The graph of \( g(x)=a\left(\frac{1}{5}\right)^{x}-5 \) passes through \( A(-2 ;-4) \) 7.1 Show that \( a=\frac{1}{25} \) 7.2 Determine the coordinates of the \( x \)-intercept of \( g \). 73 Ciren: \( h(x)=\left(\frac{1}{5}\right)^{x} \) \( 7.3 .1 \quad \) Determine the equation of \( x^{-1}(x) \), the inverse of \( h \), in the form \( y=- \) 7.3 .2 Describe the tuansformation from \( g \) to \( h \)

To solve the problem, we will go through each part step by step. ### 7.1 Show that \( a = \frac{1}{25} \) We know that the function \( g(x) = a\left(\frac{1}{5}\right)^{x} - 5 \) passes through the point \( A(-2, -4) \). This means that when \( x = -2 \), \( g(-2) = -4 \). Substituting \( x = -2 \) into the function: \[ g(-2) = a\left(\frac{1}{5}\right)^{-2} - 5 \] Calculating \( \left(\frac{1}{5}\right)^{-2} \): \[ \left(\frac{1}{5}\right)^{-2} = 5^2 = 25 \] Now substituting this back into the equation: \[ g(-2) = a(25) - 5 \] Setting this equal to -4: \[ 25a - 5 = -4 \] Adding 5 to both sides: \[ 25a = 1 \] Dividing both sides by 25: \[ a = \frac{1}{25} \] Thus, we have shown that \( a = \frac{1}{25} \). ### 7.2 Determine the coordinates of the \( x \)-intercept of \( g \) The \( x \)-intercept occurs when \( g(x) = 0 \). So we set the function equal to zero: \[ 0 = \frac{1}{25}\left(\frac{1}{5}\right)^{x} - 5 \] Adding 5 to both sides: \[ 5 = \frac{1}{25}\left(\frac{1}{5}\right)^{x} \] Multiplying both sides by 25: \[ 125 = \left(\frac{1}{5}\right)^{x} \] We can rewrite 125 as \( 5^{-3} \): \[ 5^{-3} = \left(\frac{1}{5}\right)^{x} \] Since \( \left(\frac{1}{5}\right)^{x} = 5^{-x} \), we have: \[ 5^{-3} = 5^{-x} \] This implies: \[ -3 = -x \quad \Rightarrow \quad x = 3 \] Thus, the coordinates of the \( x \)-intercept are \( (3, 0) \). ### 7.3.1 Determine the equation of \( x^{-1}(x) \), the inverse of \( h \) The function \( h(x) = \left(\frac{1}{5}\right)^{x} \). To find the inverse, we set \( y = h(x) \): \[ y = \left(\frac{1}{5}\right)^{x} \] Now we solve for \( x \): Taking the logarithm of both sides: \[ \log(y) = x \log\left(\frac{1}{5}\right) \] Since \( \log\left(\frac{1}{5}\right) = -\log(5) \): \[ \log(y) = -x \log(5) \] Rearranging gives: \[ x = -\frac{\log(y)}{\log(5)} \] Thus, the inverse function is: \[ x^{-1}(x) = -\frac{\log(x)}{\log(5)} \] ### 7.3.2 Describe the transformation from \( g \) to \( h \) The function \( g(x) = \frac{1}{25}\left(\frac{1}{5}\right)^{x} - 5 \) can be described in terms of transformations applied to \( h(x) = \left(\frac{1}{5}\right)^{x} \): 1. **Vertical Stretch**: The factor \( \frac{1}{25} \) vertically compresses the graph of \( h(x) \) by a factor of 25. 2. **Vertical Shift**: The term \( -5 \) shifts the entire graph down by 5 units. In summary, the transformation from \( h \) to \( g \) involves a vertical compression by a factor of 25 followed by a downward shift of 5 units.