Left end: \(x\to-\infty,\ y\to+\infty\)
Right end: \(x\to+\infty,\ y\to0^+\)
Connection 联系：
\(\left(\frac12\right)^x=2^{-x}\), so the decay graph is the reflection of \(y=2^x\) over the y-axis.
因为把 \(x\) 换成 \(-x\)，两张图关于 y-axis 对称。
(a) $b=0.8<1$ → Decay; y-int $(0,2)$; HA $y=0$
(b) $b=3>1$ → Growth; y-int $(0,\frac{1}{4})$; HA $y=0$
(c) $e^{-x}=\left(\frac{1}{e}\right)^x$, $b=e^{-1}\approx0.368<1$ → Decay; y-int $(0,6)$; HA $y=0$
5.2 End Behavior of Exponential Functions 端点行为
For $f(x)=a\cdot b^x$ ($a\neq0$), as $x\to+\infty$ or $x\to-\infty$ the outputs either increase without bound, decrease without bound, or get arbitrarily close to 0 (CED 2.3.A.5):
$$\lim f(x)=+\infty \qquad \lim f(x)=-\infty \qquad \lim f(x)=0$$
If $a>0$, all outputs are positive, so each end is $+\infty$ or $0$. If $a<0$, each end is $-\infty$ or $0$.
Use the properties of $f(x)=2^x$, $g(x)=\left(\dfrac{1}{2}\right)^x$, and $h(x)=-3^x$ to evaluate:
$\displaystyle\lim_{x\to+\infty}f(x)$ and $\displaystyle\lim_{x\to-\infty}f(x)$
$\displaystyle\lim_{x\to+\infty}g(x)$ and $\displaystyle\lim_{x\to-\infty}g(x)$
$\displaystyle\lim_{x\to+\infty}h(x)$ and $\displaystyle\lim_{x\to-\infty}h(x)$
(1) $f(x)=2^x$: base $b=2>1$, growth
$\displaystyle\lim_{x\to+\infty}2^x=+\infty$ $\displaystyle\lim_{x\to-\infty}2^x=0$
(2) $g(x)=\left(\frac{1}{2}\right)^x$: base $b=\frac{1}{2}<1$, decay
$\displaystyle\lim_{x\to+\infty}\left(\tfrac{1}{2}\right)^x=0$ $\displaystyle\lim_{x\to-\infty}\left(\tfrac{1}{2}\right)^x=+\infty$ Note: $\left(\frac{1}{2}\right)^x = 2^{-x}$, so it mirrors $2^x$ reflected over the y-axis.
(3) $h(x)=-3^x$: $a=-1<0$, all signs reversed
$\displaystyle\lim_{x\to+\infty}(-3^x)=-\infty$ $\displaystyle\lim_{x\to-\infty}(-3^x)=0$ HA still $y=0$, but approached from below.
5.3 Laws of Exponents 指数法则
Law 法则
Formula 公式
Example 例子
Multiplication 乘法
$b^m\cdot b^n = b^{m+n}$
$2^3\cdot2^4=2^7=128$
Division 除法
$\dfrac{b^m}{b^n}=b^{m-n}$
$\dfrac{3^5}{3^2}=3^3=27$
Power of a Power 幂的乘方
$(b^m)^n=b^{mn}$
$(2^3)^4=2^{12}$
Negative Exponent 负指数
$b^{-n}=\dfrac{1}{b^n}$ and $\dfrac{1}{b^{-n}}=b^n$
The multiplication law $b^{m+n}=b^m\cdot b^n$ shows that:
$$f(x+h) = b^{x+h} = b^h\cdot b^x = \underbrace{b^h}_{=\,a}\cdot b^x$$
A horizontal shift by $h$ is equivalent to a vertical dilation by factor $b^h$.
水平平移 $h$ 单位 ≡ 垂直乘以 $b^h$
Create the same target graph from \(f(x)=2^x\) in two different ways.
Before
Both panels begin with the parent function \(y=2^x\).
左图将图像向左移动；右图将所有 y-values 同时放大。
Method A · Horizontal Shift 水平平移
\(2^x\rightarrow2^{x+2}\): shift left 2
Point motion: \((0,1)\rightarrow(-2,1)\). x changes; y stays the same.
