Find the domain of the following rational function. h(x)=(x+3)/(x^2)-9 Select the correct choice below and , if necessary, fill in the answer box A. The domain of h(x) is restricted to square (Type your answer in interval notation.) B. The domain of h(x) has no restrictions.

To find the domain of the rational function \( h(x) = \frac{x+3}{x^2-9} \), we need to determine where the denominator is not equal to zero, since division by zero is undefined.<br /><br />The denominator of the function is \( x^2 - 9 \). We set it equal to zero and solve for \( x \):<br /><br />\[<br />x^2 - 9 = 0<br />\]<br /><br />This can be factored as:<br /><br />\[<br />(x - 3)(x + 3) = 0<br />\]<br /><br />Setting each factor equal to zero gives:<br /><br />\[<br />x - 3 = 0 \quad \Rightarrow \quad x = 3<br />\]<br />\[<br />x + 3 = 0 \quad \Rightarrow \quad x = -3<br />\]<br /><br />Thus, the values \( x = 3 \) and \( x = -3 \) make the denominator zero, so they must be excluded from the domain.<br /><br />Therefore, the domain of \( h(x) \) is all real numbers except \( x = 3 \) and \( x = -3 \). In interval notation, this is expressed as:<br /><br />\[<br />(-\infty, -3) \cup (-3, 3) \cup (3, \infty)<br />\]<br /><br />So, the correct choice is:<br /><br />A. The domain of \( h(x) \) is restricted to \( (-\infty, -3) \cup (-3, 3) \cup (3, \infty) \).