The correct answer is “This company is exponentially shrinking by \(\displaystyle 1.8%\).” The above function is an exponential function of t because \(\displaystyle t\) is an exponent of a base in the function. A linear equation would utilize \(\displaystyle t\) as a coefficient, not an exponent. The best way to differentiate between growth and shrinkage in an exponential function is to see if the exponent’s base is greater or less than \(\displaystyle 1\). If the base is \(\displaystyle 1\), there would be no change in \(\displaystyle M(t)\) regardless of the value of \(\displaystyle t\), but if the base is less than \(\displaystyle 1\), the company’s value is decreasing. \(\displaystyle 1-0.982=0.018=1.8%\) so a loss of \(\displaystyle 1.8%\) of market value annually.

The correct answer is “Plant A shows exponential growth while Plant B is linear.” Each hour doubles the height of Plant A. Doubling is a characteristic of exponential growth \(\displaystyle (2t)\). Each hour adds \(\displaystyle 5\) cm to the height of Plant B. Additive and subtractive trends are characteristic of linear growth \(\displaystyle (5t)\).

The correct answer is \(\displaystyle y=2*(2)^{x}\). The graph is curved and does not have a defined slope, so it cannot be a linear function \(\displaystyle (y=2+2x)\). Now, between our exponential functions, we can see that at \(\displaystyle x=0\), \(\displaystyle y=2\), so this must be a coefficient outside of the exponential base. This only leaves \(\displaystyle y=2*(2)^{x}\) . It might be one’s first thought to see that at \(\displaystyle x=1\), \(\displaystyle y=4\), pointing to the \(\displaystyle y=(4)x\) choice, but this equation does not work at any other value of \(\displaystyle x\).

Both exponential equations should represent curved lines while the linear equations should be straight lines. \(\displaystyle y=(2)x\)  shows exponential growth, so as \(\displaystyle x\) increases, \(\displaystyle y\) should increase. \(\displaystyle y=(\frac{1}{2})x\)  shows exponential decay, so as \(\displaystyle x\) increases, \(\displaystyle y\) should decrease. This narrows us down to Option 2 and 4. Within the linear equations, \(\displaystyle y=2x\) should be a steeper line than \(\displaystyle y=\frac{1}{2}x\) since the slope is greater, thus leaving us with Option 4 as the correct answer.

Since the total moth population is a percentage of the previous period’s population, we know this is an exponential function and not a linear one. We also know that the population has been decreasing. An exponent base of \(\displaystyle 1\) would represent no change in the population, while a base of less than one would represent a decrease, so this rules out the option with \(\displaystyle 1.07\) as the exponent base. The population has been decreasing by \(\displaystyle 7%\) so we can say the original \(\displaystyle 100%\) of the population loses \(\displaystyle 7%\), leaving us with \(\displaystyle 93%\) of the original population for every \(\displaystyle 4\) years that pass. Also, we use \(\displaystyle \frac{t}{4}\) because the population only decreases by \(\displaystyle 7%\) every \(\displaystyle 4\) years. This leaves us with \(\displaystyle M(t)=1500(0.93)^{\frac{t}{4}}\).

The correct answer is “This relationship is exponential and doubles every hour.” Since the time variable \(\displaystyle t\) is an exponent, we can narrow down our options and definitively state the function is exponential. The \(\displaystyle t\) is an exponent with a base of \(\displaystyle 2\), so as \(\displaystyle t\) grows, the number of times \(\displaystyle 2\) is multiplied by itself also grows. Each multiplication of two is a doubling of the previous hour’s population (when \(\displaystyle t=2\), \(\displaystyle C(t)=50(2)^{2}=50*2*2=200\) vs. when \(\displaystyle t=3\), \(\displaystyle C(t)=50(2)^{3}=50*2*2*2=400\)).

The correct answer is II. This is an exponential function, not a linear function. When we plug in \(\displaystyle 0\) for \(\displaystyle t\), \(\displaystyle P(t)=600(0.5)^{0}=600*1=600\). When we plug in \(\displaystyle 3\) for \(\displaystyle t\), \(\displaystyle P(t)=600(0.5)^{3}=600*0.5*0.5*0.5=75\), not \(\displaystyle 100\).

The correct answer is \(\displaystyle 41\). Maria’s apple harvest follows an exponential pattern while her peach harvest is linear. Since time is relative to any given start point, we can call 2003 the year where \(\displaystyle t=0\). Her apple harvest *grows* by \(\displaystyle 3.4%\) so when we convert our percent back to a decimal, we get an exponent base of \(\displaystyle 1+0.034=1.034\). Recall, that if her harvest was *losing* \(\displaystyle 3.4%\) of produce yearly, we would use \(\displaystyle 1-0.034=0.966\). Since her starting harvest in 2003 was \(\displaystyle 62\) bushels, we can model her total bushel count as a function of time through the following equation: \(\displaystyle A(t)=62(1.034)^{t}\).

Her peach harvest is a linear function that grows by \(\displaystyle 5\) bushels every \(\displaystyle 5\) years. Since time is measured in single years and our growth is given in periods of \(\displaystyle 5\) years, be sure to take this into account in your linear equation \(\displaystyle (\frac{t}{5})\). Since her starting harvest in 2003 was \(\displaystyle 46\) bushels, we can model her total bushel count as a function of time through the following equation: \(\displaystyle P(t)=46+5*(\frac{t}{5})\).

When we plug \(\displaystyle t=15\) into our equation from subtracting 2003 from 2018, we get \(\displaystyle 102.4\) bushels of apples and \(\displaystyle 61\) bushels of peaches. \(\displaystyle 102.4-61=41.4\).

Exponential functions are in the form \(\displaystyle y=a^{x}\) while linear are \(\displaystyle y=mx+b\). Linear functions change at a constant rate per unit interval while exponential functions change by a common ratio over equal intervals.

When \(\displaystyle x=0\) (the y-intercept), \(\displaystyle y=500\). The base of this equation is \(\displaystyle 4\). Exponential functions never have x-intercepts unless they are in the form \(\displaystyle y=a^{x}+b\).