Now we put everything together. Limits at Infinity and Horizontal Asymptotes give the long-run behavior, the first derivative gives rising and falling, and Concavity and the Second Derivative Test gives the bending. Together they pin down the shape of a graph without plotting hundreds of points.
Key idea
A sketching checklist: (1) domain and intercepts; (2) symmetry; (3) asymptotes: vertical (where the function blows up) and horizontal (limits at $\pm\infty$); (4) $f'$: increasing/decreasing and local extrema; (5) $f''$: concavity and inflection points; (6) plot the key points and join them with the right shape.
Why it works
Each step removes some ways the graph could look. The asymptotes fix the behavior at the edges. The sign of $f'$ fixes the direction on each interval, so the graph cannot wiggle more than the critical numbers allow. The sign of $f''$ fixes which way each piece bends. Once these are set, only one shape fits through the key points.
Worked example: A polynomial
Sketch $f(x) = x^3 - 3x^2$.
- Intercepts: $f(x) = x^2(x - 3)$, zero at $x = 0$ and $x = 3$. No asymptotes (a polynomial). As $x \to \infty$, $f \to \infty$; as $x \to -\infty$, $f \to -\infty$.
- $f'(x) = 3x^2 - 6x = 3x(x - 2)$: increasing on $(-\infty, 0)$ and $(2, \infty)$, decreasing on $(0, 2)$. Local max $(0, 0)$, local min $(2, -4)$.
- $f''(x) = 6x - 6$: concave down for $x < 1$, up for $x > 1$. Inflection point $(1, -2)$.
- Plot $(0, 0)$, $(1, -2)$, $(2, -4)$, $(3, 0)$ and join: rise to the peak at the origin, bend through $(1, -2)$, bottom out at $(2, -4)$, then rise.
Answer: An "S" shape with a peak at $(0, 0)$, a valley at $(2, -4)$, and an inflection point at $(1, -2)$.
Worked example: A rational function
Sketch $f(x) = \dfrac{x}{x^2 + 1}$.
- Domain: all reals (the bottom is never 0). Only intercept: $(0, 0)$. Odd symmetry: $f(-x) = -f(x)$.
- Degree of top < degree of bottom: horizontal asymptote $y = 0$ on both ends.
- $f'(x) = \frac{1 - x^2}{(x^2 + 1)^2}$: rising on $(-1, 1)$, falling outside. Local max $\left(1, \frac{1}{2}\right)$, local min $\left(-1, -\frac{1}{2}\right)$.
- $f''(x) = \frac{2x(x^2 - 3)}{(x^2 + 1)^3}$: zero at $0$ and $\pm\sqrt{3}$, changing sign each time, so three inflection points.
Answer: A smooth wave: coming in from the left just below the axis, it dips to $-\frac{1}{2}$ at $x = -1$, climbs through the origin to $\frac{1}{2}$ at $x = 1$, then decays back toward 0.
Common mistake
Plotting the critical points but ignoring concavity and asymptotes, and then joining the points with straight segments or the wrong bend. Check your sketch against each sign chart: on every interval, the direction must match $f'$ and the bend must match $f''$.