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3 years ago

PLEASE HELP

Mathematics

1 answer:

nydimaria [60]3 years ago

6 0

Answer:

y = -2(x+2)(x-1)/((x+3)(x-6)) = (-2x^2 -2x +4)/(x^2 -3x -18)

Step-by-step explanation:

A polynomial function will have a zero at x=a if it has a factor of (x-a). For the rational function to have zeros at x=-2 and x=1, the numerator factors must include (x+2) and (x-1).

For the function to have vertical asymptotes at x=-3 and x=6, the denominator of the rational function must have zeros there. That is, the denominator must have factors (x+3) and (x-6). Then the function with the required zeros and vertical asymtotes must look like ...

f(x) = (x+2)(x-1)/((x+3)(x-6))

This function will have a horizontal asymptote at x=1 because the numerator and denominator degrees are the same. In order for the horizontal asymptote to be -2, we must multiply this function by -2.

The rational function may be ...

y = -2(x +2)(x -1)/((x +3)(x -6))

If you want the factors multiplied out, this becomes

y = (-2x^2 -2x +4)/(x^2 -3x -18)

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Which function is undefined for x = 0? y=3√x-2 y=√x-2 y=3√x+2 y=√x=2

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For this case, we must indicate which of the given functions is not defined for $x = 0$

By definition, we know that:

$f (x) = \sqrt {x}$ has a domain from 0 to infinity.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. For it to be defined, the term within the root must be positive.

Thus, we observe that:

$y = \sqrt {x-2}$ is not defined, the term inside the root is negative when $x = 0$ .

While $y = \sqrt {x + 2}$ if it is defined for $x = 0.$

$f(x)=\sqrt[3]{x}$ , your domain is given by all real numbers.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. In the same way, its domain will be given by the real numbers, independently of the sign of the term inside the root.

So, we have:

$y = \sqrt [3] {x-2}$ with x = 0: $y = \sqrt [3] {- 2}$ is defined.