Linear and Exponential Functions
Introduced| English | Meaning |
|---|---|
| linear function/ˈlɪnɪə ˈfʌŋkʃn/ | A function of the form f(x) = mx + b, whose graph is a straight line. |
| exponential function/ˌekspəˈnenʃl ˈfʌŋkʃn/ | A function in which the variable appears in an exponent, such as a fixed positive base raised to that variable. |
| slope/sləʊp/ | The change in a line's vertical coordinate divided by the corresponding change in its horizontal coordinate. |
| common ratio/ˈkɒmən ˈreɪʃɪəʊ/ | The factor multiplying each term to obtain the next term in a geometric sequence. |
| domain/dəˈmeɪn/ | The set of input values for which a function is defined. |
| proportional/prəˈpɔːʃənl/ | proportional |
Start from the same known point
- At time $x=2$ hours, two models both have output 10. One adds 3 for each one-unit increase in input; the other doubles. What should each predict at $x=4$?
- Comparing starting heights alone does not answer this. A linear function 线性函数 uses a fixed additive rate, while an exponential function 指数函数 uses a fixed multiplier over equal input intervals.
Build a line from a point and a rate
- An arithmetic sequence adds the same difference at each step. A line has a constant slope 斜率 $m$: an input increase $h$ gives output change $mh$.
- From a known point $(x_i,y_i)$, write $L(x)=y_i+m(x-x_i)$. Here $(2,10)$ and slope 3 give $L(x)=10+3(x-2)$; hence $L(4)=16$. Use the input distance from the known point, not simply $3x$ added to 10.
Over equal input intervals, a nonconstant linear function changes by…
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A line has a constant slope, so equal steps in $x$ add the same amount to $y$ — like an arithmetic sequence.
A line passes through (2,10) with slope 3. Using $L(x)=10+3(x-2)$, find $L(5)$.
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$L(5)=10+3(5-2)=19$. There are three unit steps from input 2.
Build an exponential from a point and a ratio
- A geometric sequence with positive common ratio 公比 $r$ extends to real inputs by $E(x)=y_i r^{x-x_i}$. The ratio is for a one-unit input increase. For a nonconstant exponential, require $y_i\ne0$, $r>0$ and $r\ne1$.
- Through $(2,10)$ with hourly factor 2, the rule is $E(x)=10(2)^{x-2}$. It gives $E(4)=40$ and $E(2)=10$. The coefficient 10 is the output at input 2, not at input zero.
Over equal input intervals, an exponential function changes by…
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An exponential multiplies by the same factor each equal step — the continuous version of a geometric sequence.
An exponential passes through (2,10) with unit factor 2. Using $E(x)=10(2)^{x-2}$, find $E(4)$.
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$E(4)=10(2)^2=40$. The two-unit gap gives two multiplications.
Keep the interval length with the change
- For the line, equal input gaps $h$ give the same difference $mh$. For the exponential, equal gaps give the same factor $r^h$, provided the outputs are nonzero.
- In the opening example, moving from input 2 to input 4 takes two unit steps. The line gains $3(2)=6$, while the exponential has factor $2^2=4$ over the two-hour gap. Its hourly factor 2 applies to one hour.
From input 3 to input 4, the line changes from 13 to 16 and the exponential from 20 to 40. The line's difference is still 3; the exponential's ratio is still 2.
Compare additive and multiplicative change
$\begin{aligned}L(t)&=y_i+m(t-2)\\E(t)&=y_i r^{t-2}\end{aligned}$
Start at the known point (2,10), with slope 3 and unit factor 2. Compare outputs and changes over the interval from 3 to 4, then vary its length.
An exponential has unit factor 2. What is its output ratio over a two-unit input gap?
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The two-unit factor is $2^2=4$, not the unit factor 2.
Unequal input gaps need normalization
- The points $(0,4),(1,7),(3,13)$ have output differences 3 and 6. Their input gaps are 1 and 2, so both slopes are 3. Unequal raw differences do not rule out a line when the input gaps differ.
- The points $(0,2),(1,4),(3,16)$ have ratios 2 and 4 over gaps 1 and 2. Both agree with factor 2 per input unit, since $2^2=4$. A finite pattern supports a model on those observations; it does not establish the rule for all other inputs.
