Question

F(x) = 1/(x ^ 2 - 5x)​

Answer 1

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Answer 2

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Answer 3

The function you provided is F(x) = 1/(x^2 - 5x).

This is a rational function, which means it's a fraction where both the numerator and denominator are polynomials. In this case, the denominator is x^2 - 5x.

You can analyze this function as follows:

1. Find the domain of the function: To determine where the function is defined, you need to identify values of x for which the denominator is not equal to zero because division by zero is undefined. So, set the denominator equal to zero and solve for x:

x^2 - 5x = 0

Factor out an x:

x(x - 5) = 0

Now, you have two solutions:

x = 0 x - 5 = 0 x = 5

So, the function is undefined when x = 0 and x = 5. Therefore, the domain of the function is all real numbers except 0 and 5.

2. Determine the asymptotes: You can also analyze the behavior of the function as x approaches positive and negative infinity. In this case, you have a horizontal asymptote because the degree of the numerator (which is a constant) is less than the degree of the denominator (which is a quadratic polynomial).

As x approaches positive or negative infinity, the function approaches 0. So, there are horizontal asymptotes at y = 0.

3. Find any vertical asymptotes: Vertical asymptotes occur when the denominator becomes zero, but the numerator does not. In this case, you've already found that the denominator is zero at x = 0 and x = 5. Therefore, you have vertical asymptotes at x = 0 and x = 5.

4. Analyze the behavior around x = 0 and x = 5: To determine how the function behaves near these vertical asymptotes, you can examine the sign of the numerator and denominator around these points. This will help you identify the behavior of the function as it approaches the vertical asymptotes.

Near x = 0:

• As x approaches 0 from the left, both x and (x - 5) are negative, so the denominator is positive.
• As x approaches 0 from the right, x is positive, and (x - 5) is negative, so the denominator is negative.
• The numerator is always positive.

So, near x = 0, the function approaches positive infinity from the left and negative infinity from the right. There is a vertical asymptote at x = 0.

Near x = 5:

• As x approaches 5 from the left, both x and (x - 5) are negative, so the denominator is positive.
• As x approaches 5 from the right, x is positive, and (x - 5) is negative, so the denominator is negative.
• The numerator is always positive.

So, near x = 5, the function approaches positive infinity from the left and negative infinity from the right. There is a vertical asymptote at x = 5.

Overall, the function F(x) = 1/(x^2 - 5x) has a domain of all real numbers except 0 and 5, horizontal asymptotes at y = 0, and vertical asymptotes at x = 0 and x = 5. The function approaches positive and negative infinity as it gets close to these vertical asymptotes.