From a Formula to a Consistent Sketch
A polynomial is defined and continuous for every real number, but its graph can still have several important features: intercepts, turning points, changes in concavity, and different behavior at the two ends of the real line. A complete curve analysis finds these features and checks that they fit together.
The main tools are the signs of the first and second derivatives. Intercepts anchor the graph to the axes, while the degree and leading coefficient determine its end behavior. A sketch built from all of this information should be consistent on every interval—not merely pass through a few known points.
A Practical Analysis Order
For a polynomial \(f\), its domain is all real numbers, and its derivatives are also defined for all real numbers. The following sequence organizes the work:
Find the \(y\)-intercept from \(f(0)\), solve \(f(x)=0\) for any \(x\)-intercepts, and check whether the formula has even or odd symmetry.
Use the degree and leading coefficient to find the limits as \(x\) tends to positive and negative infinity.
Solve \(f'(x)=0\), divide the real line into intervals, and determine the sign of \(f'\) on each interval.
Solve \(f''(x)=0\), test the sign of \(f''\) on the resulting intervals, and confirm any inflection points by checking for a concavity change.
Plot the intercepts, critical points, and inflection points, then connect them in a way that respects the monotonicity, concavity, and end behavior.
Critical Numbers and Monotonicity
A critical number of a function is an interior number in its domain where the derivative is zero or does not exist. A polynomial’s derivative exists everywhere, so its critical numbers are exactly the real solutions of \(f'(x)=0\). A critical point on the graph has coordinates \((c,f(c))\), where \(c\) is a critical number.
Between consecutive critical numbers, the derivative is continuous and cannot change sign without being zero somewhere in that interval. Thus, testing one value of \(f'\) in each interval determines whether the function is increasing or decreasing there. A change from positive to negative indicates a local maximum; a change from negative to positive indicates a local minimum. If the sign does not change, there is no local extremum at that critical number.
Why the sign test works: Take any \(a<b\) in an interval where \(f'>0\). The Mean Value Theorem applies because a polynomial is continuous on \([a,b]\) and differentiable on \((a,b)\). It gives a number \(c\in(a,b)\) such that
Since \(f'(c)>0\) and \(b-a>0\), it follows that \(f(b)-f(a)>0\), so \(f(b)>f(a)\). This proves that \(f\) is increasing. If \(f'<0\), the same equation gives \(f(b)<f(a)\), proving that \(f\) is decreasing.
Concavity and Inflection Points
Where \(f''(x)>0\), the graph is concave up; where \(f''(x)<0\), it is concave down. For a polynomial, \(f''\) is defined everywhere. The real solutions of \(f''(x)=0\) divide the line into intervals on which its sign can be tested.
An inflection point is a point on the graph where the concavity changes. A zero of \(f''\) is a candidate, not an automatic inflection point: the sign of \(f''\) must actually change across that value. Once a change is verified at \(x=c\), the inflection point is \((c,f(c))\).
End Behavior and Special Polynomial Cases
Write a nonconstant polynomial in the form \(f(x)=a_nx^n+\cdots+a_1x+a_0\), with \(a_n\ne0\). For large \(|x|\), the leading term \(a_nx^n\) determines the end behavior because the lower-degree terms become small relative to it. If \(n\) is even, both ends go in the direction determined by the sign of \(a_n\). If \(n\) is odd, the ends go in opposite directions.
| Degree and leading coefficient | As \(x\to-\infty\) | As \(x\to\infty\) |
|---|---|---|
| Even degree, \(a_n>0\) | \(f(x)\to\infty\) | \(f(x)\to\infty\) |
| Even degree, \(a_n<0\) | \(f(x)\to-\infty\) | \(f(x)\to-\infty\) |
| Odd degree, \(a_n>0\) | \(f(x)\to-\infty\) | \(f(x)\to\infty\) |
| Odd degree, \(a_n<0\) | \(f(x)\to\infty\) | \(f(x)\to-\infty\) |
A nonconstant polynomial has no vertical or horizontal asymptotes. A polynomial of degree at least two has no slant asymptote. A constant polynomial \(f(x)=c\), however, has the horizontal asymptote \(y=c\); a degree-one polynomial is itself a line. These cases should not be hidden by a general claim about all polynomials.
