How do you determine a logarithmic rule y = a log_b(x - h) + c?

The vertical asymptote is x = h, so read h off first. At the point where x minus h equals 1 the logarithm is 0, so that point gives c. Substitute another point to find a.

Why divide the points instead of subtracting when finding a base?

In y = a b^x the parameter a is a factor, not a term. Dividing two point-equations cancels the common factor a and leaves a pure power of b, which can be solved. Subtracting would leave a inside both terms and would not isolate b.

How do you determine a logarithmic rule y = a log_b(x - h) + c?

The vertical asymptote is x = h, so read h off first. At the point where x minus h equals 1 the logarithm is 0, so that point gives c. Substitute another point to find a.

Why divide the points instead of subtracting when finding a base?

In y = a b^x the parameter a is a factor, not a term. Dividing two point-equations cancels the common factor a and leaves a pure power of b, which can be solved. Subtracting would leave a inside both terms and would not isolate b.

Year 12 Methods (Unit 3 & 4) Exponential and logarithmic functions

Determining rules for graphs of exponential andlogarithmic functions

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

Substitute both points to get two equations, then divide one by the other so that a cancels. This leaves y2 divided by y1 equals b to the power (x2 minus x1), which you solve for the base b (taking b greater than 0). Substitute b back into either equation to find a.

Q: How does a horizontal asymptote help find the rule?

For y = a b^x + c the horizontal asymptote is y = c, so the asymptote gives the value of c directly. Once c is known, substitute a point to find a, and a second point to find the base b.

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

The y-intercept gives A, because e to the power 0 is 1, so at x = 0 the value equals A. Substitute a second point and divide out A to get e to the power (k times x); then take the natural logarithm to solve for k. For example e to the power 5k equals 3 gives k equals ln 3 divided by 5.

Q: What do you read off the graph first?

Always read any asymptote first. A horizontal asymptote y = c fixes the vertical shift c of an exponential; a vertical asymptote x = h fixes the horizontal shift h of a logarithm. Only then use a point or a ratio of points for the remaining parameters.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), 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).

\(y=a\,b^x\): divide the two points to find \(b\), then back-substitute for \(a\)

\(y=a\log_b(x-h)+c\): 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 = a b^{x}

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

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

y = a b^{x} + c

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

\[y=a\,\log_{b}(x-h)+c \qquad \text{asymptote } x=h\]

y = a \log_{b} (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

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.
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 \(c\).
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 \(k\).
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, then divide to cancel \(a\).

\(4\)	\(=\)	\(a\,b^{1}\)
\(16\)	\(=\)	\(a\,b^{3}\)
\(\dfrac{16}{4}\)	\(=\)	\(b^{2}\Rightarrow b=2\)

Then \(4=a(2)\), so \(a=2\). Rule: \(y=2\,(2^{x})\).

b^{2} = 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

Read \(c\) from the asymptote, then use the point.

\(\text{asymptote } y=c\)	\(\Rightarrow\)	\(c=3\)
\(5\)	\(=\)	\(a(2^{0})+3\)
\(a\)	\(=\)	\(2\)

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 gives \(b\); the point where the log is \(0\) gives \(c\).

\(\text{asymptote } x=b\)	\(\Rightarrow\)	\(b=1\)
\((2,3):\ \log_2(1)\)	\(=\)	\(0\Rightarrow c=3\)
\((5,7):\ 7\)	\(=\)	\(2a+3\Rightarrow a=2\)

Rule: \(y=2\log_2(x-1)+3\).

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 gives \(A\); a ratio then gives \(k\) via \(\ln\).

\((0,20):\ A\)	\(=\)	\(20\)
\((5,60):\ e^{5k}\)	\(=\)	\(3\)
\(k\)	\(=\)	\(\dfrac{\ln 3}{5}\approx 0.22\)

Rule: \(y=20\,e^{0.22x}\) (to \(2\) d.p.).

k = \frac{\ln 3}{5}

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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