Curve Sketching with Derivatives

The order of work

A sketch shows the shape of a curve and the points that fix it, without plotting a table of values. The derivatives supply the shape. Work through the same list every time:

First the intercepts: set y = 0 for the x-intercepts and x = 0 for the y-intercept. Then the stationary points, from $\frac{dy}{dx}=0$dy/dx = 0, each classified as a maximum, a minimum or neither. Then the points of inflection, from the second derivative and its change of sign. Then the ends: what y does as x runs far to the right and far to the left, and any asymptotes. Finally, join the points up in a way that agrees with all of it.

The intercepts

Take $y=x^3-3x$y = x³ − 3x. Setting y = 0 gives $x^3-3x=x\left(x^2-3\right)=0$x³ − 3x = x(x² − 3) = 0, so x = 0 or $x^2=3$x² = 3. The x-intercepts are x = 0, $x=\sqrt{3}\approx 1.732$x = √3 ≈ 1.732 and $x=-\sqrt{3}\approx -1.732$x = −√3 ≈ −1.732.

Setting x = 0 gives y = 0, so the y-intercept is the origin too.

Check for symmetry as well. Replacing x by −x gives ${\left(-x\right)}^3-3\left(-x\right)=-\left(x^3-3x\right)$(−x)³ − 3(−x) = −(x³ − 3x), so the curve has rotational symmetry about the origin: whatever it does at x, it does upside down at −x.

Stationary points and inflection

$\frac{dy}{dx}=3x^2-3=3\left(x-1\right)\left(x+1\right)$dy/dx = 3x² − 3 = 3(x − 1)(x + 1), which is 0 at x = −1 and x = 1. The heights are −1 + 3 = 2 and 1 − 3 = −2.

The second derivative is 6x: −6 at x = −1, so (−1, 2) is a local maximum, and 6 at x = 1, so (1, −2) is a local minimum. The two agree with the symmetry.

6x is 0 at x = 0 and changes from negative to positive there, so the origin is a point of inflection: concave down to its left, concave up to its right. The gradient there is −3, so the curve crosses the origin falling.

The ends

For large x the $x^3$x³ term dominates: at x = 10, y = 1000 − 30 = 970. So $y\to \infty$y → ∞ as $x\to \infty$x → ∞, and by the symmetry $y\to -\infty$y → −∞ as $x\to -\infty$x → −∞. At x = 3 the curve is already at 18, and at x = −3 at −18.

Now join the pieces. Start at the bottom left, rising; cross the axis at $x=-\sqrt{3}$x = −√3; turn at the peak (−1, 2); fall through the origin with gradient −3; turn at the trough (1, −2); rise through $x=\sqrt{3}$x = √3 and off the top of the window.

The gold curve $y=x^3-3x$y = x³ − 3x, crossing the x-axis at $-\sqrt{3}$−√3, 0 and $\sqrt{3}$√3, with its peak (−1, 2) and trough (1, −2). The gold line y = −3x is the tangent at the point of inflection, the origin; both ends leave the window, rising on the right and falling on the left.

A curve that touches the axis

Take $y=x^3-6x^2+9x$y = x³ − 6x² + 9x. It factors as $x\left(x^2-6x+9\right)=x{\left(x-3\right)}^2$x(x² − 6x + 9) = x(x − 3)², so the x-intercepts are x = 0 and x = 3, and the y-intercept is 0.

The root at x = 3 is a double root: ${\left(x-3\right)}^2$(x − 3)² is never negative, so the curve touches the axis there and turns back up without crossing it. For x < 0 the factor x is negative, so the curve is below the axis; for x > 0 it is at or above it.

$\frac{dy}{dx}=3x^2-12x+9=3\left(x-1\right)\left(x-3\right)$dy/dx = 3x² − 12x + 9 = 3(x − 1)(x − 3), zero at x = 1 and x = 3. The second derivative 6x − 12 is −6 at x = 1 and 6 at x = 3, so (1, 4) is a local maximum, since 1 − 6 + 9 = 4, and (3, 0) a local minimum, on the axis, where the double root is.