Method B · Vertical Dilation 垂直伸缩
\(2^x\rightarrow4\cdot2^x\): stretch by 4
Point motion: \((0,1)\rightarrow(0,4)\). x stays the same; y is multiplied by 4.
Parent \(y=2^x\)Horizontal result \(y=2^{x+2}\)Vertical result \(y=4\cdot2^x\)
Same target graph 同一条图像：
\(2^{x+2}=2^2\cdot2^x=4\cdot2^x\)
Power Property → Horizontal Dilation = Change of Base
The power law $(b^c)^x = b^{cx}$ shows that:
$$f(cx) = b^{cx} = (b^c)^x$$
A horizontal dilation by $\frac{1}{c}$ is equivalent to changing the base to $b^c$.
水平伸缩 ≡ 改变底数
一起做 T-4 · Graph Transformations of Exponential Functions
Graph $f(x)=2^x$ and $g(x)=2^{x+3}$ on the same axes. State the transformation from $f$ to $g$.
Identify the transformation
$g(x)=2^{x+3}=f(x+3)$ → horizontal shift left by 3.
Key points mapping
Every point $(x,y)$ on $f$ maps to $(x-3,y)$ on $g$.
$f$: $(-2,\frac{1}{4}),(0,1),(1,2),(2,4)$
$g$: $(-5,\frac{1}{4}),(-3,1),(-2,2),(-1,4)$
Equivalent vertical form
$g(x)=2^{x+3}=2^3\cdot2^x=8\cdot2^x$
Also: vertical stretch by factor 8 of $f(x)$. HA still $y=0$.
一起做 T-5 · Transformation Analysis
Given $f(x)=5^{x-1}$:
(a) State its parent function.
(b) Describe the transformation from parent to $f$.
(c) Write an equivalent function $g(x)$ showing it as a vertical dilation.
(d) Describe the transformation from $g(x)$ to its parent.
(a) Parent function 母函数
$p(x)=5^x$
(b) Transformation from $p$ to $f$
$f(x)=5^{x-1}=p(x-1)$ → horizontal shift right 1 向右平移1单位
(c) Equivalent vertical form
$f(x)=5^{x-1}=5^{-1}\cdot5^x=\dfrac{1}{5}\cdot5^x$
So $g(x)=\dfrac{1}{5}\cdot5^x$ — vertical compression by $\frac{1}{5}$ 垂直压缩为 $\frac{1}{5}$ 倍
(d) From $g$ to parent $p(x)=5^x$
$g(x)=\frac{1}{5}\cdot5^x$ → vertical shrink by factor $\frac{1}{5}$
一起做 T-6 · Change of Base via Horizontal Dilation
Given $f(x)=81^x$ and parent $p(x)=3^x$, find the factor of the horizontal dilation that maps $p$ to $f$.
Rewrite $81^x$ in terms of base 3
$81=3^4$ so $81^x=(3^4)^x=3^{4x}=p(4x)$
Identify the horizontal dilation
$f(x)=p(4x)$ → horizontal compression by factor $\dfrac{1}{4}$
水平压缩为原来的 $\dfrac{1}{4}$（即 $b=4$ 意味着图像变窄到原来的 $\frac{1}{4}$）
一起做 T-7 · Reflection & Equivalent Base
Show that $f(x)=\left(\dfrac{1}{4}\right)^x$ is a reflection of $g(x)=4^x$ over the y-axis.
Conclusion
Since $f(x)=g(-x)$, the graph of $f$ is the graph of $g$ reflected over the y-axis. ✓
关于y轴的反射等价于将底数取倒数：$b\leftrightarrow\frac{1}{b}$
练习 E-3 · Transformation Practice
Given $f(x)=3^x$, write the equation for each transformation:
(a) Shift right 2, stretch vertically by 4
(b) Reflect over x-axis, shift up 5
(c) Horizontal compression by $\frac{1}{3}$ (equivalent base change)
(a) $4\cdot3^{x-2}=4\cdot3^{-2}\cdot3^x=\frac{4}{9}\cdot3^x$
(b) $-3^x+5$; HA at $y=5$
(c) Compression by $\frac{1}{3}$ → replace $x$ with $3x$: $3^{3x}=(3^3)^x=27^x$ — base changes to 27
The half-life of a radioactive substance is 20 days, and there are 5 grams initially. Write a function modeling this situation. Predict when 1 gram will remain.
