graph, find x and y intercepts and test for symmetry x=y^3

Question

graph, find x and y intercepts and test for symmetry

Answer 1

To Graph:

Create a table of ordered pairs to get a sense of what the graph looks like.

You could begin your table with a few values for x or y.  For instance,

y | -2 -1 0 1 2

x | _ _ _ _ _

Now to complete the table, substitute the values for y into the equation, and solve for x.  For y = -2, substituting -2 for y in the equation gives

x = (-2)^3, and simplifying this gives x = -8.  Now adding this value for x to the table gives

y | -2 -1 0 1 2

x | -8 _ _ _ _

After completing the rest of the table, it should look like this

y | -2 -1 0 1 2

x | -8 -1 0 1 8

and we now can graph the pairs (-8,-2), (-1,-1), (0,0), (1,1), (8,2).  Plot these ordered pairs, and you may get an idea of what the graph looks like.

If more pairs are needed it is simply a matter of repeating the process for other values to get enough pairs to recognize the graph.

x and y-intercepts:

Find the x-intercept by substituting 0 for y in the equation and solving for x.  The resulting ordered pair(s) will have the form (_,0), and they are your x-intercept(s).

Find the y-intercept by substituting 0 for x in the equation and solving for y. The resulting ordered pair(s) will have the form (0,_), and they are your y-intercept(s).

Test for symmetry:

Test for "horizontal" symmetry by replacing 'x' in every instance with '-x'.  If the resulting equation is identical to the original, then the equation has "horizontal" symmetry, also known as "even" symmetry or symmetry over the y-axis.

Test for "vertical" symmetry by replacing 'y' in every instance with '-y'.  If the resulting equation is identical to the original, then the equation has "vertical" symmetry, also known as symmetry over the x-axis.

Test for "odd" symmetry by replacing both 'x' and 'y' with '-x' and '-y', respectively.  If the resulting equation is identical to the original, then the equation has "odd" symmetry, also known as symmetry over the origin.

I hope this helps.  If you need more help, please feel free to ask and I'll be happy to assist you.

 

Answer 2

graphically, it goes through the origin (0,0). That's both the x intercept and the y intercept

x=y^3, let x=0 and y=0 then (0) = (0)^3

The graph is symmetric about the origin, (1,1) is a point on the graph and so is (-1,-1). Just change both signs of a point in quadrant I and you have another point on the curve in quadrant III 1=1^3 and -1 = (-1)^3

Graphically, draw a straight line from any point on the curve and draw it through the origin. continue the line on the other side of the origin and equidistant to the origin will be another point on the curve. (2,8) is on the curve and so is (-2,-8) each of those two points are equally distant to the origin. a straight line connecting those two points will go through the origin and the origin will be the midpoint between those two points.

Answer 3

To solve this problem, it would be preferable to solve this equation for y. Applying the cubic root on both sides, the equation now gives y=x^1/3. Now to find the x- and y- intercepts for this function, two steps must be done. The first step is to substitute 0 in y to find the x-intercept: x = (0)^1/3 = 0. This means that the graph of this function passes through the point (0,0). The next step is to substitute 0 in x to find the y-intercept: 0 = y^1/3. Applying the cubic root on both sides, it is found that y = 0. The function passes through (0,0), so both the x- and y-intercepts are the same. Finally, to use the test of symmetry, three cases have to be considered. First, find out if the function is symmetric about the x-axis. This means that if the variable signs are the same as in the original function after substituting ^-y for y and solving such equation, it is concluded that the function is symmetric about the x-axis. It is found that x = (^-y)³ => x = ^-y³. The solution is not the same as x = y³. So the function is not symmetric about the x-axis. Second, find out if the function is symmetric about the y-axis. This means that if the variable signs are the same as in the original function after substituting ^-x for x and solving such equation, it is concluded that the function is symmetric about the x-axis. It is found that ^-x = y³, which is not the same as the original function. So the function is not symmetric over the x-axis. Finally, find out if the function is symmetric about the origin. For that case, substitute ^-x and ^-y for x and y in the original function. The function has a result of ^-x = ^-y³, which is not the same as the original function, so it is not symmetric about the origin.