\[ f(x)=9 \log _{8} x \] Domain: \( \square \) Asymptote: \( \square \) \( x \)-intercept: ( \( \square \) , 0) \( y \)-intercept: ( 0 , \( \square \) Question Help: Video eBook

The domain of the function \( f(x) = 9 \log_{8} x \) consists of all positive real numbers, which means \( x > 0 \). This is because the logarithm is only defined for positive inputs. So, the domain can be expressed as \( (0, \infty) \). As for the asymptote, logarithmic functions have a vertical asymptote at the line \( x = 0 \). This is due to the fact that as \( x \) approaches zero from the right, \( f(x) \) will tend to negative infinity. So, the vertical asymptote is at \( x = 0 \). For the \( x \)-intercept, you can find this point by setting \( f(x) = 0 \): \( 0 = 9 \log_{8} x \). This implies \( \log_{8} x = 0 \), which leads to \( x = 1 \). Thus, the \( x \)-intercept is at (1, 0). To find the \( y \)-intercept, set \( x = 1 \): \[ f(1) = 9 \log_{8}(1) = 9 \times 0 = 0 \]. However, since the logarithm of zero or negative numbers is undefined, we can conclude there is no `y`-intercept in the traditional sense as it refers to a vertical crossing on a function graphed on the Cartesian plane. In conclusion, to wrap it up, we have: Domain: \( (0, \infty) \) Asymptote: \( x = 0 \) \( x \)-intercept: (1, 0) \( y \)-intercept: Not applicable (or could be (0,0) if evaluated at the neutral position).