Question

what are the domain, range, and asymptote of $h(x) = 6^x - 4$?
\\(\circ\\) domain: \\{x | x is a real number\\}; range: \\{y | y > 4\\}; asymptote: $y = 4$
\\(\circ\\) domain: \\{x | x is a real number\\}; range: \\{y | y > -4\\}; asymptote: $y = -4$
\\(\circ\\) domain: \\{x | x > -4\\}; range: \\{y | y is a real number\\}; asymptote: $y = 4$
\\(\circ\\) domain: \\{x | x > 4\\}; range: \\{y | y is a real number\\}; asymptote: $y = -4$

Explanation:

Step1: Analyze the domain of \( h(x) = 6^x - 4 \)

The function \( 6^x \) is an exponential function. For any real number \( x \), \( 6^x \) is defined (since exponential functions with positive bases are defined for all real numbers). Subtracting 4 from \( 6^x \) doesn't restrict the domain. So the domain of \( h(x) \) is all real numbers, i.e., \( \{x \mid x \text{ is a real number}\} \).

Step2: Analyze the range of \( h(x) = 6^x - 4 \)

The range of \( 6^x \) is \( \{y \mid y > 0\} \) (because exponential functions of the form \( a^x \) with \( a>1 \) have a range of positive real numbers). When we subtract 4 from \( 6^x \), we shift the graph down by 4 units. So the range of \( 6^x - 4 \) is \( \{y \mid y > 0 - 4\} = \{y \mid y > - 4\} \).

Step3: Analyze the asymptote of \( h(x) = 6^x - 4 \)

The horizontal asymptote of \( 6^x \) is \( y = 0 \) (as \( x \to -\infty \), \( 6^x \to 0 \)). When we subtract 4 from \( 6^x \), the horizontal asymptote also shifts down by 4 units. So the horizontal asymptote of \( h(x) = 6^x - 4 \) is \( y=-4 \).

Answer:

B. domain: \(\{x \mid x \text{ is a real number}\}\); range: \(\{y \mid y > -4\}\); asymptote: \(y = -4\)