From Individual Features to a Complete Sketch
A curve sketch is most reliable when it is built from evidence rather than from the appearance of a formula. A domain break divides the graph into separate pieces; intercepts locate crossings; asymptotes describe limiting behavior; and derivatives reveal how each piece moves and bends.
In this tutorial, we combine those features for a function with a vertical asymptote, a slant asymptote, two \(x\)-intercepts, and two critical points. The important task is not merely to list them. It is to check that they fit together into one consistent picture.
A Short Example: Domain, Intercepts, and Asymptotes
Worked Example: Locating the Basic Features
Consider \(g(x)=x+\frac{2}{x}-3\). The denominator requires \(x\ne 0\), so the domain is \((-\infty,0)\cup(0,\infty)\). In particular, there is no \(y\)-intercept because \(g(0)\) is undefined.
To find the \(x\)-intercepts, set \(g(x)=0\), keeping the domain restriction in mind:
Multiplying by \(x\) is valid here because \(x\ne0\). The candidate roots are \(x=1\) and \(x=2\). Substitution verifies them: \(g(1)=1+2-3=0\), and \(g(2)=2+1-3=0\). Thus the intercepts are \((1,0)\) and \((2,0)\).
As \(x\to0^+\), the term \(2/x\) tends to \(+\infty\), so \(g(x)\to+\infty\). As \(x\to0^-\), \(2/x\to-\infty\), so \(g(x)\to-\infty\). Therefore \(x=0\) is a vertical asymptote. Also, \(g(x)-(x-3)=2/x\to0\) as \(x\to\pm\infty\), so \(y=x-3\) is a slant asymptote.
The two intercepts and both asymptotes have been checked directly against the formula. Derivatives will add the information needed to determine how the graph travels between these features.
Critical Points and Concavity
A critical number is a number \(c\) in the domain of a function at which \(f'(c)=0\) or \(f'(c)\) does not exist. The requirement that \(c\) belong to the domain matters: a point where the function is not defined is not a critical number.
Where the derivative is positive, the function is increasing; where it is negative, the function is decreasing. A change from positive to negative indicates a local maximum, while a change from negative to positive indicates a local minimum. The second derivative describes concavity: \(f''(x)>0\) means concave up, and \(f''(x)<0\) means concave down.
Worked Example: Reading the Derivative Signs
For \(h(x)=x+\frac{1}{x}-3\), the domain is \(x\ne0\). Differentiating gives
The denominator \(x^2\) is positive wherever \(h\) is defined. The derivative is zero at \(x=-1\) and \(x=1\), and undefined at \(x=0\), which is not in the domain. Thus the critical numbers are \(-1\) and \(1\), not \(0\).
The numerator is positive for \(x<-1\) and \(x>1\), and negative for \(-1<x<1\). Since \(x=0\) is excluded, the derivative sign intervals are \((-\infty,-1)\), \((-1,0)\), \((0,1)\), and \((1,\infty)\). The function increases, decreases, decreases, then increases on these intervals. Therefore \(x=-1\) is a local maximum and \(x=1\) is a local minimum.
The values are \(h(-1)=-1-1-3=-5\) and \(h(1)=1+1-3=-1\). Substituting these points into the derivative confirms the stationary-point condition: \(h'(-1)=1-1=0\) and \(h'(1)=1-1=0\).
For concavity, differentiate once more: \(h''(x)=2/x^3\). This is negative when \(x<0\) and positive when \(x>0\). Thus the graph is concave down on \((-\infty,0)\) and concave up on \((0,\infty)\). There is no inflection point at zero: the function is not defined there, so there is no point on the graph at which its concavity changes.
Complete Analysis of a Mixed-Feature Function
Now assemble all of the information for one function:
The function is defined for all real \(x\) except \(x=0\). We will determine its intercepts, asymptotes, critical points, monotonicity, concavity, and end behavior, then use these facts to describe its full sketch.
Worked Example: Complete Analysis and Sketch Description
1. Domain and behavior at the domain break. The term \(1/x\) requires \(x\ne0\), so the domain is \((-\infty,0)\cup(0,\infty)\). As \(x\to0^+\), \(1/x\to+\infty\), and the other terms remain bounded, so \(f(x)\to+\infty\). As \(x\to0^-\), \(1/x\to-\infty\), so \(f(x)\to-\infty\). Hence \(x=0\) is a vertical asymptote, with opposite infinite behavior on its two sides.
2. Intercepts. There is no \(y\)-intercept because \(f(0)\) is undefined. To find the \(x\)-intercepts, solve \(f(x)=0\) for \(x\ne0\):
Both roots are positive and nonzero, so both belong to the domain. Substitution verifies each one: if \(r=(3\pm\sqrt5)/2\), then \(r^2-3r+1=0\); dividing by \(r\) gives \(r+1/r-3=0\). The two intercepts are therefore \(\bigl((3-\sqrt5)/2,0\bigr)\) and \(\bigl((3+\sqrt5)/2,0\bigr)\).
