Identify horizontal or vertical tangents

A loop with flat and steep points

The curve \(y^2 = x^2(x + 1)\) has a loop on the left. The orange line is the tangent, and its slope is \(\tfrac{dy}{dx} = \tfrac{3x^2 + 2x}{2y}\).

Drag the point around the loop. Short horizontal bars mark the top and the bottom of the loop, where the tangent is horizontal. A short vertical bar marks the far left, where the tangent is vertical.

A horizontal line has slope 0, so the numerator \(3x^2 + 2x\) is 0 at the top and the bottom. A vertical line has no slope, so the denominator \(2y\) is 0 at the far left.

At the origin, where the curve crosses itself, both readouts are 0 and the formula gives no slope.

Numerator and denominator

The slope is a fraction, \(\tfrac{dy}{dx} = \tfrac{N}{D}\), with a numerator \(N\) and a denominator \(D\).

The tangent is horizontal where \(N = 0\) and \(D \neq 0\), because the slope is 0 there. It is vertical where \(D = 0\) and \(N \neq 0\), because the slope is undefined there.

The point must also lie on the curve. A solution of \(N = 0\) or \(D = 0\) that is off the curve gives no tangent.

Notation

1. Differentiating \(y^2 = x^3 + x^2\) gives \(2y\,y' = 3x^2 + 2x\). Dividing by \(2y\) writes the slope as one fraction \(\tfrac{N}{D}\).

   \[\cy{\frac{dy}{dx}} = \frac{\cy{N}}{\cy{D}} = \frac{3x^2 + 2x}{2y}\]

2. For a horizontal tangent, solve \(N = 0\). Substitute each solution into the equation of the curve to find \(y\), and confirm that \(D \neq 0\) there.

   \[\begin{aligned} x(3x + 2) &= 0 \\ x &= 0 \text{ or } -\tfrac23 \\ x = -\tfrac23: \quad y^2 &= \tfrac49 \cdot \tfrac13 \\ y &= \pm\sqrt{\tfrac{4}{27}} \end{aligned}\]

   At \(x = 0\) the curve gives \(y = 0\), so \(D = 0\) too and the test fails. At \(x = -\tfrac23\), \(D = 2y\) is not 0.

3. For a vertical tangent, solve \(D = 0\). Substitute into the equation of the curve to find \(x\), and confirm that \(N \neq 0\) there.

   \[\begin{aligned} y &= 0 \\ x^2(x + 1) &= 0 \\ x &= -1 \text{ or } 0 \end{aligned}\]

   At \((-1, 0)\), \(N = 3 - 2 = 1\), which is not 0, so the tangent is vertical. At \((0, 0)\), \(N = 0\) too.

When it fails

Each test needs the other part of the fraction to be nonzero. At the origin, both \(3x^2 + 2x\) and \(2y\) are 0, so the fraction is \(\tfrac00\) and gives no slope.

The curve crosses itself there and has two tangent lines, shown in red, with slopes 1 and \(-1\). Neither is horizontal or vertical.

Example 1: An ellipse

Find the points where the curve \(x^2 + 4y^2 - 4x = 0\) has a horizontal or a vertical tangent.

1. Differentiate both sides with respect to \(x\), and solve for \(\cy{\tfrac{dy}{dx}}\).

   \[\begin{aligned} 2x + 8y\,\cy{y'} - 4 &= 0 \\ \cy{y'} &= \frac{2 - x}{4y} \end{aligned}\]

2. For a horizontal tangent, the numerator is 0, so \(x = 2\). Substitute into the equation of the curve to find \(y\). The denominator \(4y\) is not 0 at these points, so the tangent is horizontal at \((2, 1)\) and \((2, -1)\).

   \[\begin{aligned} 4 + 4y^2 - 8 &= 0 \\ y &= \pm1 \end{aligned}\]

3. For a vertical tangent, the denominator is 0, so \(y = 0\). Substitute into the equation of the curve to find \(x\). The numerator \(2 - x\) is not 0 at these points, so the tangent is vertical at \((0, 0)\) and \((4, 0)\).

   \[\begin{aligned} x^2 - 4x &= 0 \\ x &= 0 \text{ or } 4 \end{aligned}\]

Example 2: No horizontal tangent

Find the points where the curve \(y^3 + 2xy = 16\) has a horizontal or a vertical tangent.

1. Differentiate both sides with respect to \(x\), and solve for \(\cy{\tfrac{dy}{dx}}\). The term \(2xy\) needs the product rule.

   \[\begin{aligned} &3y^2\,\cy{y'} + 2y + 2x\,\cy{y'} = 0 \\ &\cy{y'} = -\frac{2y}{3y^2 + 2x} \end{aligned}\]

2. A tempting wrong move: The numerator is 0 when \(y = 0\), so the tangent is horizontal wherever \(y = 0\).

   \[-2y = 0 \;\Rightarrow\; y = 0\]

   Substitute \(y = 0\) into the equation of the curve. This gives \(0 = 16\), which is false. The curve never meets the red line \(y = 0\).

3. The curve has no point with \(y = 0\), so it has no horizontal tangent.

4. For a vertical tangent, the denominator is 0, so \(x = -\tfrac{3y^2}{2}\). Substitute this into the equation of the curve. The numerator \(-2y = 4\) is not 0 there, so the tangent is vertical at \((-6, -2)\).

   \[\begin{aligned} y^3 - 3y^3 &= 16 \\ y^3 &= -8 \\ y &= -2, \quad x = -6 \end{aligned}\]

Practice

Find the points where the curve \(y^2 = x^3 - 3x\) has a horizontal tangent.

Set the numerator equal to 0. Test each solution in the equation of the curve, because the figure has no curve over one of them.

Questions

Questions 1 and 2 use the curve \(x^2 - xy + y^2 = 3\), where \(\tfrac{dy}{dx} = \tfrac{y - 2x}{2y - x}\). Short horizontal bars mark points with a horizontal tangent, and short vertical bars mark points with a vertical tangent.

1. At the point \((1, 2)\), the tangent is horizontal. Which statement explains this?

   • The numerator \(y - 2x\) is 0, and the denominator \(2y - x\) is 3.
   • The denominator \(2y - x\) is 0.
   • Both the numerator and the denominator are 0.
   • The slope is 2, because \(y = 2\).
   Answer

   The numerator \(y - 2x\) is 0, and the denominator \(2y - x\) is 3.

2. How many points of the curve have a vertical tangent, according to the figure?

   • 2
   • 1
   • 4
   • 0
   Answer

   2

3. Which points of the curve \(x^2 + y^2 - 2x = 3\) have a vertical tangent?

   • \((-1, 0)\) and \((3, 0)\)
   • \((1, 2)\) and \((1, -2)\)
   • \((1, 0)\)
   • \((0, \sqrt3)\) and \((0, -\sqrt3)\)
   Answer

   \((-1, 0)\) and \((3, 0)\)

Summary

You set the numerator of \(\tfrac{dy}{dx}\) to 0 for a horizontal tangent, and the denominator to 0 for a vertical tangent.

If the other part of the fraction is not 0, this works because a horizontal line has slope 0 and a vertical line has none.

You use it for a curve given by an equation in \(x\) and \(y\), and you keep only the solutions on the curve.

This is one of four skills in topic 5.12 of Which & Why: AP Calculus.