The points (0,4),(1,7),(3,13) have unequal output changes. Which interpretation is correct?
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The slopes are $(7-4)/(1-0)=3$ and $(13-7)/(3-1)=3$. Compare changes per unit input.
A sequence and a function have different domains
- A sequence may use only $n=1,2,3,\ldots$, even when its formula extends to a function with real inputs. An arithmetic sequence starting at 4 with difference 6 has $a_n=4+6(n-1)$; the corresponding line is $A(x)=4+6(x-1)$.
- The line gives $A(1.5)=7$, but that sequence has no term numbered 1.5. The domain 定义域 comes from the object being modeled: a function's mathematical domain can also be restricted by a physical context.
A sequence defined only at indices 1, 2, 3, ... has a term at index 1.5 whenever its formula can be evaluated there.
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A formula extension does not enlarge the stated sequence domain. The corresponding function may accept 1.5.
Ask what the percentage is taken from
- A fixed percentage of the current positive amount gives proportional 成比例 change. Starting at 100, multiplying by 1.1 each period gives $100(1.1)^t$ and output 121 at $t=2$.
- Adding 10% of the original 100 each period instead adds a fixed 10: $100+10t$, with output 120 at $t=2$. A percentage in the description does not by itself identify an exponential. A zero percentage gives a constant rather than a nonconstant exponential.
A fixed percentage of the current positive amount gives ____ change.
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Proportional change means each step scales with the current value — grow 10% of whatever you have now.
A positive population that doubles over every one-year interval follows a linear pattern of change.
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False. Doubling means multiplying by 2 each year, a constant factor over equal intervals, which is exponential. Linear models add a constant amount.
Start at 100 and add 10% of the original 100 each period. Which model describes this?
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10% of the original 100 is a fixed 10 each period. Taking 10% of the current amount would give the exponential instead.
State the conditions behind a growth claim
- If the exponential has a positive starting value and factor greater than 1, it eventually exceeds any fixed linear function as the input increases without bound. A finite plotted window may not reach the crossing, and this comparison of formulas is not a guarantee about a real forecast.
- Decay, a zero starting value or a negative starting value do not give that positive-growth conclusion. For example, $-10(2)^{x-2}$ becomes more negative. With a zero start all outputs are zero, and an output ratio would be the undefined expression $0/0$.
Select all true statements.
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The first,second and fourth statements hold with their stated assumptions. Equal nonzero additive changes give a line, not a nonconstant exponential.
The function $-10(2)^{x-2}$ must eventually exceed the growing line $x+1$ as x increases.
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The exponential becomes more negative while the line increases. The positive-start condition is essential.
Label the observations and the limit
- The lily-pad counts $10,20,40,80$ on days $d=0,1,2,3$ are consistent with $P(d)=10(2)^d$. Calling the first day 1 instead gives $P(d)=10(2)^{d-1}$. The exponent counts changes from the labeled starting day.
- The pond has finite space, so continued doubling is a limited model. Check additional observations and the context before forecasting; a short list of counts does not establish unlimited growth.
Check the input intervals, the starting point and the reference amount for any percentage. Constant differences or ratios across a few observations suggest a model; they do not prove a unique rule or justify every future prediction.
From a known point, use $y_i+m(x-x_i)$ for additive change and $y_i r^{x-x_i}$ for multiplicative change. Across an input gap $h$, compare $mh$ with $r^h$. Preserve the sequence's index domain and qualify any claim about eventual positive growth.
Explain the fifth term and the matching line
- An arithmetic sequence has first term $a_1=4$ and common difference 6. Find $a_5$, explaining how many additions are needed.
- Give the corresponding linear function through $(1,4)$ and explain why evaluating it at a fractional input does not create a new term of this sequence.
An arithmetic sequence has $a_1=4$ and common difference 6. Find $a_5$ and give the corresponding linear function through (1,4). Why does its value at a fractional input not give a term of this sequence?
Use the input distance from the known first term,then preserve the stated index domain.
An arithmetic sequence has first term 4 and common difference 6. What is its fifth term?
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Four additions of 6 reach 28. Other options — 10 (the second term); 6 (the common difference).