Why These Function Families Have Predictable Shapes
Exponential and logarithmic graphs are built from a small number of familiar shapes. Their transformations can shift, stretch, reflect, or reverse those shapes, but their derivatives and end behavior still follow consistent patterns. Recognizing those patterns makes a curve analysis more efficient and helps catch errors that a quick sketch might hide.
For a complete analysis, identify the domain and intercepts, determine where the function increases or decreases, determine its concavity, and describe its end behavior. Then use those facts together to sketch the graph. The focus here is on exponential and logarithmic functions and their basic transformations.
Transformed Exponential Functions
Consider the family
The parameters \(A\) and \(c\) can reverse the graph's direction or concavity; \(h\) shifts it horizontally, and \(D\) shifts it vertically. The exponential expression is defined for every real \(x\), so the domain is \(\mathbb{R}\). Since \(b^{c(x-h)}>0\), the function never equals \(D\). As \(x\) moves toward the end where the exponential term tends to zero, the graph approaches the horizontal asymptote \(y=D\).
To read the end behavior, note that \(b^{c(x-h)}=e^{L(x-h)}\). If \(L>0\), the exponential term approaches zero as \(x\to-\infty\) and grows without bound as \(x\to\infty\). If \(L<0\), those directions are reversed. When \(A<0\), the unbounded term tends to \(-\infty\) rather than \(+\infty\); the horizontal asymptote remains \(y=D\).
Worked Example: An Increasing, Concave-Up Exponential
Worked Example: An Increasing, Concave-Up Exponential
Analyze \(f(x)=2^{x-1}-2\), including its intercepts, monotonicity, concavity, asymptote, and end behavior.
Domain and asymptote. The exponential is defined for every real \(x\), so the domain is \(\mathbb{R}\). Since \(2^{x-1}\to 0\) as \(x\to-\infty\), the horizontal asymptote is \(y=-2\).
Intercepts. At \(x=0\),
Thus the \(y\)-intercept is \((0,-\frac32)\). For the \(x\)-intercept, solve:
The \(x\)-intercept is \((2,0)\).
Monotonicity and concavity.
The function is increasing and concave up on \(\mathbb{R}\). Neither derivative is zero, so there are no critical numbers or inflection points.
End behavior and check. As \(x\to-\infty\), \(f(x)\to-2\) from above; as \(x\to\infty\), \(f(x)\to\infty\). The point \((2,0)\) is consistent with the increasing curve: the function is below zero at \(x=0\) and reaches zero at \(x=2\). Substitution confirms the intercept: \(f(2)=2^1-2=0\).
Transformed Logarithmic Functions
A useful general form is
The logarithm requires its argument to be positive, so the domain is precisely the set of \(x\) satisfying \(q(x-h)>0\). If \(q>0\), this means \(x>h\); if \(q<0\), it means \(x<h\). The boundary \(x=h\) is excluded from the domain and is a vertical asymptote: the logarithm's argument approaches zero from above as \(x\) approaches \(h\) from within the domain.
For the first derivative, the chain rule gives \(A\cdot \frac{q}{q(x-h)}=\frac{A}{x-h}\); the factor \(q\) cancels. Differentiating once more gives \(-A/(x-h)^2\). At the vertical asymptote, the argument tends to \(0^+\), so its logarithm tends to \(-\infty\). Consequently, \(g(x)\) tends to \(-\infty\) there if \(A>0\), and \(+\infty\) if \(A<0\). Toward the unbounded end of the domain, the argument tends to \(+\infty\), so the logarithm tends to \(+\infty\); the resulting function tends to \(+\infty\) if \(A>0\), and \(-\infty\) if \(A<0\).
Worked Example: A Decreasing Logarithm with a Vertical Asymptote
Worked Example: A Decreasing Logarithm with a Vertical Asymptote
Analyze \(g(x)=-\ln(x-2)+1\).
Domain and asymptote. The argument must be positive:
Thus the domain is \((2,\infty)\), and \(x=2\) is a vertical asymptote.
