Find the zeros of f(x) =x^3+2x^2-3x

0, 1, -3

Step-by-step explanation:

f(x) = x^3 +2x^2 -3x

f(x) = x(x^2 +2x-3)

f(x) = x(x-1)(x+3)

to find zeroes, plug into f(x) and solve for each value of x. doing this gives 0, 1, and -3

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.