Half-life model
$A(t)=5\cdot\left(\dfrac{1}{2}\right)^{t/20}$
每20天减少一半，所以 $t/20$ 是经过的半衰期数量。
$\$12{,}000$ is invested for 7 years at 5% annual interest, compounded continuously. What is the balance at the end of 7 years?
Continuous compounding formula
$A=Pe^{rt}$, where $P=12000$, $r=0.05$, $t=7$
Calculate
$A=12000\cdot e^{0.05\times7}=12000\cdot e^{0.35}$
$e^{0.35}\approx1.419068$ (keep the full calculator value; do not round early)
$A\approx\boxed{\$17{,}028.81}$
练习 E-4 · Car Depreciation 汽车折旧
Christopher buys a new car for $\$32{,}750$. The car depreciates by about 11.75% per year. Write an equation to model this situation. Determine the value of the car after 4 years.
Model: $V(t)=32750(1-0.1175)^t=32750(0.8825)^t$
After 4 years: $V(4)=32750\times(0.8825)^4\approx32750\times0.606539\approx\$19{,}864.16$
练习 E-5 · Exponential Model from Rate
Determine the exponential function that has an initial value of 15 and increases at a rate of 12% each year. Then find the value after 10 years.
$f(t)=15(1.12)^t$
After 10 years: $f(10)=15\times(1.12)^{10}=15\times3.1058\approx46.59$
Model from Two Points 由两个点建模
If $f(x)=ab^x$ passes through $(x_1,y_1)$ and $(x_2,y_2)$, divide the two equations to eliminate $a$:
$$\frac{ab^{x_2}}{ab^{x_1}}=\frac{y_2}{y_1}\ \Longrightarrow\ b^{\,x_2-x_1}=\frac{y_2}{y_1}.$$
Then substitute back for $a$. Point-ratio form: $f(x)=y_1\cdot b^{\,x-x_1}$ (like point-slope form for lines). 两式相除消去 $a$，求出 $b$ 后再回代求 $a$。
一起做 T-14 · Two Points, Neither at x = 0 两点都不在 x = 0
An exponential function $f(x)=ab^x$ satisfies $f(2)=12$ and $f(5)=96$. Find $a$ and $b$, and write $f$ in point-ratio form.
Back-substitute $a\cdot2^2=12\Rightarrow a=3$, so $f(x)=3\cdot2^x$.
Point-ratio form $f(x)=12\cdot2^{\,x-2}$ (check: $12\cdot2^{-2}=3$ ✓).
$f(x)=3\cdot2^x=12\cdot2^{x-2}$
练习 E-6 · Town Population 人口模型
A town had 8,400 people in 2012 and 10,200 people in 2018. Assume the population grows exponentially. Let $t$ be years since 2012.
(a) Find a model $P(t)$. (b) What is the annual percent growth rate? (c) Predict the population in 2025. (d) Would it be reasonable to use the model to predict the population in 2150? Explain.
(a) $b^6=\dfrac{10200}{8400}\Rightarrow b=\left(\dfrac{10200}{8400}\right)^{1/6}\approx1.032889$, so $P(t)\approx8400(1.032889)^t$.
(b) About $3.289\%$ per year.
(c) $t=13$: $P(13)\approx12{,}793$ people.
(d) No. 2150 is far outside the data ($2012$–$2018$); a town cannot grow by the same percent forever (limited land, resources). Predictions should stay close to the data's domain (CED 2.5.B.3).
Equivalent Forms: Changing the Time Unit 换时间单位的等价形式
The base tells the growth factor per one unit of input. Rewrite with the power property to get the factor for a different unit (CED 2.5.B.2):
$f(d)=2^d$ (doubles every day) $=\left(2^7\right)^{d/7}=128^{\,d/7}$ (multiplies by 128 every week).
(a) An account grows by 6% per year: $A(t)=1000(1.06)^t$, $t$ in years. Rewrite to find the growth factor and percent change per month and per decade.
(b) The half-life model $A(t)=5\left(\tfrac12\right)^{t/20}$ ($t$ in days) from T-12: by what percent does the amount decrease each day?
