.
It cannot assign y and -y for the same x.
2. I assume that means symmetry about the y=x line.
For that symmetry, if a point (x,y) is part of the graph, (y,x) must be part of the graph too.
That means that swapping the x and y variables you get the same equation.
Some functions could do that, but the only polynomial that can do that is the degree 1 polynomial f(x)=x or y=x.
If one of the variables has an exponent that the other does not have,
swapping variables cannot possibly yield the same equation.

f%28x%29=-2x%5E3%2B4x%5E2-2x%2B1 is a polynomial in x .
Polynomials in x have graphs like
graph%28300%2C300%2C-10%2C10%2C-10%2C10%2C0.05%28x%2B5%29%28x-1%29%28x-3%29%2B1%29 and graph%28300%2C300%2C-10%2C10%2C-10%2C10%2C0.05%28x%2B5%29%28x%2B1%29%28x-2%29%28x-6%29%29 .
With any luck,
a polynomial of even degree may be symmetrical with respect to the y-axis,
or a polynomial of odd degree may be symmetrical about the origin.