Here we have this function:
[tex]y=\frac{x+3}{x^2+8x+15}[/tex]
Remember that, to find a hole, we could follow the steps as follows:
Before putting the rational function into lowest terms, we should factor the numerator and denominator. If there is the same factor in the numerator and denominator, there is a hole. Set this factor equal to zero and solve. The solution is the x-value of the hole.
We could factor the denominator asking for two numbers, whose sum is 8 and its multiplication is 15. These numbers are 5 and 3. Now, put these numbers in brackets as follows:
[tex]x^2+8x+15=(x+5)(x+3)[/tex]
Now, our function can be written as:
[tex]y=\frac{x+3}{x^2+8x+15}=\frac{(x+3)}{(x+3)(x+5)}[/tex]
Notice that the factor "x+3" is in the numerator and the denominator, so there is a hole in:
[tex]\begin{gathered} x+3=0 \\ x=-3 \end{gathered}[/tex]
Now, we're asked to find the point where the hole is. So, we could replace x=-3 in the re-written function. This is:
[tex]y=\frac{1}{x+5}=\frac{1}{-3+5}=\frac{1}{2}[/tex]
Therefore, the hole is located at the point ( -3, 1/2).
And, the correct answer is actually A.