(a) Monthly $t$ years $=12t$ months: $A=1000\left(1.06^{1/12}\right)^{12t}\approx1000(1.004868)^{12t}$ → about $0.487\%$ per month (not $6\%\div12=0.5\%$).
(a) Per decade $A=1000\left(1.06^{10}\right)^{t/10}\approx1000(1.790848)^{t/10}$ → about $79.085\%$ per decade (not $60\%$).
(b) Per day $A=5\left(0.5^{1/20}\right)^t\approx5(0.965936)^t$ → decreases by about $3.406\%$ per day.
A patient takes 200 mg of a medicine. Every 4 hours, 30% of the amount in the body is eliminated: $A(t)=200(0.7)^{t/4}$, $t$ in hours.
(a) Find the hourly decay factor and the percent eliminated each hour. (b) Find $A(10)$. (c) Rewrite $A$ with a factor per day (24 h) and interpret. (d) What domain makes sense?
(a) $A(t)=200\left(0.7^{1/4}\right)^t\approx200(0.914691)^t$: about $8.531\%$ eliminated per hour.
(b) $A(10)=200(0.7)^{2.5}\approx81.993$ mg.
(c) $A(t)=200\left(0.7^{6}\right)^{t/24}\approx200(0.117649)^{t/24}$: after each full day, only about $11.765\%$ remains (about $88.235\%$ eliminated).
(d) $t\ge0$, from when the dose is taken until the next dose (the model ignores new doses).
Shifted Exponential Models $y=ab^x+k$ 带水平渐近线的指数模型
If the outputs level off toward some value $k\neq0$, the outputs themselves are not proportional, but the values of $y-k$ are. Subtract (or add) a constant first, then check ratios (CED 2.5.A.2). The model is
$$y=a\,b^x+k,\qquad \text{horizontal asymptote } y=k.$$
典型情境：冷却（趋近室温）、药物达到稳定浓度、学习曲线。先减去"趋近的值" $k$，再看比值是否恒定。
一起做 T-16 · Cooling Coffee 咖啡冷却（calculator 可用）
A cup of coffee cools in a 20 °C room. Its temperature $T$ (°C) is recorded every 5 minutes.
$t$ (min)
0
5
10
15
20
$T$ (°C)
90
62
45.2
35.12
29.072
(a) Show that $T$ itself is not exponential but $T-20$ is. (b) Write a model $T(t)$. (c) Find $\displaystyle\lim_{t\to\infty}T(t)$ and interpret. (d) Predict $T(30)$. (e) When does the coffee reach 25 °C?
(b) Model $T(t)=20+70(0.6)^{t/5}$ (initial difference 70, factor 0.6 per 5 min, shifted up 20).
(c) Long run $(0.6)^{t/5}\to0$, so $\displaystyle\lim_{t\to\infty}T(t)=20$: the coffee approaches room temperature (HA $y=20$).
(d) $T(30)=20+70(0.6)^6\approx23.266$ °C.
(e) Graph $Y_1=20+70(0.6)^{X/5}$ and $Y_2=25$; intersect: $t\approx25.831$ min.
$T(t)=20+70(0.6)^{t/5}$; $\lim T=20$ °C; $T(30)\approx23.266$ °C; 25 °C at $t\approx25.831$ min.
练习 E-8 · Find the Hidden Constant 找出隐藏的常数
$x$
0
1
2
3
4
$f(x)$
5
7
11
19
35
(a) Show that $f$ is neither linear nor exponential as given. (b) Find a constant $k$ so that $f(x)-k$ is proportional over equal intervals. (c) Write a model for $f$ and find $\displaystyle\lim_{x\to-\infty}f(x)$.
(a) Differences $2,4,8,16$ are not constant; ratios $\tfrac75,\tfrac{11}{7},\dots$ are not constant.
(b) The differences double, a hint of an exponential pattern. Try $k=3$: $f(x)-3=2,4,8,16,32$, ratio 2 ✓.
(c) $f(x)=2\cdot2^x+3=2^{x+1}+3$; $\displaystyle\lim_{x\to-\infty}f(x)=3$ (HA $y=3$).
Exponential Regression with Technology 指数回归
TI-84 · ExpReg
STAT → 1: Edit: inputs in L1, outputs in L2 (outputs must be positive).
STAT → CALC → 0: ExpReg. Xlist: L1, Ylist: L2, Store RegEQ: Y1 (VARS → Y-VARS → Function → Y1).
