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When looking at the rational function f of x equals the quantity x minus one times the quantity x plus two times the quantity x plus four all divided by the quantity x plus one times the quantity x minus two times the quantity x minus four , Sue and Ed have two different thoughts. Sue says that the graph of the function has zeros at x = −1, x = 2, and x = 4. Ed says that the function is undefined at those x values and discontinuities are created in the graph . Who is correct? Justify your reasoning by explaining the difference between zeros and discontinuities. (10 points)

Respuesta :

As per the given question, the expression of the function [tex]f(x)[/tex] is as:

[tex]f(x)=\frac{(x-1)(x+2)(x+4)}{(x+1)(x-2)(x-4)}[/tex]

Now, as per the definition of zeros, the zero is that value of x which when plugged into the function should make the function zero (it is also called "making the function vanish"). In our case, plugging in "zero values" of x should make the numerator zero without making the denominator zero. Now, if we plug in the given values of x, which are x=-1, x=2 and x=4 in the function we see that the numerator does not become a zero but the denominator does.

Thus, x=-1, x=2 and x=4 are not the zeros of the function and therefore, Sue is wrong. However, we do have discontinuities at the aforementioned values of x because the denominator becomes a zero at those points and as per the definition of a discontinuity the denominator should be a zero at that point. Therefore, Ed is correct.

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