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The Tangent Function

Lesson ~11 min read 8 MCQs

In simple terms: In simple terms, the tangent function describes the slope of a line at any angle on a circle, creating a unique repeating graph with vertical breaks called asymptotes.

Why this matters

Imagine you're operating a lighthouse on the coast of Maine on a foggy night. Your light rotates, casting a powerful beam. Let's say the coastline is a perfectly straight north-south line. As your beam sweeps from pointing east out to sea, toward the north, it hits the coastline farther and farther away.

When the beam is just slightly angled north, it hits the coast nearby. As you turn it more, it hits the coast miles away. And what happens when your beam points exactly parallel to the coast? It never hits it at all—the light travels on forever into the fog.

That relationship between the angle of your light and the position where it hits the coast is exactly what the tangent function describes. We're going to explore how this idea of slope and angles creates a totally different kind of trigonometric graph, one with dramatic breaks and an infinite reach.

The tangent function relates an angle to the slope of a line, illustrating how it can become infinitely large.

Concept overview

flowchart TD
    A[Start with Parent Function y = tan(θ)] --> B{Identify a, b, c, d in<br>y = a tan(b(θ - c)) + d};
    B --> C[Calculate New Period<br>P = π / |b|];
    B --> D[Find Phase Shift<br>Shift right by c];
    B --> E[Find Vertical Stretch<br>Multiply y-values by a];
    B --> F[Find Vertical Shift<br>Shift up by d];
    C --> G[Determine New Asymptotes<br>Apply shift 'c' to<br>compressed asymptotes];
    D --> G;
    E --> H[Plot Key Points];
    F --> H;
    G --> I[Sketch one cycle of the<br>transformed curve];
    H --> I;
    I --> J[Repeat the pattern<br>every P units];

Core explanation

Hello! I’m Saavi, and I’m so glad you’re here. Today we’re tackling the tangent function. It might seem a little strange compared to its cousins, sine and cosine, but once you understand its origin story, it all clicks into place.

From Unit Circle to Slope

Remember how for any angle θ on the unit circle, the point (x, y) on the circle gives us cos(θ) = x and sin(θ) = y? The tangent function asks a different question: what is the slope of the terminal ray for that angle θ?

Slope is "rise over run," which in this case is y / x. Since y = sin(θ) and x = cos(θ), we get the most important identity for tangent:

Visualizing tan(θ) as the slope (y/x) of the terminal ray on the unit circle.

tan(θ) = sin(θ) / cos(θ)

This simple ratio explains everything about the tangent graph.

Think about it:

  • At θ = 0, the point is (1, 0). The slope y/x is 0/1 = 0. So, tan(0) = 0.
  • At θ = π/4 (or 45°), the point is (√2/2, √2/2). The slope y/x is (√2/2) / (√2/2) = 1. So, tan(π/4) = 1.
  • At θ = π/2 (or 90°), the point is (0, 1). The slope y/x is 1/0. We can't divide by zero! The slope is undefined.

This "undefined" result is the key. It’s not an error; it’s a feature. It tells us that at θ = π/2, the graph of the tangent function has a vertical asymptote. This is a vertical line that the graph approaches but never, ever touches.

The tangent function's vertical asymptotes occur where cos(θ) = 0, making the slope undefined.

Key Features of the Parent Graph: y = tan(θ)

Let's sketch out the parent function based on what we know.

  1. 1
    Period
    How often does the pattern repeat? For sine and cosine, it's 2π. But for tangent, it's just π. Why? The slope of the ray at angle θ is the exact same as the slope of the ray at θ + π (180° later), because they form a single straight line. The pattern of slopes repeats every half-circle, not every full circle.
  2. 2
    Asymptotes
    We saw an asymptote at θ = π/2 because cos(π/2) = 0. Where else will this happen? Anywhere cosine is zero. This occurs at π/2, 3π/2, 5π/2, and also in the negative direction at -π/2, -3π/2, etc. We can generalize this: asymptotes occur at θ = π/2 + kπ, where k is any integer.
  3. 3
    Shape and Concavity
    Between any two consecutive asymptotes, the tangent function is always increasing. It starts from negative infinity, crosses the x-axis at the midpoint, and heads up toward positive infinity.
    • The graph is concave down on the left half of the interval (e.g., from -π/2 to 0).
    • It passes through a point of inflection at the midpoint (e.g., at (0,0)).
    • It becomes concave up on the right half (e.g., from 0 to π/2).

This gives us that classic repeating "S" curve, separated by asymptotes.

Transforming the Tangent Function

Just like with other functions, we can transform the parent graph y = tan(θ) using the general form:

y = a · tan(b(θ - c)) + d

(Note: Your book might use θ + c, which means the shift is -c. The logic is the same, just watch the signs! We'll stick with the θ - c form here, where c is the direct shift value.)

Let's break down what each parameter does:

  • d: Vertical Shift This is the most straightforward. The d value slides the entire graph up (if d > 0) or down (if d < 0). The points of inflection, which used to be on the x-axis, are now on the line y = d.