The calculator gives $y=a\cdot b^x$. Report $a$ and $b$ to 3 decimal places, but use the stored Y1 for predictions.
一起做 T-17 · App Users 应用用户数（calculator 可用）
The number of users $U$ (millions) of an app $t$ years after launch:
$t$
0
1
2
3
4
5
$U$
3.1
4.4
6.5
9.3
13.8
19.9
(a) Explain why an exponential model is reasonable. (b) Find the exponential regression. (c) Interpret the base. (d) Predict $U(7)$ and comment on its reliability.
(a) Successive ratios $1.419,\ 1.477,\ 1.431,\ 1.484,\ 1.442$ are roughly constant → roughly proportional growth over equal intervals.
(b) ExpReg: $U(t)\approx3.066(1.453)^t$.
(c) The number of users is multiplied by about 1.453 each year, i.e. grows by about $45.3\%$ per year.
(d) $U(7)\approx41.983$ million. $t=7$ is beyond the data (0–5), so this is an extrapolation; growth this fast cannot continue forever.
$U(t)\approx3.066(1.453)^t$; about 45.3% growth per year; $U(7)\approx41.983$ million (extrapolation).
5.6 Competing Function Model Validation 模型适合性验证
When several model types (linear, quadratic, exponential) could fit a data set, AP validates a model with its residual plot (CED 2.6.B.1), together with context (CED 2.6.A.2):
Residual plot — the model is appropriate if the residuals appear without pattern.
Context — does the model's behavior (e.g. proportional growth, a maximum, a limit) make sense?
$R^2$ — extra information only; the CED does not use it to validate a model.
① Coefficient of Determination $R^2$ 决定系数
$R^2$ measures the proportion of variance in the data explained by the model.
$R^2=1$ → perfect fit; $R^2=0$ → model explains nothing.
一起做 T-18 · Which Model Fits? 哪个模型合适（calculator 可用）
A bacteria culture is measured (thousands of cells) every hour:
$x$ (h)
0
1
2
3
4
5
6
$y$
2.0
2.9
4.6
6.7
10.3
15.1
23.4
(a) Find linear, quadratic, and exponential regressions. (b) Compute residuals for the linear and exponential models and compare the residual plots. (c) Which model is most appropriate? Justify. (d) Compare the three predictions at $x=10$.
Exponential: no clear pattern ✓ (note the much smaller scale)
(c) Choose
The linear residuals are positive, then negative, then positive: a U-shaped pattern, so the linear model is not appropriate. The exponential residuals are small and scattered without pattern, and bacteria grow proportionally (ratios $\approx1.5$ per hour). The exponential model is most appropriate. (The quadratic residuals, $-0.486$ to $0.707$, are larger and less random.)
(d) Predictions at x = 10
Linear $\approx32.861$, quadratic $\approx62.593$, exponential $\approx119.514$ thousand. The models agree on the data but disagree badly outside it, which is why validation matters.
Exponential $y\approx1.977(1.507)^x$: its residual plot has no pattern and it matches proportional growth.
一起做 T-19 · Choose the Right Model from Context
For each dataset description, determine whether a linear, quadratic, or exponential model is most appropriate. Justify.
Fuel efficiency (mpg) of a vehicle at different speeds — increases then decreases.
U.S. population from 1900–2011.
Heights of females based on humerus bone length.
(1) Fuel efficiency vs. speed
Increases then decreases → single maximum → Quadratic model 二次函数
Parabolic shape fits this scenario.
(2) U.S. population 1900–2011
Population grows rapidly over time, successive ratios are approximately constant → Exponential model 指数函数
$P(t)=a\cdot b^t$ where $t$ = years since 1900.
(3) Female height vs. humerus length
Bone length and height are roughly proportional — a relatively straight trend → Linear model 线性函数
练习 E-9 · Residual Interpretation 残差图解读
A linear regression model is applied to a dataset. The residual plot shows a clear U-shaped (curved) pattern. What does this tell you, and what should you do?
A U-shaped residual plot indicates a non-random pattern, meaning the linear model is NOT appropriate for this data.
残差图呈U形说明线性模型不适合——预测误差有规律，不是随机分布。
Action 建议： Try a quadratic or exponential model instead, and check if the new residual plot shows random scatter.
