Find the intercepts of y=x^4-10x^2-96
I can't seem to find them. I set y to zero to find x and set x to zero to find y. Thanks.
When x = 0, y = -96, so the y-intercept is (0,-96).
When y = 0, you have the quartic equation $\displaystyle 0 = x^4 - 10x^2 - 96$. Quartics are not easy to solve but you can let $\displaystyle z = x^2$, so
$\displaystyle 0 = z^2 - 10z - 96$. This factors to $\displaystyle 0 = (z-16)(z+6) \Rightarrow z = x^2 = 16, -6 \Rightarrow x = \pm 4, \pm \sqrt{6}i$. Discard the imaginary solutions since the x-axis only takes on real values. Therefore the x-intercepts are (-4,0) and (4,0).