Let's break down the question step by step: ### **Given:** - Initially, nobody has seen the poster: \(N(0) = 0\) - After 3 days, 10,500 students have seen it: \(N(3) = 10,500\) - Eventually, all students (12,500) will have seen it: horizontal asymptote \(k = 12,500\) - Function form: \(N(x) = \frac{Ca^x}{1 + a^x} + k\) But from the context, the function is likely to be of the **logistic growth** form, which is commonly used in such scenarios: \[ N(x) = \frac{L}{1 + Ce^{-kx}} \] But let's use the given form, which seems to be: \[ N(x) = \frac{Ca^x}{1 + a^x} + k \] Let's proceed step by step. --- ## **Step 1: Sketch the graph** - At \(x = 0\), \(N(0) = 0\) - As \(x \to \infty\), \(N(x) \to 12,500\) (horizontal asymptote) - At \(x = 3\), \(N(3) = 10,500\) --- ## **Step 2: Model the function \(N(x)\) and find the equation** Let’s substitute \(N(0) = 0\): \[ N(0) = \frac{C \cdot a^0}{1 + a^0} + k = \frac{C}{2} + k = 0 \implies C = -2k \] So, the function becomes: \[ N(x) = \frac{-2k a^x}{1 + a^x} + k \] As \(x \to \infty\), \(a^x \to \infty\): \[ N(\infty) = \frac{-2k a^x}{a^x} + k = -2k + k = -k \] But this does not give us the correct horizontal asymptote (\(k=12,500\)). So let's check another approach. Let’s use a more typical **logistic growth model**: \[ N(x) = \frac{L}{1 + Ce^{-kx}} \] Where: - \(L = 12,500\) (asymptote) - \(N(0) = 0\) - \(N(3) = 10,500\) #### At \(x = 0\): \[ N(0) = \frac{12,500}{1 + C} = 0 \implies C \to \infty \] But this would not make sense unless the model is shifted. Alternatively, let’s use this form: \[ N(x) = \frac{L}{1 + Ce^{-kx}} \] Given two points, we can solve for \(C\) and \(k\): #### At \(x = 0\): \[ N(0) = \frac{12,500}{1 + C} = 0 \implies C \to \infty \] Again, problem. Alternatively, use the following logistic form: \[ N(x) = \frac{L}{1 + Ae^{-kx}} \] At \(x = 0\), \(N(0) = 0\): \[ 0 = \frac{12,500}{1 + A} \implies 1 + A \to \infty \] So let's use another form. Let's try: \[ N(x) = L(1 - e^{-kx}) \] At \(x=0\), \(N(0) = 0\) At \(x \to \infty\), \(N(x) \to L\) At \(x=3\), \(N(3) = 10,500\): \[ N(3) = 12,500(1 - e^{-3k}) = 10,500 \] \[ 1 - e^{-3k} = \frac{10,500}{12,500} = 0.84 \] \[ e^{-3k} = 1 - 0.84 = 0.16 \] \[ -3k = \ln 0.16 \] \[ k = -\frac{\ln 0.16}{3} \] \[ k = -\frac{-1.832581}{3} \approx 0.61086 \] So, the function is: \[ N(x) = 12,500(1 - e^{-0.61086 x}) \] --- ## **Step 3: Find x such that \(N(x) = 40,000\)** But note that the asymptote is 12,500, so \(N(x)\) can never reach 40,000, which may be a typo. If the question expects a different model or if the asymptote is actually 125,000, please clarify. **Assuming the intent is to find \(x\) such that \(N(x) = 12,000\):** \[ N(x) = 40,000 \] \[ 12,500(1 - e^{-0.61086x}) = 40,000 \] \[ 1 - e^{-0.61086x} = \frac{40,000}{12,500} = 3.2 \] \[ - e^{-0.61086x} = 3.2 - 1 = 2.2 \] \[ e^{-0.61086x} = -2.2 \] But \(e^{-0.61086x}\) cannot be negative. This confirms that with an asymptote of 12,500, you can never reach 40,000. If you're meant to find \(x\) for another value (like 11,000, for example), substitute that value in. Let’s check for 11,000 instead: \[ 12,500(1 - e^{-0.61086x}) = 11,000 \] \[ 1 - e^{-0.61086x} = \frac{11,000}{12,500} = 0.88 \] \[ e^{-0.61086x} = 0.12 \] \[ -0.61086x = \ln{0.12} \] \[ -0.61086x = -2.1202635 \] \[ x = \frac{2.1202635}{0.61086} \approx 3.471 \] --- ## **Summary of Steps and Final Answer** ### 1. **Graph**: The function is an exponential growth curve starting at \(N(0)=0\), approaching asymptote \(N(x)=12,500\). ### 2. **Function**: \[ N(x) = 12,500(1 - e^{-0.61086x}) \] ### 3. **Find x for N(x) = 40,000**: **It is not possible** because the maximum is 12,500. If you want a different value, substitute it into the equation above and solve as shown. --- #### **If you need the value for a different number of students, let me know!** If you can clarify the 40,000 value, I can recalculate for you.