Overestimate or Underestimate? 高估还是低估
Residual $=$ actual $-$ predicted.
Residual > 0: actual is above the prediction → the model underestimates. 残差为正 → 模型低估
Residual < 0: actual is below the prediction → the model overestimates. 残差为负 → 模型高估
Depending on the context, one kind of error may be safer than the other (CED 2.6.B.2). 根据情境判断：哪种误差代价更小？
练习 E-10 · Interpreting Errors 解读误差
(a) Using the exponential model from the bacteria example, does it overestimate or underestimate at $x=5$? At $x=6$? By how much?
(b) A city models how much road salt it needs for winter. Is it better for the model to overestimate or underestimate? Explain.
(c) A pharmacist models the maximum safe dose of a medicine for a patient's weight. Should the model err toward overestimating or underestimating the safe dose?
(a) At $x=5$ the residual is $-0.272$: the model overestimates by about 0.272 thousand cells. At $x=6$ the residual is $0.233$: it underestimates by about 0.233 thousand.
(b) Overestimate: running out of salt during a storm is worse than having some left over.
(c) Underestimate the safe dose: giving too much medicine is dangerous, while a slightly smaller dose is safer.
R-1Power Property — 幂的性质与水平伸缩改编自 Practice Exam 3 · MCQ Part A #7
The function $f$ is given by $f(x)=3^{(2x)}$. Which of the following describes the graph of $f$ in the $xy$-plane and gives an equivalent form of $f(x)$?
(A) A vertical dilation of the graph of $y=3^x$ by a factor of $2$; $f(x)=9^x$
(B) A horizontal dilation of the graph of $y=3^x$ by a factor of $2$; $f(x)=6^x$
(C) A horizontal dilation of the graph of $y=3^x$ by a factor of $\tfrac12$; $f(x)=9^x$
(D) A horizontal dilation of the graph of $y=3^x$ by a factor of $\tfrac12$; $f(x)=6^x$
R-3Model from Two Points — 由两点求指数函数改编自 Practice Exam 2 · MCQ Part A #11
The figure shows the graph of an exponential function $f$ of the form $f(x)=ab^x$. The coordinates of two points on the graph are labeled. What is the $y$-coordinate of the point on the graph where $x=0$?
R-4Half-Life Model — 半衰期模型改编自 2024 AP 真题（亚太卷）· MCQ Part A #4
A radioactive substance has a half-life of $8$ days. A sample contains $50$ grams of the substance at time $t=0$. Which of the following is an appropriate model for $A(t)$, the number of grams of the substance remaining after $t$ days?
R-5Reading Residuals — 残差：高估还是低估改编自 Practice Exam 2 · MCQ Part A #27
A regression model $f$ is constructed for a data set. The residual at $x=2$ is $-0.8$, and the residual at $x=5$ is $2.1$. Which of the following statements is true?
(A) $f$ overestimates at $x=2$ and underestimates at $x=5$; the error is greater at $x=5$.
(B) $f$ underestimates at $x=2$ and overestimates at $x=5$; the error is greater at $x=5$.
(C) $f$ overestimates at $x=2$ and underestimates at $x=5$; the error is greater at $x=2$.
(D) $f$ underestimates at $x=2$ and overestimates at $x=5$; the error is greater at $x=2$.
Part B选择题 · 可用计算器 Multiple Choice · Calculator Allowed
R-6Exponential Regression — 指数回归改编自 Practice Exam 2 · MCQ Part B #9
$t$ (months)
$0$
$1$
$2$
$3$
$4$
$5$
$U(t)$ (thousands)
$2.0$
$2.9$
$4.4$
$6.3$
$9.8$
$14.1$
The table gives the number of users of a new app, in thousands, $t$ months after it was launched. An exponential regression is used to model these data. What is the number of users, in thousands, at $t=7$ predicted by the exponential regression model?
R-7How Long to Grow? — 增长所需时间改编自 2024 AP 真题（亚太卷）· MCQ Part B #9
The population of a town is modeled by $P(t)=25{,}000(1.032)^t$, where $t$ is the number of years since 2020. According to the model, how many years does it take for the population to increase from $40{,}000$ to $60{,}000$?
