QUESTION 4 Given: \( f(x)=a^{x}-1 \) for \( a>0 . \mathrm{B}\left(2 ; \frac{-5}{9}\right) \) is a point on \( f \). \( 4.1 \quad \) Calculate the value of \( a \). \( 4.2 \quad \) Write down the range of \( f \). \( 4.3 \quad \begin{array}{l}\text { Sketch the graph of } f . \text { Clearly show the intercepts with the axes and asymptotes, } \\ \text { if any. }\end{array} \) It is further given that C is a point on \( f \) at \( y=\frac{19}{8} \). Determine the coordinates of \( \mathrm{C}^{\prime} \), the image of C , when C is reflected about the line \( y=x \).

We are given the function   f(x) = a^x − 1  with a > 0 and that the point B(2, –5/9) lies on f. Later, a point C on f has y = 19/8, and we need to find the coordinates of its reflection C′ in the line y = x. ───────────────────────────── Step 4.1: Find a Since B(2, –5/9) lies on f, the function satisfies   f(2) = a² − 1 = –5/9. Solve for a²:   a² = –5/9 + 1 = –5/9 + 9/9 = 4/9. Taking the positive square root (since a > 0):   a = 2/3. ───────────────────────────── Step 4.2: Determine the Range of f The function is now   f(x) = (2/3)^x − 1. Observations: • The exponential term (2/3)^x is always positive. • As x → ∞, (2/3)^x → 0, so f(x) tends toward –1. • As x → –∞, (2/3)^x → ∞, so f(x) tends toward ∞. Thus, f never actually reaches –1 but can get arbitrarily close to it. Therefore, the range of f is   (–1, ∞). ───────────────────────────── Step 4.3: Sketch the Graph and Reflect Point C 1. Sketching the Graph: • y-intercept: When x = 0, f(0) = (2/3)^0 − 1 = 1 − 1 = 0. So, the graph passes through (0, 0). • Axis intercept: x-intercept occurs when f(x) = 0, which we have already found to be at x = 0. • Horizontal asymptote: As x → ∞, f(x) → –1, so the line y = –1 is a horizontal asymptote. The graph of f is a decreasing exponential curve (since 0 < 2/3 < 1) with a horizontal asymptote y = –1, passing through (0, 0). 2. Finding the Coordinates of C: We are given that point C on f has a y-coordinate of 19/8. This means   f(x) = (2/3)^x − 1 = 19/8. Solve this equation for x:   (2/3)^x = 19/8 + 1 = (19 + 8)/8 = 27/8. Notice that 27/8 can be written as (3/2)^3 because:   (3/2)^3 = 27/8. Also recognize that (2/3)^x = (3/2)^(–x). Therefore, we have:   (3/2)^(–x) = (3/2)^3  ⟹  –x = 3  ⟹  x = –3. So the coordinates of C are (–3, 19/8). 3. Reflecting C about the Line y = x: To reflect any point (p, q) about the line y = x, simply switch the coordinates. Thus, the reflection of C(–3, 19/8) is:   C′ = (19/8, –3). ───────────────────────────── Final Answers: 4.1. a = 2/3. 4.2. The range of f is (–1, ∞). 4.3. The graph of f includes:   • The y-intercept at (0, 0).   • The x-intercept at (0, 0).   • A horizontal asymptote at y = –1.   Point C on f is (–3, 19/8), and its reflection about the line y = x is C′ = (19/8, –3).