4. Draw a sketch graph of \( f(x)=\left(\frac{1}{3}\right)^{x} \). 4.1 Give the domain and range of \( f \). 4.2 For which values of \( x \) is \( f(x)>0 \) ? 4.3 Give the coordinates of any three points on \( f \). 4.4 Give the equation of \( g \) if \( g \) is formed when the graph of \( f \) is moved 3 units down. 4.5 Give the equation of \( h \) if \( h \) is formed when the graph of \( f \) is moved 3 units left. 4.6 Give the equation of \( k \) if \( k \) is formed when the graph of \( f \) is moved 2 units upwards and 3 un to the right. 4.7 Give the domain and range of \( g \). 4.8 Give the domain and range of \( h \). 4.9 Give the domain and range of \( k \). 4.10 Give the asymptotes of \( f, g, h \) and \( k \).
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Beyond the Answer
To sketch the graph of \( f(x) = \left(\frac{1}{3}\right)^{x} \), imagine a smoothly descending curve that approaches the x-axis but never quite touches it. As x increases, the function values decrease, showcasing an exponential decay. The domain of \( f \) is all real numbers, while the range is all positive real numbers (\( f(x) > 0 \)). For \( f(x) > 0 \), all x-values suffice, reinforcing that this function never reaches or dips below zero. Three points on \( f \) could be (0, 1), (1, \(\frac{1}{3}\)), and (-1, 3). If we move \( f \) down by 3 units to form \( g \), the equation becomes \( g(x) = \left(\frac{1}{3}\right)^{x} - 3 \). For \( h \), moving \( f \) left by 3 units results in \( h(x) = \left(\frac{1}{3}\right)^{(x + 3)} \). The equation for \( k \), with shifts of 2 units up and 3 units to the right, would be \( k(x) = \left(\frac{1}{3}\right)^{(x - 3)} + 2 \). The domain of \( g \) and \( h \) remains all real numbers, while the range of \( g \) is \( (-3, \infty) \), the same for \( h \) as this function only shifts left without affecting its positive nature. The range of \( k \), however, is \( (2, \infty) \). For asymptotes, they are all \( y = 0 \) for \( f \), \( g \) and \( h \); for \( k \), the asymptote shifts to \( y = 2 \).
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