horizontal tangents to y = x^4 - 8x^2 + 2, if any

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R.W.

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Determine the points (if any) at which the function has a horizontal tangent line:

. . .y = x^4 - 8x^2 + 2

I know that functions like "1/x^2" don't have any horizontal tangent lines, because the graph is never horizontal. Those are easy.

For this one, I just tried plugging values in for x. I really have no clue what to do. Could someone please help? I know the derivative is:

. . .y' = 4x^3 - 16x

PS: Does anyone know any sites that give sample problems to try out? I would like to practice for my quiz on Monday. Thank you!
 
Re: horizontal tangents

Where is y'=0?
Horizontal lines have slope 0, RIGHT?
 
oh you're right they do have 0 slope. i was thinking that 0 slope=undefined, but it is just horizontal, thanks.
 
Find the derivative, set to 0 and solve for x.

A horizontal slope is slope 0. That's why you set the derivative equal to 0 and solve for x.

\(\displaystyle \L\\4x^{3}-16x=0\)

\(\displaystyle \L\\4x(x^{2}-4)\)

\(\displaystyle \L\\x=0 \;\ and \;\ \pm{2}\)

That's where the slope is equal to 0(horizontal).
 
i need to find the y coordinates too, so i just plug in those x values into y prime right?
 
R.W. said:
i need to find the y coordinates too, so i just plug in those x values into y prime right?
Click to expand...
To find points on the original curve, you would need to plug x-values into the equation for the original curve, not the derivative of that curve.

Eliz.
 
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