Worked Example: A Cubic with Two Turning Points
Worked Example: A Cubic with Two Turning Points
Analyze and sketch \(f(x)=x^3-3x\).
Intercepts and symmetry. Factor to find the \(x\)-intercepts:
Thus the intercepts are \((-\sqrt{3},0)\), \((0,0)\), and \((\sqrt{3},0)\). Also, \(f(0)=0\), confirming the \(y\)-intercept. Since \(f(-x)=-f(x)\), the function is odd and its graph has origin symmetry.
End behavior. The leading term is \(x^3\), an odd-degree term with positive coefficient. Therefore \(f(x)\to-\infty\) as \(x\to-\infty\), and \(f(x)\to\infty\) as \(x\to\infty\).
Critical numbers and monotonicity.
The critical numbers are \(x=-1\) and \(x=1\). Testing the intervals gives \(f'(x)>0\) on \((-\infty,-1)\), \(f'(x)<0\) on \((-1,1)\), and \(f'(x)>0\) on \((1,\infty)\). The function increases, then decreases, then increases. At the critical numbers,
So \((-1,2)\) is a local maximum and \((1,-2)\) is a local minimum.
Concavity and inflection point.
For \(x<0\), \(f''(x)<0\), so the graph is concave down. For \(x>0\), \(f''(x)>0\), so it is concave up. The sign changes at \(x=0\), and \(f(0)=0\); therefore \((0,0)\) is an inflection point.
Sketch check. From left to right, the curve rises from the lower left to \((-1,2)\), then falls through the origin, changing from concave down to concave up, reaches \((1,-2)\), and rises to the upper right. This order agrees with all three intercepts and both end limits. As a calculation check, differentiating \(x^3-3x\) gives \(3x^2-3\), and differentiating again gives \(6x\), as used above.
Worked Example: An Even Quartic with Two Inflection Points
Worked Example: An Even Quartic with Two Inflection Points
Analyze \(g(x)=x^4-4x^2\).
Intercepts and symmetry. Factor:
The \(x\)-intercepts are \((-2,0)\), \((0,0)\), and \((2,0)\); \(g(0)=0\) is also the \(y\)-intercept. Since \(g(-x)=g(x)\), the function is even and symmetric about the \(y\)-axis.
End behavior. The leading term is \(x^4\), so both ends rise: \(g(x)\to\infty\) as \(x\to\pm\infty\).
Critical numbers and monotonicity.
The critical numbers are \(-\sqrt{2}\), \(0\), and \(\sqrt{2}\). The derivative signs on the four intervals are, in order, negative, positive, negative, and positive. Thus \(g\) decreases on \((-\infty,-\sqrt{2})\), increases on \((-\sqrt{2},0)\), decreases on \((0,\sqrt{2})\), and increases on \((\sqrt{2},\infty)\). The corresponding critical values are
There are local minima at \((-\sqrt{2},-4)\) and \((\sqrt{2},-4)\), and a local maximum at \((0,0)\).
Concavity and inflection points.
Solving \(g''(x)=0\) gives \(x=\pm\sqrt{2/3}\). For \(|x|>\sqrt{2/3}\), \(g''(x)>0\); for \(|x|<\sqrt{2/3}\), \(g''(x)<0\). Concavity changes at both candidates. At either value, \(x^2=2/3\) and \(x^4=4/9\), so
The inflection points are \((-\sqrt{2/3},-20/9)\) and \((\sqrt{2/3},-20/9)\).
Sketch check. The curve is symmetric about the \(y\)-axis, rises at both ends, touches the axis at its local maximum \((0,0)\), and has equal minima at \(x=\pm\sqrt{2}\). Its concavity is up on the two outer intervals and down between the inflection points. These features agree with the intercepts and derivative signs. Differentiating \(g\) gives \(4x^3-8x\), and differentiating again gives \(12x^2-8\), verifying the derivatives used.
