Horizontal and Vertical Tangents

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Key Ideas Easy Medium Hard

\(\dfrac{dy}{dx} = 0\) horizontal, denominator 0 vertical

only when the other part is not 0

What to Remember

Solve for dy/dx before you sort the tangents

Differentiate the relation, then write

\(\dfrac{dy}{dx} = \dfrac{\text{numerator}}{\text{denominator}}\)

Horizontal when the top is zero and the bottom is not

A horizontal tangent has slope 0. Set the numerator equal to 0, keep the point on the curve, and check that the denominator is not 0 there.

Vertical when the bottom is zero and the top is not

A vertical tangent has an undefined slope of that kind. Set the denominator equal to 0, keep the point on the curve, and check that the numerator is not 0.

The curve increases where the slope is positive

After you have \(dy/dx\), solve \(dy/dx > 0\) on the curve. Both numerator and denominator carry signs.

Level 1

Easy

A circle or a sideways parabola. Name the horizontal and vertical points.

Example

The circle \((x - 2)^2 + y^2 = 4\) can be written \(x^2 + y^2 - 4x = 0\). Find \(dy/dx\).

\(2x + 2y\,\dfrac{dy}{dx} - 4 = 0\)

\(\dfrac{dy}{dx} = \dfrac{2 - x}{y}\)

Answer: \(\dfrac{2 - x}{y}\)

On \(x^2 + y^2 - 4x = 0\), find the points with a horizontal tangent.

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\(\dfrac{dy}{dx} = \dfrac{2 - x}{y}\)

Horizontal when \(2 - x = 0\) and \(y \neq 0\).

\(x = 2\)

On the circle, \(y^2 = 4\), so \(y = \pm 2\).

Answer: \((2, 2)\) and \((2, -2)\)

On the same circle, find the points with a vertical tangent.

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Vertical when \(y = 0\) and the numerator is not 0.

\(x^2 - 4x = 0\), so \(x = 0\) or \(x = 4\).

At both points the numerator \(2 - x\) is not 0.

Answer: \((0, 0)\) and \((4, 0)\)

For \(y^2 = x\), find \(dy/dx\) and the vertical tangent.

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\(2y\,\dfrac{dy}{dx} = 1\)

\(\dfrac{dy}{dx} = \dfrac{1}{2y}\)

The numerator is never 0, so there is no horizontal tangent.

The denominator is 0 at \((0, 0)\), which is on the curve. Vertical tangent there.

Answer: vertical tangent at \((0, 0)\)

On \(xy = 4\), is there a horizontal or vertical tangent?

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\(\dfrac{dy}{dx} = -\dfrac{y}{x} = -\dfrac{4}{x^2}\)

The slope is negative at every point on the curve.

It is never 0 and never undefined on the curve.

Answer: neither; the curve is always decreasing

On \(x^2 - y^2 = 1\), where are the vertical tangents?

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\(\dfrac{dy}{dx} = \dfrac{x}{y}\)

Vertical when \(y = 0\) and \(x \neq 0\).

On the curve, \(x^2 = 1\), so the points are \((1, 0)\) and \((-1, 0)\).

Answer: \((1, 0)\) and \((-1, 0)\)

Level 2

Medium

Sort every point where the numerator or the denominator is zero.

Example

For \(x^2 + xy + y^2 = 3\), find \(dy/dx\).

\(2x + y + x\,\dfrac{dy}{dx} + 2y\,\dfrac{dy}{dx} = 0\)

\(\dfrac{dy}{dx}(x + 2y) = -(2x + y)\)

\(\dfrac{dy}{dx} = -\dfrac{2x + y}{x + 2y}\)

Answer: \(-\dfrac{2x + y}{x + 2y}\)

Find the points on \(x^2 + xy + y^2 = 3\) where the tangent is horizontal.

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Numerator 0: \(y = -2x\).

Denominator at those points: \(x + 2(-2x) = -3x\), which is not 0 when \(x \neq 0\).

Substitute \(y = -2x\) into the curve.

\(x^2 + x(-2x) + 4x^2 = 3\)

\(3x^2 = 3\)

\(x = \pm 1\), so the points are \((1, -2)\) and \((-1, 2)\).

Answer: \((1, -2)\) and \((-1, 2)\)

Find the points on \(x^2 + xy + y^2 = 3\) where the tangent is vertical.

