Question

Please help me out here with precalc.

I want the answers for the holes of the function when solved and the end behaviors when x is going to negative and positive.
I solved the rest I just don’t know the holes for this function.

Answer 1

To find the value of x for which f(x) = 0, we can set the numerator of the function equal to zero and solve for x:

x^2 - x + 6 = 0

Using the quadratic formula, we get:

x = [1 ± sqrt(1 - 4(1)(6))] / 2

x = [1 ± sqrt(-23)] / 2

Since the discriminant is negative, there are no real solutions to this equation. Therefore, the function f(x) has no real zeros.

To find the limit of f(x) as x approaches positive or negative infinity, we can divide both the numerator and denominator of the function by x^2:

f(x) = (1 - 1/x + 6/x^2) / (1 - 9/x^2)

As x approaches infinity, both 1/x and 6/x^2 approach zero, while 9/x^2 approaches zero faster than 1/x^2. Therefore, the limit of the denominator is 1, and the limit of the numerator is 1. Thus, we have:

Limit of f(x) as x approaches positive infinity = 1/1 = 1

Similarly, as x approaches negative infinity, both 1/x and 6/x^2 approach zero, while 9/x^2 approaches zero faster than 1/x^2. Therefore, the limit of the denominator is 1, and the limit of the numerator is 1. Thus, we have:

Limit of f(x) as x approaches negative infinity = 1/1 = 1

Therefore, the limit of f(x) as x approaches infinity (either positive or negative) is 1.

Answer 2

`f(x) = \frac{ {x}^(2) - x + 6}{ {x}^(2) - 9 } = ((x - 3)(x + 2))/((x - 3)(x + 3))`

There is a hole at x = 3.

As x-->negative infinity, f(x)-->1.

As x-->infinity, f(x)-->1.