Worked Example: A Quadratic with One Turning Point
Worked Example: A Quadratic with One Turning Point
Analyze \(h(x)=2x^2-8x+6\).
Intercepts. Factor the expression:
The \(x\)-intercepts are \((1,0)\) and \((3,0)\). The \(y\)-intercept is \((0,6)\). Completing the square also gives \(h(x)=2(x-2)^2-2\), which identifies the vertex \((2,-2)\).
End behavior. The leading term \(2x^2\) has even degree and positive coefficient, so \(h(x)\to\infty\) at both ends.
Monotonicity and extrema.
The only critical number is \(x=2\). For \(x<2\), \(h'(x)<0\); for \(x>2\), \(h'(x)>0\). Therefore \(h\) decreases on \((-\infty,2)\) and increases on \((2,\infty)\), with a local minimum at \(h(2)=2(4)-16+6=-2\). This is also the absolute minimum because both ends tend to infinity.
Concavity.
The graph is concave up everywhere. The second derivative never changes sign, so there are no inflection points.
Sketch check. The graph falls from the upper left through \((0,6)\) and \((1,0)\) to the vertex \((2,-2)\), then rises through \((3,0)\) to the upper right, remaining concave up throughout. Substitution confirms the vertex value: \(h(2)=-2\). Differentiating \(2x^2-8x+6\) gives \(4x-8\), and its derivative is \(4\).
Constant and Linear Cases
A constant polynomial \(f(x)=c\) has \(f'(x)=0\) for every real \(x\), and its graph is the horizontal line \(y=c\). It has no intervals of strict increase or decrease and no change in concavity. Under the usual limit definition, \(y=c\) is also its horizontal asymptote.
A degree-one polynomial \(f(x)=mx+b\), with \(m\ne0\), has constant derivative \(m\) and second derivative \(0\). It increases everywhere if \(m>0\), decreases everywhere if \(m<0\), and has no inflection point because its concavity does not change. Its graph is the line itself.
Common Pitfall: Confusing Intercepts with Critical Numbers
The equation \(f(x)=0\) finds \(x\)-intercepts. The equation \(f'(x)=0\) finds critical numbers of a polynomial. These are different questions.
For example, for \(q(x)=x^2-4\), setting \(q(x)=0\) gives \(x=-2\) and \(x=2\), the intercepts. A mistaken analysis might call these critical numbers. But
The actual critical number is \(0\), and \(q(0)=-4\). The points \((-2,0)\) and \((2,0)\) are intercepts, not critical points. Always match the equation to the feature being sought.
Putting the Evidence Together
A reliable sketch is a conclusion drawn from the whole analysis. Before drawing, check that the features agree: a function cannot be increasing on an interval where its derivative is negative, and a concavity change requires a change in the sign of the second derivative. The plotted points should also fit the end behavior and the intercepts.
- Use \(f(x)=0\) for \(x\)-intercepts and \(f(0)\) for the \(y\)-intercept.
- Use \(f'(x)=0\) to locate polynomial critical numbers, then test derivative signs.
- Use \(f''(x)=0\) to find possible inflection numbers, then verify a sign change.
- Use the leading term to determine how the two ends of the graph behave.
- Plot each important point with its actual function value before connecting the curve.
Check Your Understanding
Use derivative signs and end behavior where appropriate. Give exact values, and explain how each feature supports a sketch.
- For \(p(x)=x^3-12x\), find the intercepts, critical numbers, and intervals of increase and decrease.
- For the same polynomial, find its concavity intervals and inflection point.
- What are the end limits of \(r(x)=-2x^4+x\)?
- For \(s(x)=x^2+2x+5\), determine its critical point, concavity, and whether it has real \(x\)-intercepts.
- Why does solving \(f''(x)=0\) alone not establish that a point is an inflection point?