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Denominator 0: \(y = -\dfrac{x}{2}\).

Substitute into the curve.

\(x^2 + x\left(-\dfrac{x}{2}\right) + \dfrac{x^2}{4} = 3\)

\(\dfrac{3}{4}x^2 = 3\)

\(x^2 = 4\)

Points: \((2, -1)\) and \((-2, 1)\).

The numerator \(2x + y\) is 3 at \((2, -1)\) and \(-3\) at \((-2, 1)\), so it is not 0.

Answer: \((2, -1)\) and \((-2, 1)\)

Write the tangent line to \(x^2 + xy + y^2 = 3\) at \((1, -2)\).

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That point has a horizontal tangent.

\(y = -2\)

Answer: \(y = -2\)

On the circle \(x^2 + y^2 - 4x = 0\), where is the slope positive?

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\(\dfrac{dy}{dx} = \dfrac{2 - x}{y}\)

The slope is positive when \(2 - x\) and \(y\) have the same sign.

Upper half and \(x < 2\), or lower half and \(x > 2\).

Answer: upper half left of \(x = 2\), and lower half right of \(x = 2\)

Which points on \(x^2 + y^2 = 25\) have a horizontal tangent?

A. \((0, 5)\) and \((0, -5)\)

B. \((5, 0)\) and \((-5, 0)\)

C. \((3, 4)\) only

D. no points

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\(\dfrac{dy}{dx} = -\dfrac{x}{y}\)

Horizontal when \(x = 0\) and \(y \neq 0\).

The points on the circle are \((0, 5)\) and \((0, -5)\).

Answer: A. \((0, 5)\) and \((0, -5)\)

Level 3

Hard

A cusp needs a second look, and a slope at a point that is neither horizontal nor vertical.

Example

The curve \(y^2 = x^3\) has \(\dfrac{dy}{dx} = \dfrac{3x^2}{2y}\). At \((0, 0)\), both the numerator and the denominator are 0. What is the tangent there?

The slope formula does not decide the tangent when both parts are 0.

For \(x \ge 0\), \(y = x^{3/2}\) or \(y = -x^{3/2}\).

\(\dfrac{d}{dx} x^{3/2} = \dfrac{3}{2} x^{1/2}\)

As \(x \to 0^+\), that derivative approaches 0. The lower half approaches 0 as well.

The tangent line at the origin is horizontal: \(y = 0\).

Answer: horizontal tangent \(y = 0\)

Find the slope of \(x^2 + xy + y^2 = 3\) at \((0, \sqrt{3})\).

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\(\dfrac{dy}{dx} = -\dfrac{2x + y}{x + 2y}\)

Substitute \((0, \sqrt{3})\).

\(-\dfrac{\sqrt{3}}{2\sqrt{3}}\)

\(-\dfrac{1}{2}\)

Answer: \(-\dfrac{1}{2}\)

On \(x^2 + xy + y^2 = 3\), explain why \((1, -2)\) is a horizontal tangent and not a vertical one.

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At \((1, -2)\), the numerator \(2x + y = 0\).

The denominator \(x + 2y = 1 - 4 = -3 \neq 0\).

Slope 0, so the tangent is horizontal.

Answer: horizontal, because the numerator is 0 and the denominator is not

On \(xy = 4\), show that \(dy/dx\) is always negative.

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\(\dfrac{dy}{dx} = -\dfrac{4}{x^2}\)

\(x^2 > 0\) at every point of the curve, since \(x \neq 0\).

The slope is negative everywhere on the curve. The relation is decreasing on \((0, \infty)\) and on \((-\infty, 0)\).

Answer: always negative

For \(y^2 = x\), which description of \((0, 0)\) is correct?

A. vertical tangent

B. horizontal tangent

C. no tangent, because both parts of the slope are 0

D. slope 1

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\(\dfrac{dy}{dx} = \dfrac{1}{2y}\)

At \((0, 0)\) the denominator is 0 and the numerator is 1.

The tangent is vertical.

Answer: A. vertical tangent

On the circle \(x^2 + y^2 - 4x = 0\), find \(dy/dx\) at \((2, 2)\) and name the tangent line.

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\(\dfrac{dy}{dx} = \dfrac{2 - x}{y}\)

Substitute \((2, 2)\).

\(0\)

Horizontal tangent \(y = 2\).

Answer: \(y = 2\)

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