Intercepts. There is no \(y\)-intercept because \(x=0\) is outside the domain. For the \(x\)-intercept, solve:
Therefore the \(x\)-intercept is \((2+e,0)\). Substitution verifies it: \(-\ln(e)+1=-1+1=0\).
Monotonicity and concavity.
The graph decreases and is concave up throughout its domain. There are no critical numbers or inflection points in the domain.
End behavior and check. As \(x\to2^+\), \(\ln(x-2)\to-\infty\), so \(g(x)\to+\infty\). As \(x\to\infty\), \(g(x)\to-\infty\). The decreasing curve therefore crosses the \(x\)-axis once, at \(x=2+e\), in agreement with the intercept calculation.
Worked Example: A Logarithm Increasing on a Left-Side Domain
Worked Example: A Logarithm Increasing on a Left-Side Domain
Analyze \(r(x)=\ln(3-x)\), paying particular attention to the side of its vertical asymptote on which the graph lies.
Domain and asymptote. The logarithm's argument must be positive:
The domain is \((-\infty,3)\), and the vertical asymptote is \(x=3\).
Intercepts. At \(x=0\), \(r(0)=\ln 3\), so the \(y\)-intercept is \((0,\ln 3)\). For the \(x\)-intercept:
Thus the \(x\)-intercept is \((2,0)\). Checking gives \(r(2)=\ln 1=0\).
Monotonicity and concavity. Here \(r(x)=\ln\bigl((-1)(x-3)\bigr)\), so \(A=1\) and \(h=3\). Direct differentiation gives
The graph is decreasing and concave down on its entire domain.
End behavior. As \(x\to3^-\), \(3-x\to0^+\), so \(r(x)\to-\infty\). As \(x\to-\infty\), \(3-x\to\infty\), so \(r(x)\to\infty\). These limits agree with a decreasing curve on the left side of \(x=3\).
Common Pitfall: Ignoring the Logarithm's Argument
A common error is to differentiate \(\ln(x-2)\) as \(1/x\), or to treat it as if it were defined for every real \(x\). Both errors ignore the function inside the logarithm. The chain rule gives
The condition \(x>2\) is essential: it is the domain, not an optional restriction. A derivative sign chart or sketch must not include points where the original function is undefined.
Putting the Features Together
For these function families, the domain and derivative patterns sharply constrain the graph. A nonconstant transformed exponential is defined on all real numbers, has one horizontal asymptote, and has a fixed direction and concavity. A transformed logarithm is defined on one side of a vertical asymptote, and its derivative and concavity signs remain fixed throughout that domain. Neither family has an inflection point or an interior critical number under the nonzero-parameter conditions used here.
Use a consistent checklist when sketching:
- Find the domain from the function's defining expression.
- Locate the horizontal asymptote for an exponential or vertical asymptote for a logarithm.
- Calculate the intercepts, checking that each candidate lies in the domain.
- Use \(f'\) to determine increasing or decreasing behavior and \(f''\) to determine concavity.
- Use limits to place the graph relative to its asymptote and describe its unbounded end.
These steps connect algebraic transformations to the graph's overall shape. In particular, end behavior and derivative signs provide independent checks: a proposed sketch that rises toward an asymptote at the wrong end, or turns despite having a derivative of fixed sign, cannot be correct.
Check Your Understanding
For each question, use domains, derivatives, and end behavior to justify your conclusions.
- For \(f(x)=3^{x+1}+2\), state the domain, horizontal asymptote, monotonicity, and concavity.
- For \(g(x)=2\ln(x+4)\), find the domain and vertical asymptote, then determine its monotonicity and concavity.
- For \(r(x)=-\ln(5-x)\), identify the domain and describe both end limits.
- Why does \(A b^{c(x-h)}+D\), with the nonzero parameters specified in this tutorial, have no critical numbers?
- Find the \(x\)-intercept of \(y=\ln(x-1)-2\), and verify it by substitution.
