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Year 12 Maths - Methods (Unit 3 & Unit 4) Power, exponential and logarithmic functions

Determining rules for graphs of exponential and logarithmic functions

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Theory

Determining the rule of an exponential function \(y=a\,b^x\ (+c)\) or a logarithmic function \(y=a\log_b(x-h)+c\) from a graph, a table or two or three points means solving for the unknown parameters. Read any asymptote first, use the \(y\)-intercept or a point for \(a\), and use a ratio of two points (exponential) or the point where the log is \(0\) for the remaining parameter.

An exponential function \(y=a\,b^x\) (with \(b>0\)) has two unknowns: the coefficient \(a\) (its \(y\)-intercept, since \(b^0=1\)) and the base \(b\) (the growth or decay factor). Add a constant to get \(y=a\,b^x+c\), whose graph has the horizontal asymptote \(y=c\). A logarithmic function \(y=a\log_b(x-h)+c\) has a vertical asymptote at \(x=h\).

To determine the rule is to find those parameters from what the graph shows. Each parameter is tied to one readable feature: an asymptote fixes the shift (\(c\) or \(h\)); the \(y\)-intercept or a substituted point fixes \(a\); and a ratio of two points (for \(y=a\,b^x\)) or a logarithm law (for the log form) fixes the base or rate.

The same idea uses base \(e\): for \(y=A\,e^{kx}\) the \(y\)-intercept gives \(A\), and a second point with the natural logarithm gives \(k\). This is the “find the equation” counterpart to sketching a graph from a known rule.

Key idea. Read the asymptote first (\(y=c\) or \(x=h\)); the \(y\)-intercept or a point gives \(a\); a ratio of two points gives the base \(b\) (divide, do not subtract).
Exponential y=a b^x through two pointsAn increasing exponential curve passing through the marked points (1,4) and (3,16); dividing the two point-equations gives the base b. x y (1,4) (3,16)
\(y=a\,b^x\): divide the two points to find \(b\), then back-substitute for \(a\)
Logarithmic y=a log_b(x-h)+c with vertical asymptoteAn increasing logarithmic curve with a dashed vertical asymptote at x=h; the asymptote gives the horizontal shift h. x y x=h
\(y=a\log_b(x-h)+c\): the vertical asymptote gives \(h\); a point where \(x-h=1\) gives \(c\)

An exponential rule from two points (divide to cancel \(a\)):

\[y=a\,b^{x} \qquad \frac{y_2}{y_1}=b^{\,x_2-x_1} \quad(b>0)\]
y=abx

An exponential with a vertical shift (the asymptote gives \(c\)):

\[y=a\,b^{x}+c \qquad \text{asymptote } y=c\]
y=abx+c

A logarithmic rule (the vertical asymptote gives \(h\)):

\[y=a\,\log_{b}(x-h)+c \qquad \text{asymptote } x=h\]
y=alogb(x-h)+c
Base \(e\). For \(y=A\,e^{kx}\), the \(y\)-intercept gives \(A\); a second point gives \(e^{kx}\), then take \(\ln\): \(e^{5k}=3\Rightarrow k=\dfrac{\ln 3}{5}\).