R-8Shifted Exponential Model — 带水平渐近线的指数模型改编自 Practice Exam 3 · MCQ Part B #12
The temperature of a cup of tea, in degrees Celsius, $t$ minutes after it is poured is modeled by $T(t)=a(0.92)^t+b$, where $a$ and $b$ are constants. If $T(0)=90$ and $\displaystyle\lim_{t\to\infty}T(t)=20$, for what value of $t$ is $T(t)=40$?
R-9Exponential Model from Two Data Points — 由两个数据建指数模型改编自 Practice Exam 1 · FRQ #2
A city began tracking its public electric-vehicle charging stations in 2018. In 2019 $(t=1)$ there were $84$ stations, and in 2024 $(t=6)$ there were $210$ stations. The number of stations can be modeled by $P(t)=ab^t$, where $t$ is the number of years since 2018. A calculator is allowed.
(a) (i) Use the given data to write two equations that can be used to find the values of $a$ and $b$. (ii) Find the values of $a$ and $b$ as decimal approximations. (b) (i) Find the average rate of change of the number of stations, in stations per year, from $t=1$ to $t=6$. (ii) Use the average rate of change from (i) to estimate the number of stations at $t=4$. (iii) Is the estimate in (ii) greater than or less than the value predicted by the model $P$? Explain your reasoning using characteristics of the model. (c) The model $P$ can be rewritten as $P=a\cdot c^{m}$, where $m$ is the number of months since 2018. Find the value of $c$, and interpret it as a percent change in context.
(a) (i) $ab=84$，$ab^6=210$。 (ii) 两式相除：$b^5=\dfrac{210}{84}=2.5$，$b=2.5^{1/5}\approx1.201$；$a=\dfrac{84}{b}\approx69.934$。
R-10Which Model Fits Better? — 用残差比较模型改编自 2024 AP 真题（亚太卷）· MCQ Part A #20
$x$ (days)
$0$
$1$
$2$
$3$
$4$
$5$
$y$ (cm²)
$3.1$
$4.4$
$6.5$
$9.4$
$13.6$
$20.1$
The table gives the area $y$, in square centimeters, of a mold colony $x$ days after a scientist began observing it. A linear regression gives $g(x)=3.3x+1.267$, and an exponential regression gives $f(x)=3.066(1.454)^x$. A calculator is allowed.
(a) Using the table, explain why an exponential model is more appropriate for these data than a linear model. (b) Find the residual of the exponential model $f$ at $x=4$. Does $f$ overestimate or underestimate the area at $x=4$? (c) The residual plots for the two models are shown. Which residual plot belongs to the linear model? Based on the residual plots, explain why $f$ is the more appropriate model. (d) Use $f$ to predict the area at $x=8$. Give one reason this prediction is less reliable than a prediction at $x=2.5$.
A patient takes a $400$-milligram dose of a medicine. The amount of the medicine in the patient's body $t$ hours later is modeled by $D(t)=400(0.85)^t$. A calculator is allowed.
(a) By what percent does the amount of medicine decrease each hour? Find the half-life of the medicine, to the nearest hundredth of an hour. (b) The patient takes a second $400$-milligram dose at $t=6$. For $t\ge6$, the total amount in the body is $T(t)=400(0.85)^t+400(0.85)^{t-6}$. Show that $T(t)$ can be written in the form $k(0.85)^t$, and find the value of $k$. (c) Find the total amount of medicine in the body at $t=8$. (d) The medicine is no longer effective when the total amount falls below $100$ milligrams. After the second dose, at what time $t$ does this first happen?
Directions: Solve each part without a calculator. Show the work that leads to your answers.
(a) Solve $4^{x+1}=8^{x-1}$ for $x$. (b) Rewrite $g(x)=3\cdot2^{(2x-1)}$ in the form $g(x)=ab^x$, where $a$ and $b$ are constants. (c) The graph of $p(x)=16\cdot2^x$ is a vertical dilation of the graph of $y=2^x$. The graph of $p$ is also a horizontal translation of the graph of $y=2^x$. Describe this translation, and write $p(x)$ in the form $2^{(x-h)}$. (d) The function $h$ is given by $h(x)=5\cdot2^x-3$. Solve $h(x)=37$, find $\displaystyle\lim_{x\to-\infty}h(x)$, and state the range of $h$.
Axiom Learning·AP Precalculus
