• Jul 12th 2010, 03:16 PM
jamesk486
What are the coordinates of the points at which the slope of the function with the equation y= x^4 -8x^2 is equal to zero?

• Jul 12th 2010, 03:45 PM
pickslides
Quote:

Originally Posted by jamesk486
What are the coordinates of the points at which the slope of the function with the equation y= x^4 -8x^2 is equal to zero?

This is not a quadratic, its a quartic!

Anyhow I think you are trying to find $0 = x^4-8x^2$

First take out a common factor of $x^2$ giving

$0 = x^2(x^2-8)$ then using the difference of 2 squares.

$0 = x^2(x-\sqrt{8})(x+\sqrt{8})\implies x= 0, 0,\pm\sqrt{8}$
• Jul 12th 2010, 03:47 PM
pickslides
This is a quartic function, not a quadratic! You can solve it by taking out a common factor then using the difference of 2 squares.

$0= x^4-8x^2$

$0= x^2(x^2-8)$

$0= x^2(x-\sqrt{8})(x+\sqrt{8})$

$x=0,0,\pm\sqrt{8}$
• Jul 13th 2010, 04:30 AM
HallsofIvy
This is the problem with not showing any work (one of them, anyway)! It is impossible to be sure exactly what the problem is or what methods you have available to do the problem.

If the problem really is "find where the slope is 0", there are two methods you could use:

1) Set the derivative equal to 0: $4x^3- 16x= 0$. That is easily solvable.

But I suspect, because this is posted in "Pre-algebra and algebra", not even "Precalculus", that you do not know about the derivative of a function. Instead

2) Think of $x^4- 8x^2$ as a "bi-quadratic": $(x^2)^2- 8(x^2)$- that is, as a quadratic in $x^2$: $y^2- 8y$ with $y= x^2$. A quadratic function, with graph a parabola, has slope 0 at its vertex and you can find that by completing the square: $y^2- 8y+ 16- 16= (y- 4)^2- 16$. Now it is easy to find the "y" value where the parabola "turns" and from that the x value(s) where the slope of the original quartic is 0.