Exponential functionsIB MYP Maths Extended: Revision notes

An exponential function has the variable in the power, such as $y=a^x$ with $a>0$ and $a\ne1$. All graphs of this form pass through $(0,1)$ because $a^0=1$, and $y$ is always positive, so the curve stays above the $x$-axis. If $a>1$ the graph rises steeply to the right (exponential growth), e.g. $y=2^x$ gives $1,2,4,8,\ldots$ for $x=0,1,2,3$. If $0<a<1$ the graph falls (exponential decay), e.g. $y=0.5^x$ gives $1,0.5,0.25,\ldots$. The $x$-axis is a horizontal asymptote, $y=0$: the curve gets closer and closer to it but never touches it. For $a>1$ this happens on the left (very negative $x$), and for $0<a<1$ it happens on the right.

Use the index laws to evaluate points on the graph: $a^{-n}=\frac{1}{a^n}$, so $3^{-2}=\frac19$ and $2^{-3}=\frac18$. The graph of $y=\left(\frac12\right)^x$ is the same as $y=2^{-x}$, a reflection of $y=2^x$ in the $y$-axis. To find the base from a point, substitute it. If $y=a^x$ passes through $(3,64)$, then $a^3=64$, so $a=4$.

Many real situations follow $y=a\times b^x$, where $a$ is the starting value and $b$ is the growth factor. If a quantity grows by $r\%$ each period, $b=1+\frac{r}{100}$. If it decays by $r\%$, $b=1-\frac{r}{100}$. A car worth \24,000$that loses 15$V=24,000\times0.85^t$. A population of 800 growing 5$P=800\times1.05^t$, so after 6 years$P=800\times1.05^6\approx1072$. A model is exponential if the ratio of successive values is constant. For$500,600,720,864$the ratios are all$1.2$, so$N=500\times1.2^t$.

With compound interest, interest is added to the balance, so next year's interest is earned on the new, larger amount. For an amount $P$ at $r\%$ per year for $n$ years: $A=P\left(1+\frac{r}{100}\right)^n.$ Example: \2000$at 4$A=2000\times1.04^5=$2433.31$. **Simple interest** adds the same amount every year,$A=P+\frac{Pr}{100}n$, which is a straight line (linear growth). Compound interest is exponential, so it starts slower than a larger simple rate but eventually overtakes it.

Method 1: write both sides with the same base. $3^x=27=3^3$ gives $x=3$, and $3^x=\frac1{27}=3^{-3}$ gives $x=-3$. Method 2: use technology when the bases cannot be matched. To solve $500\times1.2^t=5000$, graph $y=500\times1.2^t$ and $y=5000$ and find the intersection ($t=12.6$), or use a table or solver. Check by substituting back. In a real context, think about what the answer means: if you want the first whole year the value passes a target, round up to the next whole number.

To model data, enter it in a table or spreadsheet, plot it and check that the pattern looks exponential (successive ratios roughly constant). A calculator or graphing software can then fit the curve $y=a\times b^x$ by exponential regression. Then use the model to predict, but remember its limits: the model assumes the growth factor stays constant, and predictions far beyond the data (extrapolation) become less reliable. Real populations run out of space or food, and car values depend on condition as well as age. When you evaluate a model, compare predicted values with real ones and comment on how close they are.