For each of the rational functions find: a. domain b. holes c. vertical asymptotes d. horizontal asymptotes e. oblique asymptotes f. y-intercept g. x-intercepts

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

domain: all real numbers except holes and vertical asymptotes

vertical asymptotes where the denominator is zero and the numerator is not zero

holes where the denominator and numerator are both zero

horizontal asymptotes at the x-extremes. When x gets huge, you basically have

f(x) = x^2/x^2 = 1

I'm sure you can handle the rest. Get stuck? Come on back and show how far you got.

To find the domain of a rational function, we need to identify any values of x that make the denominator equal to zero. In this case, the denominator is (x^2 - x - 6). So, the values of x that make the denominator zero are:

(x - 3)(x + 2) = 0

Setting each factor equal to zero, we get:

x - 3 = 0 --> x = 3
x + 2 = 0 --> x = -2

Therefore, the domain of the function is all values of x except x = 3 and x = -2.

a. Domain: All real numbers except x = 3 and x = -2.

To identify any holes in the function, we simplify the rational expression and see if any factors cancel out. In this case, when we simplify the function, we get:

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

We notice that the factor (x + 2) appears in both the numerator and denominator. Therefore, we can cancel this factor out:

f(x) = (x - 1) / (x - 3)

From this, we can see that there is a hole in the graph at x = -2.

b. Hole: At x = -2

To find the vertical asymptotes, we look for values of x that would make the denominator equal to zero, but are not canceled out by factors in the numerator. In this case, the denominator factors are (x - 3)(x + 2), and the numerator does not have any matching factors. Therefore, the values that make the denominator zero, but are not canceled out, are:

x = 3

So, the vertical asymptote is x = 3.

c. Vertical Asymptote: x = 3

To find the horizontal asymptotes, we take the limit of the function as x approaches positive or negative infinity. In this case, the degree of the numerator and the denominator are both 2, so we compare the coefficients of the highest-degree terms. The highest-degree terms are x^2 in both the numerator and the denominator. Therefore, we take the ratio of the leading coefficients:

Horizontal asymptote = 1/1 = 1

d. Horizontal Asymptote: y = 1

Oblique asymptotes occur when the degree of the numerator is one greater than the degree of the denominator. In this case, the degree of the numerator (2) is not one greater than the degree of the denominator (2). Therefore, there are no oblique asymptotes.

e. Oblique Asymptote: None

To find the y-intercept, we set x = 0 and evaluate the function. Plugging x = 0 into the function, we get:

f(0) = (0^2 + 0 - 2) / (0^2 - 0 - 6) = -2 / -6 = 1/3

So, the y-intercept is (0, 1/3).

f. Y-intercept: (0, 1/3)

To find the x-intercepts, we set the numerator equal to zero and solve for x. In this case:

x^2 + x - 2 = 0

Factoring or using the quadratic formula, we find:

(x - 1)(x + 2) = 0

Setting each factor equal to zero, we get:

x - 1 = 0 --> x = 1
x + 2 = 0 --> x = -2

Therefore, the x-intercepts are (1, 0) and (-2, 0).

g. X-intercepts: (1, 0) and (-2, 0)

To find the domain of a rational function, we need to determine the values of x for which the function is defined. In this case, the denominator cannot be equal to zero since it would result in division by zero, which is undefined.

a. Domain:
To find the domain, we set the denominator (x^2 - x - 6) equal to zero and solve for x:
x^2 - x - 6 = 0

Factorizing, we get:
(x - 3)(x + 2) = 0

Setting each factor equal to zero, we have:
x - 3 = 0 or x + 2 = 0
x = 3 or x = -2

Therefore, the domain of the function is all real numbers except x = 3 and x = -2.

b. Holes:
To find any holes in the rational function, we look for common factors between the numerator (x^2 + x - 2) and the denominator (x^2 - x - 6). If there are common factors, they can cancel out, resulting in holes.

In this case, there are no common factors, so there are no holes in the rational function.

c. Vertical Asymptotes:
Vertical asymptotes occur when the denominator of a rational function is zero and the numerator is not zero. In this case, we already found the values of x that make the denominator (x^2 - x - 6) equal to zero: x = 3 and x = -2.

Therefore, the vertical asymptotes of the function are x = 3 and x = -2.

d. Horizontal Asymptotes:
To find the horizontal asymptote, we compare the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at y = 0. If the degrees are equal, then the horizontal asymptote is the ratio of the leading coefficients. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

In this case, the degrees of the numerator (2) and the denominator (2) are equal. Therefore, to find the horizontal asymptote, we compare the leading coefficients of the numerator (1) and the denominator (1):

The horizontal asymptote is at y = leading coefficient of the numerator / leading coefficient of the denominator, which is y = 1/1, or y = 1.

e. Oblique Asymptotes:
Oblique asymptotes occur when the degree of the numerator is exactly one greater than the degree of the denominator.

In this case, the degrees of the numerator (2) and the denominator (2) are the same, so there are no oblique asymptotes.

f. Y-intercept:
To find the y-intercept, we substitute x = 0 into the function and solve for y:

f(0) = (0^2 + 0 - 2) / (0^2 - 0 - 6)
f(0) = -2 / -6
f(0) = 1/3

Therefore, the y-intercept of the function is (0, 1/3).

g. X-intercepts:
To find the x-intercepts, we set the numerator (x^2 + x - 2) equal to zero and solve for x:

x^2 + x - 2 = 0

Factorizing, we get:
(x + 2)(x - 1) = 0

Setting each factor equal to zero, we have:
x + 2 = 0 or x - 1 = 0
x = -2 or x = 1

Therefore, the x-intercepts of the function are (-2, 0) and (1, 0).