3. End behavior and the slant asymptote. Subtract the line \(y=x-3\) from the function:
Since \(1/x\to0\) as \(x\to\pm\infty\), the slant asymptote is \(y=x-3\) at both ends. More specifically, \(1/x<0\) for \(x<0\), so the graph lies below this line on the negative branch; for \(x>0\), it lies above the line.
4. Critical points and monotonicity. The first derivative is
Its zeros are \(x=-1\) and \(x=1\). The derivative is positive on \((-\infty,-1)\) and \((1,\infty)\), and negative on \((-1,0)\) and \((0,1)\). Thus \(f\) increases to a local maximum at \(x=-1\), decreases from there toward the left side of the vertical asymptote, decreases from the right side of the asymptote to a local minimum at \(x=1\), and then increases.
The critical values are \(f(-1)=-1-1-3=-5\) and \(f(1)=1+1-3=-1\). The derivative check gives \(f'(-1)=1-1=0\) and \(f'(1)=1-1=0\). The derivative changes from positive to negative at \(-1\), confirming a local maximum there; it changes from negative to positive at \(1\), confirming a local minimum there.
5. Concavity. The second derivative is \(f''(x)=2/x^3\). It is negative for every \(x<0\), so the negative branch is concave down. It is positive for every \(x>0\), so the positive branch is concave up. There is no inflection point: the sign change occurs across \(x=0\), which is outside the domain, and \(f''\) has no zero in the domain.
6. Assemble the sketch. On the negative branch, the graph approaches the slant asymptote as \(x\to-\infty\), rises to the local maximum \((-1,-5)\), then falls to \(-\infty\) as \(x\to0^-\). Because the branch’s maximum is \(-5\), it has no \(x\)-intercept. On the positive branch, the graph begins at \(+\infty\) as \(x\to0^+\), decreases to the local minimum \((1,-1)\), and then rises to \(+\infty\). Its two \(x\)-intercepts lie on either side of \(x=1\): \((3-\sqrt5)/2<1<(3+\sqrt5)/2\). The branch is above the \(x\)-axis before the first root and after the second, and below the axis between them.
These descriptions also determine the range. The negative branch takes values \((-\infty,-5]\), and the positive branch takes values \([-1,\infty)\). Therefore the range is \((-\infty,-5]\cup[-1,\infty)\).
Why the Sign Chart Justifies the Shape
A sign chart is not a drawing technique based on guesswork. On any interval where \(f\) is continuous and differentiable, the Mean Value Theorem says that for \(a<b\) in that interval, some \(c\in(a,b)\) satisfies \(f(b)-f(a)=f'(c)(b-a)\). Since \(b-a>0\), a positive derivative throughout the interval forces \(f(b)>f(a)\); a negative derivative forces \(f(b)<f(a)\). This is why derivative signs establish increasing and decreasing behavior.
The domain break matters to this argument. The Mean Value Theorem cannot be applied across \(x=0\), because the function is not continuous on any interval containing zero. Accordingly, the two branches must be analyzed separately. The asymptote limits then describe how each branch approaches the excluded point, while the derivative signs determine its movement on each side.
Likewise, the asymptote claim follows from its defining comparison: a line \(y=mx+b\) is an asymptote at an end when the difference between the function and that line tends to zero there. Here the exact difference is \(1/x\), which tends to zero as \(x\to\pm\infty\). This gives a direct justification, not merely a visual guess.
Common Pitfall: Treating an Excluded Point as a Critical Point
A tempting but incorrect statement is: “The derivative is undefined at \(x=0\), so \(x=0\) is a critical point of \(f\).” For this function, that conclusion is wrong. A critical number must be in the function’s domain, and \(f(0)\) does not exist. The correct critical numbers are \(-1\) and \(1\), where \(f'(x)=0\).
The distinction has consequences for the sketch. Zero is a vertical asymptote that separates two branches; it is not a point on either branch. It cannot be a local extremum or an inflection point. In contrast, \((-1,-5)\) and \((1,-1)\) are points on the graph and are genuine local extrema.
A Practical Order for a Complete Sketch
For a function with several interacting features, a consistent order helps prevent omissions:
Mark excluded points before interpreting derivative information.
Verify roots by substitution and determine behavior near domain breaks and at the ends.
Compare the function with candidate vertical or linear asymptotes using limits or differences.
Use the signs of \(f'\) and \(f''\) to establish monotonicity, extrema, and concavity.
Draw each branch so it respects every verified point, sign, limit, and asymptote.
This order is useful because each item constrains the possible graph. If a proposed sketch crosses an asymptote, places a turning point outside the domain, or contradicts a derivative sign, the sketch is inconsistent even if it looks plausible at first.
Check Your Understanding
Use the analysis of \(f(x)=x+\frac{1}{x}-3\) to answer these questions.
- Why is \(x=0\) a domain break but not a critical number?
- Which limit establishes the vertical asymptote on the right side of zero?
- Why does the negative branch have no \(x\)-intercept?
- What derivative sign change identifies the local minimum?
- Why is \(x=0\) not an inflection point, even though concavity differs on either side?
- What calculation verifies the slant asymptote \(y=x-3\)?
