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Analyzing cubic functions: extrema, inflection, and graphs

2026-10-22 huynhvietkhanh Comments: 0

Classify cubic graphs by derivative roots and leading-coefficient sign, then calculate extrema, the inflection point, and intercepts.

High schoolFunctions, Graphs & Tangents

Content updated: October 22, 2026

The first derivative determines a cubic's local extrema; the second derivative locates its inflection point and center of symmetry. Its graph must agree with derivative signs and both limits at infinity.

Objectives: classify cubic functions, describe their variation, and graph them using calculated points and signs.

Prerequisites: derivatives, signs of quadratic polynomials, local extrema, and limits at infinity.

On this page: Definition · Method · Examples · Practice · Further learning

1. The function and its two ends

Consider $f(x)=ax^3+bx^2+cx+d$, with $a\ne0$. Its domain is $\mathbb R$. If $a>0$, the left limit is $-\infty$ and the right limit $+\infty$. If $a<0$, the signs reverse. This follows because $ax^3$ dominates the lower-degree terms at infinity.

There is no vertical asymptote, since a polynomial is continuous and finite at every input. There is also no horizontal or oblique line asymptote: $f(x)-(mx+n)$ remains a cubic polynomial with nonzero leading coefficient, so cannot tend to 0 at infinity.

2. Six cases from the derivative and leading coefficient

We have $f'(x)=3ax^2+2bx+c$. Set $\delta=b^2-3ac$, the derivative's reduced discriminant. For $\delta>0$, the derivative has roots $x_1<x_2$ and changes sign at each. If $a>0$, the function increases–decreases–increases, with a local maximum at $x_1$ and local minimum at $x_2$. If $a<0$, it decreases–increases–decreases, with the local minimum first and maximum second.

If $\delta=0$, the derivative is 0 at $x_I=-b/(3a)$ but does not change sign; there are no local extrema. The function is strictly increasing on $\mathbb R$ for $a>0$ and strictly decreasing for $a<0$. Indeed, with $t=x-x_I$, it has the form $f(x_I)+at^3$ in this case.

If $\delta<0$, the derivative never vanishes and always has the sign of $a$. The function is again strictly increasing or decreasing on $\mathbb R$. The three cases for $\delta$ combined with two signs of $a$ give six types. A derivative of 0 at a point does not alone establish a local extremum.

Since $f''(x)=6ax+2b$, the inflection input is $x_I=-b/(3a)$, where the second derivative changes sign. Set $t=x-x_I$; expansion gives $f(x_I+t)=f(x_I)+at^3+\lambda t$ for some real $\lambda$. The part depending on $t$ is odd, so $f(x_I+t)+f(x_I-t)=2f(x_I)$; hence $I=(x_I;f(x_I))$ is a center of symmetry.

3. A complete analysis

The function y = 2x³ − 9x² + 12x − 4

The domain is $\mathbb R$; the positive leading coefficient gives left limit $-\infty$ and right limit $+\infty$. The derivative is $f'(x)=6x^2-18x+12=6(x-1)(x-2)$, positive for $x<1$ or $x>2$ and negative for $1<x<2$.

The function increases on $(-\infty;1)$ and $(2;+\infty)$ and decreases on $(1;2)$. The value $f(1)=1$ gives local maximum point $(1;1)$; $f(2)=0$ gives local minimum point $(2;0)$. Its variation, read from left to right, is $-\infty\nearrow1\searrow0\nearrow+\infty$.

The correct second derivative is $f''(x)=12x-18=6(2x-3)$. Thus $x_I=3/2$, $f(3/2)=1/2$, and the center of symmetry is $I=(3/2;1/2)$. There are no asymptotes.

For graphing, add the intercepts: $f(0)=-4$ and $f(x)=(2x-1)(x-2)^2$. The graph crosses the horizontal axis at $(1/2;0)$ and is tangent to it at $(2;0)$. Draw through these points while retaining the three monotonicity intervals and symmetry about $I$. The diagram below is calculated from the formula, rather than inferred by eye.

0.0 0.5 1.0 1.5 2.0 2.5 3.0 x −4 −2 0 2 4 y (1, 1) (2, 0) I (3/2, 1/2) y = 2x³ − 9x² + 12x − 4
The graph is calculated from the function formula; the two local extrema and center of symmetry are marked. Only a finite part of the graph is shown.

4. Practice

  1. Find the local extremum points and center of symmetry of $x^3-3x$.
  2. The derivative of $-x^3$ is 0 at 0. Is there a local extremum there?
  3. Find $m$ such that $x^3+mx$ has two local extrema.
Hints and answers
  1. The derivative $3(x-1)(x+1)$ gives local maximum $(-1;2)$ and local minimum $(1;-2)$. The inflection point and center of symmetry are $(0;0)$.
  2. No. The function strictly decreases on all of $\mathbb R$, and its derivative $-3x^2$ does not change sign at 0; this is an inflection point with a horizontal tangent.
  3. The derivative $3x^2+m$ has two distinct roots exactly when $m<0$. For $m=0$, there is only a stationary point, with no local extremum.

5. Common mistakes

Do not use a second-derivative root as an extremum input. Do not count a double derivative root as two extrema. Retain the correct intercepts, signs at both ends, and maximum–minimum order when drawing. The coefficient $a$ must be nonzero in the cubic classification.

6. Further learning

Graphs of even quartic functions (English lesson) · Graphs of linear fractional functions (English lesson) · Counting equation solutions using graphs (English lesson) · Derivatives (English lesson).

Original exercise material on Scribd.

GeoGebra graphing tool from the original lesson.

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