6x − 12 is 0 at x = 2 and changes from negative to positive, so (2, 2) is a point of inflection, since 8 − 24 + 18 = 2. The gradient there is 12 − 24 + 9 = −3.

At the ends the $x^3$x³ term wins: the value is 10.125 at x = 4.5 and −6.125 at x = −0.5, so $y\to \infty$y → ∞ on the right and $y\to -\infty$y → −∞ on the left.

The gold curve $y=x{\left(x-3\right)}^2$y = x(x − 3)². It rises through the origin to the peak (1, 4), falls through the point of inflection (2, 2), touches the x-axis at its trough (3, 0), and rises off the top; on the left it falls off the bottom.

A curve with an asymptote

Take $y=\frac{x}{x^2+1}$y = x/(x² + 1). The denominator is never 0, so the curve is defined for every x. y = 0 only when x = 0, so the origin is the only intercept. Replacing x by −x changes the sign of y, so the curve is symmetric about the origin, as $x^3-3x$x³ − 3x is.

By the quotient rule, $\frac{dy}{dx}=\frac{1-x^2}{{\left(x^2+1\right)}^2}$dy/dx = (1 − x²)/(x² + 1)², which is 0 at x = 1 and x = −1. The second derivative is $\frac{d^{2}y}{d{x}^2}=\frac{2x\left(x^2-3\right)}{{\left(x^2+1\right)}^3}$d²y/dx² = 2x(x² − 3)/(x² + 1)³: $-\frac{1}{2}$−1/2 at x = 1, so $\left(1,\frac{1}{2}\right)$(1, 1/2) is a local maximum, and $\frac{1}{2}$1/2 at x = −1, so $\left(-1,-\frac{1}{2}\right)$(−1, −1/2) is a local minimum.

The second derivative is 0 at x = 0 and $x=\pm \sqrt{3}$x = ±√3, and changes sign at each: at x = −2, −1, 0.5 and 2 it is −0.032, 0.5, −1.408 and 0.032. So there are three points of inflection: the origin, and $\left(\pm \sqrt{3},\pm \frac{\sqrt{3}}{4}\right)$(±√3, ±√3/4), at heights about $\pm 0.433$±0.433.

Far out, the $x^2$x² in the denominator outgrows the x on top: $y=\frac{10}{101}\approx 0.099$y = 10/101 ≈ 0.099 at x = 10 and $\frac{100}{10001}\approx 0.01$100/10001 ≈ 0.01 at x = 100. So $y\to 0$y → 0 as $x\to \infty$x → ∞, from above, and as $x\to -\infty$x → −∞, from below. The x-axis is a horizontal asymptote. A curve that levels off is not always running off to infinity.

The gold curve $y=\frac{x}{x^2+1}$y = x/(x² + 1), with the vertical scale stretched: a maximum at $\left(1,\frac{1}{2}\right)$(1, 1/2), a minimum at $\left(-1,-\frac{1}{2}\right)$(−1, −1/2), points of inflection at the origin and at $x=\pm \sqrt{3}$x = ±√3, and both ends closing in on the x-axis.

The usual mistakes

Mixing up the intercepts. y = 0 gives the x-intercepts; x = 0 gives the y-intercept.

Setting y = 0 to find the stationary points. That gives where the curve crosses the axis; flat points come from $\frac{dy}{dx}=0$dy/dx = 0.

Drawing a crossing at a double root. $x{\left(x-3\right)}^2$x(x − 3)² touches the axis at x = 3 and turns back.

Leaving out the ends. A peak and a trough fit both $y=x^3-3x$y = x³ − 3x and $y=3x-x^3$y = 3x − x³, in opposite orders; the sign of the $x^3$x³ term decides which end rises.