To find the vertical asymptotes and holes of the rational function
[tex]\[ h(x) = \frac{x+3}{x(x+2)} \][/tex]
we need to analyze the factors of the numerator and the denominator separately.
1. Identify the vertical asymptotes:
Vertical asymptotes occur where the denominator of the function is equal to zero, provided that these points do not also cancel out with the numerator (which would be holes). To find vertical asymptotes, set the denominator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x(x + 2) = 0 \][/tex]
This equation factors into:
[tex]\[ x = 0 \quad \text{or} \quad x + 2 = 0 \][/tex]
Solving these equations gives us:
[tex]\[ x = 0 \quad \text{or} \quad x = -2 \][/tex]
Therefore, the vertical asymptotes are at [tex]\( x = 0 \)[/tex] and [tex]\( x = -2 \)[/tex].
2. Determine if there are any holes:
Holes occur when a factor in the denominator can be canceled out by a factor in the numerator. To check for holes, we factor both the numerator and the denominator and see if there are any common factors.
- The numerator [tex]\( x + 3 \)[/tex] is already factored.
- The denominator is [tex]\( x(x + 2) \)[/tex].
Since [tex]\( x + 3 \)[/tex] does not factor further and does not share any common factors with the denominator, there are no common factors to cancel out.
Therefore, there are no holes in the function.
Summarizing:
- The vertical asymptotes are at [tex]\( x = -2 \)[/tex] and [tex]\( x = 0 \)[/tex].
- There are no holes in the graph of the function.
So, the correct choice is:
A. The vertical asymptote(s) is(are) [tex]\(x = -2, 0\)[/tex]. There are no holes.
