What are the y and x intercepts of this problem: x^3(x-1)^3(x+1)^2?

Question

I'm trying to graph this problem, but its hard, is there any way i can rearrange it so it doesn't look so intimidating?

Answer 1

It's actually in the best possible form.

The x-intercepts are anything that will make it 0.

Those would be 0 (for the x^3), 1 (for the (x-1)^3), and -1 (for (x+1)^2)

And for the y-intercept, just plug in x = 0, and you'll see that the x^3 out front makes the whole thing 0, so the y-intercept is also 0.

The REAL trick is figuring out what the graph does BETWEEN the intercepts.

One way to get an idea is to divide the number line into regions:

x < -1
-1 < x < 0
0 < x < 1
x > 1

If x<-1, x^3 will be negative, (x-1)^3 will be negative, and (x+1)^2 will be positive (anything squared is positive), so the function will be positive (above the x-axis).

If -1 < x < 0, x^3 will be negative, (x-1)^3 will be negative, and (x+1)^2 will be positive, so the function is STILL positive...this means that the function will dip down and "kiss" the x-axis and then start going back up, the way y=x^2 does.

For 0 < x < 1, x^3 is positive, (x-1)^3 is negative, and (x+1)^2 is positive, so the function will be negative (below the x-axis) in this region.

Finally, for x > 1, all of the factors are positive, so the function will again be positive (and go to infinity).

Answer 2

y = x^3 (x - 1)^3 (x+1)^2
x intercepts means that x = 0 => y = 0
y intercepts means that y = 0 => x = 0 or x = 1 or x =-1

Answer 3

If y= is in front of this mess, it will resemble a graph of y= x^8 except y=0 at x=-1,0,+1 and y has slightly negative values between those points, kind of a W shape instead of a U.

Answer 4

Try multiplying out the second and third terms. Combine like terms and solve. This is assuming the equation is set equal to y. To find the y-intercepts, set y to zero and solve for x. To find the x-intercepts, set x equal to zero and solve for y. Sounds simple, but I see that you have third degree polynomials, which will keep you from using the quadratic formula, unless you can simplify some x's out of the equation, to get to a second degree polynomial.

Answer 5

x= 0 ===> y= 0
y= 0 ===> x= 0 or (x-1)^3 = 0 or (x+1)^2 = 0
x-1 = 0 ===> x= 1
x+1 = 0 ===> x= -1

Answer 6

there are 3-X intercept. they are 0, 1, -1. sub these values into the equation to get the respective y intercept.

Answer 7

the x intercepts are the roots of that equation.

As it is in factors you find them quickly x=0 , x=1 , x= -1 . these three numbers zero one of the factors? so , you have intercepts on axis x

The intercept on y is easy if you put x=0 the first factor is also 0.

So the origin of coodinates is on the curve

Answer 8

there's none (i think)---cuz there's no equal sign...