Let f(x) = x^{4}- 8x^{2 }, where x is a real number.

Anyone?

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- Apr 7th 2012, 04:57 AMDeeRoseintercepts
Let f(x) = x

^{4}- 8x^{2 }, where x is a real number.

Anyone? - Apr 7th 2012, 05:03 AMa tutorRe: Find the co-ordinates of the points where the graph of f(x) cuts the x-axis
Factorise x^4 - 8x^2=0

- Apr 7th 2012, 05:04 AMDeeRoseRe: Find the co-ordinates of the points where the graph of f(x) cuts the x-axis
Thanks. I knew it was simple, I just wanted to make sure :P

- Apr 9th 2012, 02:04 PMExplanationCentralRe: intercepts
The graph of f(x) will intersect the x axis at y = 0. So 0 = x^4 - 8x^2 = x^2(x^2-8).

This gives you x = 0, x = (8)^(1/2) and x = -(8)^(1/2).