The graph of \( g(x)=a\left(\frac{1}{5}\right)^{x}-5 \) passes throughi \( (-2 ;-4) \). 7.1 Show that \( a=\frac{1}{25} \) 7.2 Determine the coordinates of the \( x \) intere \( E^{4} f_{8} \), 7.3 Given: \( h(\tau)=\left(\frac{1}{5}\right)^{x} \) 7.3.1 Determine the cquation of \( x^{-1}(x) \), the inverse of \( h \), in the forks \( y=- \) 7.3.2 Describe the transformation from. g to h .

We are given that   g(x) = a·(1/5)^x – 5 and that the point (–2, –4) lies on the graph of g. We also have a related function   h(x) = (1/5)^x. Below are the steps to solve each part. ───────────────────────────── Step 7.1. Show that a = 1/25 Since (–2, –4) is on the graph of g, we substitute x = –2 and g(–2) = –4 into the equation:   –4 = a·(1/5)^(–2) – 5. Recall that (1/5)^(–2) = 5² = 25. Thus, we have   –4 = a·25 – 5. Now, add 5 to both sides:   –4 + 5 = 25a   1 = 25a. Finally, divide both sides by 25:   a = 1/25. ───────────────────────────── Step 7.2. Determine the coordinates of the x-intercept of g The x-intercept is found by setting g(x) = 0. That is, solve   0 = a·(1/5)^x – 5. We now substitute a = 1/25 into the equation:   0 = (1/25)·(1/5)^x – 5. Add 5 to both sides:   (1/25)·(1/5)^x = 5. Multiply both sides by 25 to isolate (1/5)^x:   (1/5)^x = 125. Recall that 125 = 5³. Also note that (1/5)^x = 5^(–x). Therefore,   5^(–x) = 5³. Since the bases are the same, equate the exponents:   –x = 3 ⟹ x = –3. Thus, the x-intercept is at (–3, 0). ───────────────────────────── Step 7.3. Inverse of h and the transformation from g to h 7.3.1. Find the inverse function h⁻¹ The function is given by   h(x) = (1/5)^x. It is often useful to write this as   h(x) = 5^(–x). To find the inverse, start with   y = 5^(–x). Swap x and y:   x = 5^(–y). Now solve for y by taking the logarithm with base 5 on both sides:   log₅(x) = –y ⟹ y = –log₅(x). Thus, the inverse function is   h⁻¹(x) = –log₅(x). 7.3.2. Describe the transformation from g to h Recall that   g(x) = a·(1/5)^x – 5  with a = 1/25  so that g(x) = (1/25)·(1/5)^x – 5,   h(x) = (1/5)^x. We can express g in terms of h as follows:   g(x) = (1/25)·h(x) – 5. Thus, compared to h, the graph of g is obtained by: 1. Vertically compressing h by a factor of 1/25. 2. Then shifting the resulting graph downward by 5 units. Equivalently, starting with g, if one “undoes” these transformations (that is, first shifting upward by 5 units and then stretching vertically by a factor of 25), one obtains h. ───────────────────────────── Summary of Answers 7.1 a = 1/25. 7.2 The x-intercept of g is at (–3, 0). 7.3.1 The inverse of h is h⁻¹(x) = –log₅ x. 7.3.2 The graph of g is obtained from the graph of h by a vertical compression by a factor of 1/25 followed by a downward shift of 5 units. (Conversely, h can be obtained from g by first shifting upward 5 units and then stretching vertically by a factor of 25.) Any answer equivalent to the above is correct.