How to determine the rule from a graph or points

  1. Asymptote first. A horizontal asymptote \(y=c\) gives the vertical shift of an exponential; a vertical asymptote \(x=h\) gives the horizontal shift of a logarithm.
  2. Use the \(y\)-intercept or an easy point. For \(y=a\,b^x\) the \(y\)-intercept is \(a\) (since \(b^0=1\)); for a log, the point where \(x-h=1\) makes \(\log_b(1)=0\), giving \(c\).
  3. Ratio for the base. With two points on \(y=a\,b^x\), divide the equations so \(a\) cancels: \(\dfrac{y_2}{y_1}=b^{\,x_2-x_1}\); solve for \(b>0\). For base \(e\), take \(\ln\) to find \(k\).
  4. Back-substitute and check. Put the values back and verify the rule reproduces every given point.
Table clue. If \(x\) increases in equal steps and successive \(y\)-values share a constant ratio, the data is exponential and that ratio is the base \(b\); the value at \(x=0\) is \(a\).
Example 1 — \(y=a\,b^x\) from two points
The points \((1,4)\) and \((3,16)\) lie on \(y=a\,b^x\) (\(b>0\)). Find \(a\) and \(b\).
Solution
Substitute both points into \(y=a\,b^x\):
\(4\)\(=\)\(a\,b^{1}\)
\(16\)\(=\)\(a\,b^{3}\)
Divide the equations to cancel \(a\):
\(\dfrac{16}{4}\)\(=\)\(\dfrac{a\,b^{3}}{a\,b^{1}}\)
\(4\)\(=\)\(b^{2}\)
\(b\)\(=\)\(2\quad(b>0)\)
Back-substitute \(b=2\) to find \(a\):
\(4\)\(=\)\(a(2)\)
\(a\)\(=\)\(2\)
\(\therefore\) Rule: \(y=2\,(2^{x})\)
b2=4
Example 2 — \(y=a\,b^x+c\) with an asymptote
\(y=a\,(2^x)+c\) has horizontal asymptote \(y=3\) and passes through \((0,5)\). Find the rule.
Solution
Asymptote — read \(c\) off the horizontal asymptote \(y=c\):
\(c\)\(=\)\(3\)
Substitute the point \((0,5)\) to find \(a\):
\(5\)\(=\)\(a(2^{0})+3\)
\(5\)\(=\)\(a(1)+3\)
\(a\)\(=\)\(5-3=2\)
\(\therefore\) Rule: \(y=2\,(2^{x})+3\)
c=3
Example 3 — Logarithmic rule
\((2,3)\) and \((5,7)\) lie on \(y=a\log_2(x-b)+c\), with vertical asymptote \(x=1\). Find \(a,b,c\).
Solution
Asymptote — the vertical asymptote \(x=b\) gives \(b\):
\(b\)\(=\)\(1\)
Use \((2,3)\), where \(x-b=1\) makes the log \(0\):
\(3\)\(=\)\(a\log_2(2-1)+c\)
\(3\)\(=\)\(a\log_2(1)+c\)
\(3\)\(=\)\(a(0)+c\)
\(c\)\(=\)\(3\)
Use \((5,7)\) to find \(a\):
\(7\)\(=\)\(a\log_2(5-1)+3\)
\(7\)\(=\)\(a\log_2(4)+3\)
\(7\)\(=\)\(2a+3\)
\(2a\)\(=\)\(4\)
\(a\)\(=\)\(2\)
\(\therefore\) Rule: \(y=2\log_2(x-1)+3\)
Graph of y=2 log_2(x-1)+3Increasing logarithmic curve with dashed vertical asymptote x=1 through the points (2,3) and (5,7). x y x=1 (5,7) (2,3)
a=2
Example 4 — \(y=A\,e^{kx}\) (base \(e\))
\(y=A\,e^{kx}\) passes through \((0,20)\) and \((5,60)\). Find \(A\) and \(k\).
Solution
The \(y\)-intercept \((0,20)\) gives \(A\) (since \(e^{0}=1\)):
\(20\)\(=\)\(A\,e^{k(0)}\)
\(20\)\(=\)\(A(1)\)
\(A\)\(=\)\(20\)
Substitute \((5,60)\) and divide by \(A\):
\(60\)\(=\)\(20\,e^{5k}\)
\(e^{5k}\)\(=\)\(3\)
Take \(\ln\) of both sides to solve for \(k\):
\(5k\)\(=\)\(\ln 3\)
\(k\)\(=\)\(\dfrac{\ln 3}{5}\)
\(\approx\)\(0.22\)
\(\therefore\) Rule: \(y=20\,e^{0.22x}\) (to \(2\) d.p.)
k=ln35

Common pitfalls

Divide the points, do not subtract. For \(y=a\,b^x\), dividing two point-equations cancels \(a\) and leaves \(\dfrac{y_2}{y_1}=b^{\,x_2-x_1}\). Subtracting leaves \(a\) in both terms and does not isolate \(b\).
Read \(c\) from the asymptote first. For \(y=a\,b^x+c\), the horizontal asymptote is \(y=c\). If you substitute a point before finding \(c\), you will get the wrong \(a\).
A log's vertical asymptote is an \(x\)-value. For \(y=a\log_b(x-h)+c\) the asymptote is \(x=h\), a horizontal shift — not a \(y\)-value. Also a base must satisfy \(b>0\).

Frequently asked questions

How do you find the rule y = a b^x from two points?

Substitute both points, then divide so \(a\) cancels: \(\dfrac{y_2}{y_1}=b^{\,x_2-x_1}\). Solve for \(b>0\), then back-substitute for \(a\).

How does a horizontal asymptote help find the rule?

For \(y=a\,b^x+c\) the asymptote is \(y=c\), so it gives \(c\) directly. Then a point gives \(a\) (and a second point gives \(b\)).

How do you find A and k for y = A e^(kx)?

The \(y\)-intercept gives \(A\) (since \(e^0=1\)). A second point gives \(e^{kx}\); take \(\ln\), e.g. \(e^{5k}=3\Rightarrow k=\dfrac{\ln 3}{5}\).

How do you determine a logarithmic rule?

The vertical asymptote is \(x=h\), so read \(h\) first. The point where \(x-h=1\) makes \(\log_b(1)=0\), giving \(c\); another point gives \(a\).

Why divide instead of subtracting?

\(a\) is a factor, not a term. Dividing cancels the common factor and leaves a pure power of \(b\) to solve; subtracting does not isolate \(b\).

What do you read off the graph first?

Any asymptote: \(y=c\) fixes an exponential's vertical shift, \(x=h\) fixes a logarithm's horizontal shift. Then use points for \(a\) and the base.

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