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Tangents parallel to a line: equal slopes and distinct lines

2020-01-31 huynhvietkhanh Comments: 0

Find every tangent with a given direction, exclude a coincident line when distinct parallelism is required, and count distinct equations.

High schoolFunctions, Graphs & Tangents

Content updated: October 22, 2026

Two nonvertical lines with equal slopes have the same direction. To be distinct parallel lines, their constant terms must differ. After finding contacts using derivatives, check whether a resulting tangent coincides with the given line.

Objectives: find every tangent with the specified direction, filter coincident lines, and count distinct lines rather than just contact points.

Prerequisites: slopes, line equations, and tangents with a given slope.

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

1. The parallelism convention

For $d:y=kx+b$ and $\Delta:y=kx+\beta$, the lines have the same direction. They are distinct and parallel exactly when $\beta\ne b$; if $\beta=b$, they coincide. Here a request for “parallel” tangents means distinct parallel lines, so a coincident line is excluded.

A tangent at $x_0$ with finite derivative is nonvertical. A given vertical line, such as $x=a$, has no finite slope, so do not use $f'(x)=k$ with a real number $k$. Vertical tangents, when present, require separate treatment.

Contact points considered here are interior points of the domain at which $f$ is differentiable with a finite derivative.

2. Find the direction, then check the position

For a nonvertical line $y=kx+b$, solve $f'(x_0)=k$ on the domains of the function and derivative. Calculate $y_0=f(x_0)$ and form the tangent

$$\Delta:y=k(x-x_0)+y_0=kx+\beta,\qquad\beta=y_0-kx_0.$$

Accept it when $\beta\ne b$. If distinct inputs yield the same $\beta$, they give the same line, counted once. Retain domain conditions throughout; an input outside the domain cannot be a contact point.

If a problem explicitly means “the same direction, allowing coincidence,” list the coincident line separately. Stating the convention clarifies an answer when the given line is already tangent to the graph.

These steps are necessary and sufficient under the stated differentiability assumptions. Every required tangent has derivative $k$ at its contact point. Conversely, each allowed solution of $f'(x_0)=k$ produces a tangent with slope $k$, and the intercept test ensures that it is distinct from the given line.

3. Worked examples

Example 1. A parabola and a line of slope 2

Let $f(x)=x^2-2x-1$ and $d:y=2x-1$. We have $f'(x)=2x-2$. Equating to 2 gives $x_0=2$, with height $f(2)=-1$. The tangent is $y=2(x-2)-1=2x-5$. Its constant term $-5$ differs from $-1$ for $d$, so the lines are distinct and parallel.

Example 2. Two tangents to a cubic

Let $f(x)=x^3+3x-1$ and $d:y=6x-1$. The equation $3x^2+3=6$ gives $x=1,-1$, with $f(1)=3$ and $f(-1)=-5$. The lines are $y=6x-3$ and $y=6x+1$, both different from $d$ because their constant terms differ from $-1$. Accept both.

Example 3. Equal slope but excluded

Let $f(x)=x^2$ and $d:y=2x-1$. The condition $f'(x)=2$ gives $x=1$, with height 1. The tangent is $y=2(x-1)+1=2x-1$, precisely $d$. Under the distinct-parallel convention, no tangent meets the requirement; if only the same direction is requested, this one line is included.

−1 0 1 2 3 4 x −4 −2 0 2 4 6 8 y M (2, −1) Two distinct parallel lines f(x) = x² − 2x − 1 d: y = 2x − 1 Δ: y = 2x − 5
Illustration of Example 1. The dashed line d and tangent Δ have the same slope 2 and different constant terms.

4. Practice

  1. Find the tangents to $y=x^3$ that are distinct and parallel to $d:y=3x+2$.
  2. Find the tangents to $y=(x^2-1)^2$ parallel to $d:y=2$. How many contact points and distinct lines are there?
  3. Can the slope of $x=1$ be assigned a real number and then used in $f'(x)=k$?
Hints and answers
  1. The contact inputs are $x=\pm1$. The same-direction tangents are $y=3x-2$ and $y=3x+2$. The latter coincides with $d$ and is excluded; only $y=3x-2$ remains.
  2. The derivative equation $4x(x^2-1)=0$ gives contacts at $-1,0,1$. The outer points give the same line $y=0$, while the middle point gives $y=1$. Both differ from $d$, so there are three contact points but two distinct lines.
  3. No. A vertical line has no finite slope; this method applies only to $y=kx+b$.

5. Common mistakes

Equal slopes alone do not ensure distinct parallelism. Compare constant terms after finding a tangent. Do not count a line twice when it has multiple contacts. Do not assign a real slope to a vertical line.

6. Further learning

The tangent at a point (English lesson) · Tangents with a given slope (English lesson) · Tangents perpendicular to a given line (English lesson) · Tangents passing through a point (English lesson).

Original lesson video (Vietnamese-language video)

Original lesson video (Vietnamese-language video)

Original lesson video (Vietnamese-language video)

Additional examples and practice

The following material extends the examples and practice above.

Additional worked example. Three contacts, two lines

For $f(x)=(x^2-1)^2$, find tangents parallel to $d:y=3$. The slope is $0$, and

$$f^{\prime}(x)=4x(x^2-1)=0\quad\Longleftrightarrow\quad x\in\{-1,0,1\}.$$

The corresponding ordinates are $0,1,0$. The contact points at $x=-1$ and $x=1$ have the same tangent $y=0$; the contact at $x=0$ gives $y=1$. Both are distinct from $y=3$. There are three contact points but only two distinct tangent lines.

Further practice

  1. Find tangents to $y=x^2$ parallel to $y=4x+1$.
  2. Find tangents to $y=x^2$ parallel to $y=2x-1$, using the convention that parallel lines are distinct.
  3. Find tangents to $y=x^3$ parallel to $y=-x$.
Show answers and hints
  1. $2x_0=4$ gives $x_0=2$, $y_0=4$, and $y=4x-4$.
  2. The only slope candidate has $x_0=1$, giving $y=2x-1$. It coincides with the given line, so there is no distinct parallel tangent.
  3. The slope equation $3x_0^2=-1$ has no real solution. There is no such tangent.

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