Question

learning goal from lesson 13.2 i can determine the effect on the graph of replacing f(x) by f(x)+k, kf(x), f(kx), and f(x + k) for specific values of k (both positive and negative). i can determine the translation value k, given a graph for slides, shifts, and stretches. i can explain the translation effects on the graph of a function using technology. lesson reflection (circle one) starting... getting there... got it! lesson 13.2 checkpoint once you have completed the above problems and checked your solutions, complete the lesson checkpoint below. complete the lesson reflection above circling your current understanding of the learning goal. 1. which function is represented by the graph to the left? a. f(x)=-(\frac{1}{2})^{x + 2}-3 b. f(x)=-(\frac{1}{2})^{x - 3}+2 c. f(x)=-(\frac{1}{2})^{x - 2}-3 2. a population p is initially 500 people. the population decreases every 32% per year. determine when the population will be left with 1 person.

Explanation:

1.

Step1: Identify the horizontal and vertical asymptote

The horizontal asymptote of the graph is $y = - 3$. For an exponential - type function of the form $y=a\cdot b^{x - h}+k$, the horizontal asymptote is $y = k$. So $k=-3$.

Step2: Use a point on the graph

We have a point $(-2,-4)$ on the graph. Substitute $x=-2$ and $y = - 4$ into the function $y=-(\frac{1}{2})^{x - h}-3$.

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Since $1 = (\frac{1}{2})^0$, then $-2 - h=0$, and $h=-2$.
The function is $f(x)=-(\frac{1}{2})^{x + 2}-3$.

Step1: Write the population - decay formula

The population - decay formula is $P = P_0(1 - r)^t$, where $P_0$ is the initial population, $r$ is the rate of decay, and $t$ is the time. Here, $P_0 = 500$, $r=0.32$, and $P = 1$. So the equation becomes $1 = 500(1 - 0.32)^t$, which simplifies to $1 = 500\times0.68^t$.

Step2: Solve for $t$

First, rewrite the equation as $\frac{1}{500}=0.68^t$.
Take the natural logarithm of both sides: $\ln(\frac{1}{500})=\ln(0.68^t)$.
Using the property of logarithms $\ln(a^b)=b\ln(a)$, we get $\ln(1)-\ln(500)=t\ln(0.68)$.
Since $\ln(1) = 0$, then $t=\frac{-\ln(500)}{\ln(0.68)}$.
We know that $\ln(500)\approx6.2146$ and $\ln(0.68)\approx - 0.3857$.

$$ t=\frac{-6.2146}{-0.3857}\approx16.11 $$

Answer:

A. $f(x)=-(\frac{1}{2})^{x + 2}-3$

2.