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4x^2+1 is greater than or equal to 4x
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\[4x ^{2}+1 \ge 4x\]
What is your question @emmyvirkus?
I need to express the solution in interval notation
solution set***
OK, so interval notation, not to be confused with a coordinate on the Cartesian coordinate plane will look like (x, y) or [x, y) or (x, y] etc. Where we use brackets to say we are including the point's and parentheses when we are not including the points. Are you with me so far?
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Yes I understand that part :)
Let's make this simpler.
yes i'm with you!
4x^2 - 4x [>/=] -1 We can arrive at this by moving around our terms. |dw:1405200827685:dw|
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