## anonymous 5 years ago at what points on the curve does y = x^3 slope at 45 degrees?

1. anonymous

The gradient function is the derivative of your equation. You'd have,$y=x^3 \rightarrow y'=3x^2$Now, the slope of a line is equal to the tangent that line makes with the positive horizontal axis. A slope of 45 degrees is then,$m=\tan 45 ^o=1$Since y' gives the gradient at any x, you have then,$y'=1 \rightarrow 3x^2=1$You then solve for x.

2. anonymous

$x=\pm \frac{1}{\sqrt{3}}$

3. anonymous

oh i see now, i didn't realize that a slope of 45 degrees meant tan 45. thankx a million lokisan!

4. anonymous

np :)