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How do you find the zeros of #f(x)=(x^2-3x-4)/(x^2-x-12)#?

Answer 1

Ethan Adkins

# x=-1#

Factorising and simplifying f(x) as follows.

#f(x)=(x^2-3x-4)/(x^2-x-12)=(cancel((x-4))(x+1))/(cancel((x-4))(x-3))=(x+1)/(x-3)#

with exclusion x ≠ 4

The zeros of f(x) are the values of x which make f(x) equal zero. That is f(x)=0.

For #f(x)=(x+1)/(x+3)#

The denominator of f(x) cannot be zero as this would make f(x) undefined. The numerator is the only part of the rational function that can equal zero.

#rArrx+1=0rArrx=-1" is the zero of f(x)"#

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Answer 2

Nathan Axel

To find the zeros of the function f(x) = (x^2-3x-4)/(x^2-x-12), we need to set the numerator equal to zero and solve for x.

Setting x^2-3x-4 = 0, we can factor the quadratic equation as (x-4)(x+1) = 0.

Next, we set the denominator x^2-x-12 = 0 and factor it as (x-4)(x+3) = 0.

Therefore, the zeros of the function f(x) are x = 4 and x = -1.

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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.