How do you identify all asymptotes or holes for #g(x)=(x+3)+5/(x+2)#?

Question

Ariana Alvarez · Accepted Answer

Ther is an asymptote at X=-2

Do you understand why as the divisor gets smaller and smaller the answer gets bigger and bigger. As the divisor approaches zero the value of the function approaches infinity...an asymptote on the graph. With this function we will get an asymptote when X+2=0 or when X=-2

Greyson Brown · Answer

undefined To identify the asymptotes and holes for the function $ g(x) = \frac{x+3}{x+2} $, follow these steps:

1. **Vertical Asymptote**: Set the denominator equal to zero and solve for $ x $. Any value of $ x $ that makes the denominator zero will result in a vertical asymptote. In this case, $ x = -2 $ gives us a vertical asymptote.

2. **Horizontal Asymptote**: If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is the $ x $-axis ($ y = 0 $). If the degree of the numerator is equal to the degree of the denominator, divide the leading coefficients to find the horizontal asymptote. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. Here, since the degrees of the numerator and denominator are the same (both are 1), the horizontal asymptote is the ratio of the leading coefficients, which is $ y = 1 $.

3. **Oblique Asymptote** (if applicable): If the degree of the numerator is one more than the degree of the denominator, perform long division to find the oblique asymptote. Since this isn't the case here, there's no oblique asymptote.

4. **Hole**: Check if there's a factor common to both the numerator and the denominator that can be canceled out. If so, determine the value(s) of $ x $ that would make the function undefined after canceling out the common factor. These are the $ x $-values of the hole(s). In this function, both the numerator and the denominator share a common factor of $ (x + 3) $, so there's a hole at $ x = -3 $.

So, for the function $ g(x) = \frac{x+3}{x+2} $, the vertical asymptote is $ x = -2 $, the horizontal asymptote is $ y = 1 $, and there's a hole at $ x = -3 $.

Sadie Ackerman · Answer

undefined To identify all asymptotes or holes for $ g(x) = \frac{(x+3)+5}{x+2} $, follow these steps:

1. Check for Vertical Asymptotes:
   - Vertical asymptotes occur where the denominator equals zero and the numerator does not. So, set $ x+2 = 0 $ and solve for $ x $. Here, $ x = -2 $. Therefore, there is a vertical asymptote at $ x = -2 $.

2. Check for Holes:
   - To check for holes, factor the numerator and denominator and cancel out common factors. Here, the numerator is $ (x+3) + 5 $ and the denominator is $ x + 2 $.
   - Simplify the expression: $ g(x) = \frac{x + 8}{x + 2} $.
   - Notice that both the numerator and denominator share a common factor of $ (x + 2) $. Canceling this common factor leaves $ g(x) = x + 8 $.
   - There is no longer a denominator in this expression, which means there are no holes in the graph of $ g(x) $.

Therefore, the identified asymptote is a vertical asymptote at $ x = -2 $, and there are no holes in the graph of $ g(x) $.