  • a: Vertical Stretch & Reflection The a value stretches or compresses the graph vertically. If |a| > 1, the graph gets steeper, faster. If 0 < |a| < 1, it gets flatter. For example, the point that used to be at (π/4, 1) is now at (π/4, a). If a is negative, the graph is reflected across the x-axis; instead of increasing, it will decrease between its asymptotes.

  • c: Phase Shift (Horizontal Shift) The c value slides the graph left or right. In the form tan(θ - c), the graph shifts right by c. In the form tan(θ + c), it shifts left by c. This is where most students slip up. The shift is opposite the sign you see. This shift moves everything—the points of inflection and the asymptotes.

  • b: Horizontal Stretch & Period Change This is the other big point of confusion. The b value affects the period. It compresses or stretches the graph horizontally. The new period is calculated as: New Period = π / |b|

    This is critical: The original period is π, not 2π. Using 2π is a very common mistake carried over from sine and cosine. If |b| > 1, the period gets shorter (the graph is compressed). If 0 < |b| < 1, the period gets longer (the graph is stretched).

By identifying a, b, c, and d, you can take the simple parent function and systematically apply these changes to find the new graph.

The parent function y = tan(θ) showing its period of π and asymptotes at odd multiples of π/2.

Worked examples

Let's walk through a couple of examples together. The key is to be methodical and identify each transformation one by one.

Example 1

Vertical Stretch and Shift

Problem: Analyze and graph the function g(θ) = 2 tan(θ) - 1.

Step 1: Identify the parameters. Comparing g(θ) = 2 tan(θ) - 1 to the general form y = a tan(b(θ - c)) + d, we can see:

  • a = 2 (Vertical stretch by a factor of 2)
  • b = 1 (No change in period, since Period = π/|1| = π)
  • c = 0 (No phase shift)
  • d = -1 (Vertical shift down by 1)

Step 2: Analyze the effect on key features.

  • Center Line
    The vertical shift d = -1 moves the points of inflection from y = 0 down to y = -1. So, the point that was at (0, 0) is now at (0, -1).
  • Asymptotes
    Since b = 1 and c = 0, the asymptotes don't move. They remain at θ = π/2 + kπ.
  • Stretched Points
    The vertical stretch a = 2 affects the "quarter-way" points. The parent function has points at (π/4, 1) and (-π/4, -1).
    • The stretch changes (π/4, 1) to (π/4, 2).
    • The vertical shift then moves this point down 1, to (π/4, 1).
    • Let's do the other one: Stretch changes (-π/4, -1) to (-π/4, -2). The shift moves it to (-π/4, -3).

Step 3: Sketch the graph.

  1. Draw your vertical asymptotes at ...-π/2, π/2, 3π/2....
  2. Plot the new center point at (0, -1).
  3. Plot the stretched and shifted points at (π/4, 1) and (-π/4, -3).
  4. Draw a smooth, increasing curve through these points that approaches the asymptotes. Repeat the pattern for other periods.
Example 2

Period Change and Phase Shift

Problem: Find the period, phase shift, and asymptotes for f(θ) = tan(2(θ - π/4)).

Step 1: Identify the parameters.

  • a = 1 (No vertical stretch)
  • b = 2 (This will change the period)
  • c = π/4 (This is a phase shift to the right)
  • d = 0 (No vertical shift)

Step 2: Calculate the new period. The formula is Period = π / |b|. Here, b = 2, so the new period is π / 2. The function will complete a full cycle in half the horizontal space as the parent function.

Step 3: Determine the phase shift. The c value is π/4. Since the form is (θ - c), this is a shift of π/4 to the right. This means the center of a cycle, which used to be at θ = 0, is now at θ = π/4.

Step 4: Find the new asymptotes. This is where students often get tripped up. You have to apply the transformations to the parent asymptotes. The parent asymptotes are at θ = -π/2 and θ = π/2 (for one cycle).

  1. 1
    Horizontal Compression
    The b=2 value compresses everything by a factor of 1/2. So, the asymptotes would move to θ = (-π/2) * (1/2) = -π/4 and θ = (π/2) * (1/2) = π/4.
  2. 2
    Phase Shift
    Now, apply the shift of π/4 to the right.
    • New left asymptote: -π/4 + π/4 = 0.
    • New right asymptote: π/4 + π/4 = π/2.

So, one full cycle of this new graph is squeezed between θ = 0 and θ = π/2. The center of this cycle is at θ = π/4, which matches our phase shift. The general formula for the asymptotes is θ = 0 + k(π/2) and θ = π/2 + k(π/2), or more simply, θ = kπ/2 for any integer k.

Transforming y = tan(θ) to g(θ) = 2 tan(θ) - 1, showing vertical stretch and shift.

Try it yourself

Ready to try one on your own? Take your time and work through the steps we just practiced.

Problem 1: Consider the function f(θ) = -tan(θ + π/2) + 1.

  • What is the period?
  • Describe the horizontal and vertical shifts.
  • Is the function increasing or decreasing between its asymptotes? Why?
  • Where is the new center point of the cycle that is usually centered at (0,0)?

Problem 2: A transformed tangent function has a period of 4π, a vertical asymptote at θ = 2π, and passes through the point (0, 3). What could be an equation for this function?

The graph of f(θ) = -tan(θ + π/2) + 1, demonstrating reflection, phase shift, and vertical shift.
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