How do you find the restrictions of #(x^3-2x^2-8x)/(x^2-4x)#?

Answer 1

Eva Abramson

#x!=0 and x!= 4#
These values will make the denominator 0.

The restrictions in this case will be any value(s) which make(s) the denominator equal to 0.

You can have 0 as a numerator, but not as a denominator.

What will make #x^2 - 4x=0#?

#x(x-4) = 0" "larr# factorise

#x = 0 " or " x = 4#

These are the restrictions.

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Answer 2

Jace Anderson

To find the restrictions of the given rational function (x^3-2x^2-8x)/(x^2-4x), we need to determine the values of x for which the denominator is equal to zero.

Setting the denominator equal to zero, we have x^2-4x = 0. Factoring out an x, we get x(x-4) = 0.

This equation is satisfied when x = 0 or x = 4.

Therefore, the restrictions of the given rational function are x ≠ 0 and x ≠ 4.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.