Given: \( h(x)=-2.3^{x}+2 \) (a) Write down the equation of the asymptote. (b) Draw the graph of \( h(x) \). (c) Find the range of \( h(x) \). (d) Find the domain of \( h(x) \). (e) For which values of \( x \) is \( h(x) \geqslant-16 \) ? (f) For

We are given the function   h(x) = –2·3ˣ + 2 and asked to answer several parts. (Note: In the expression h(x) = –2.3ˣ + 2 the “–2” is a factor multiplying 3ˣ.) Let’s go through the parts one by one. ─────────────────────────────── Part (a): Equation of the Horizontal Asymptote For exponential functions of the form h(x) = A·3ˣ + k the horizontal asymptote is y = k provided the exponential term tends to zero in one direction. In our function   h(x) = –2·3ˣ + 2, as x → –∞, we have 3ˣ → 0 so h(x) → –2·0 + 2 = 2. Thus, the horizontal asymptote is   y = 2. ─────────────────────────────── Part (b): Sketching the Graph of h(x) To sketch the graph of h(x) = –2·3ˣ + 2, consider these key points and features: 1. Horizontal Asymptote: y = 2. 2. y-intercept: Set x = 0.   h(0) = –2·3⁰ + 2 = –2·1 + 2 = 0.   So the graph passes through (0, 0). 3. Behavior as x → ∞: When x increases, 3ˣ grows rapidly. Multiplying by –2 makes the term very large in magnitude but negative. Thus,   h(x) → –∞. 4. Behavior as x → –∞: As mentioned, 3ˣ → 0 so   h(x) → 2.   The graph approaches the line y = 2 from below or above depending on the transformation. Here, note that for x very negative, 3ˣ is positive but very small; then –2·(small positive) is very small and negative, so h(x) is just a little less than 2. (In fact, h(x) = 2 – (a small positive number).) 5. Additional point: It’s often good to have another point. Let x = 1:   h(1) = –2·3¹ + 2 = –6 + 2 = –4.   So (1, –4) is on the graph. When drawing the graph:  • Start with the horizontal asymptote y = 2.  • Plot the point (0, 0) and (1, –4).  • Notice the curve is decreasing because the factor –2 makes the exponential function “flip” vertically.  • As x increases, the function falls steeply toward –∞.  • As x decreases, the curve approaches y = 2 without ever touching it. ─────────────────────────────── Part (c): Finding the Range of h(x) Because 3ˣ is always positive (for every real x, 3ˣ > 0) and –2·3ˣ is always negative, h(x) is always less than 2. In detail:  – For any real x, 3ˣ > 0 ⇒ –2·3ˣ < 0;  – Thus, h(x) = 2 + (a negative number) < 2. Also, as x → ∞, h(x) → –∞. Therefore, the range is all y‐values strictly less than 2. That is,  Range: (–∞, 2). ─────────────────────────────── Part (d): Finding the Domain of h(x) The function h(x) = –2·3ˣ + 2 is defined for every real number x because the exponential function 3ˣ is defined for all real x.  Domain: (–∞, ∞). ─────────────────────────────── Part (e): For Which Values of x Is h(x) ≥ –16? We need to solve the inequality:   –2·3ˣ + 2 ≥ –16. Step 1. Subtract 2 from both sides:   –2·3ˣ ≥ –18. Step 2. Divide both sides by –2. (Remember: dividing by a negative number reverses the inequality.)   3ˣ ≤ 9. Step 3. Express 9 as a power of 3:   9 = 3². So the inequality becomes:   3ˣ ≤ 3². Since the exponential function 3ˣ is increasing, we can take logarithms or simply equate the exponents:   x ≤ 2. Thus, the solution to the inequality is   x ≤ 2. ─────────────────────────────── Part (f): (Incomplete Statement) It seems that part (f) of your question was not fully included. Could you please provide the complete details or the full wording of part (f) so that I can help you with that part as well? ─────────────────────────────── Summary of Answers: (a) Horizontal asymptote: y = 2. (b) Sketch: The graph has a y-intercept at (0, 0), passes through (1, –4), decreases steeply as x increases (approaching –∞), and approaches the horizontal line y = 2 as x → –∞. (See the explanation above for details on how to sketch.) (c) Range: (–∞, 2). (d) Domain: (–∞, ∞). (e) h(x) ≥ –16 for all x such that x ≤ 2. Please provide the complete text for part (f) so I can continue.