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Determining Trig Functions Given Their Graphs
This lesson shows how to work backward from a trig graph to its equation. You will learn to identify amplitude, period, midline, phase shift, and reflection, then combine them into a sine or cosine formula, with fully worked examples.
The four features you need to read off a graph
Any sine or cosine graph can be written in the general form \(y = A\sin(B(x - C)) + D\) or \(y = A\cos(B(x - C)) + D\). Instead of guessing, you read each of the four constants directly from the picture:
There is also a fifth thing to watch for that is not a separate number: whether the curve is reflected over its midline. If a curve that "should" start by rising instead starts by falling at the same point, the leading coefficient \(A\) is negative.
Step-by-step process
- Find the midline \(D\): average the maximum and minimum \(y\)-values, \(D = \dfrac{\max + \min}{2}\).
- Find the amplitude: \(A = \dfrac{\max - \min}{2}\). Keep it positive for now.
- Find the period: measure the horizontal distance for one complete cycle, then solve \(\)period\( = \dfrac{2\pi}{B}\) for \(B\).
- Decide between sine and cosine as your starting shape, based on where the curve crosses the midline or hits an extreme value.
- Find the phase shift \(C\) by comparing where your chosen parent function normally starts to where the actual graph starts.
- Check whether the curve is flipped upside down; if so, make \(A\) negative.
Worked example 1: a sine curve
Suppose a graph crosses its midline at \(y = 0\) while rising, reaches a maximum of \(2\), a minimum of \(-2\), and completes one full cycle over a length of \(2\pi\).
- Midline: \(D = \dfrac{2 + (-2)}{2} = 0\)
- Amplitude: \(A = \dfrac{2 - (-2)}{2} = 2\)
- Period is \(2\pi\), so \(B = \dfrac{2\pi}{2\pi} = 1\)
- The curve rises through the midline at \(x = 0\), exactly where an ordinary sine curve does, so no shift is needed and there is no reflection.
Putting the pieces together gives \(y = 2\sin(x)\).
Worked example 2: a reflected, shifted cosine curve
Now suppose a graph has a maximum of \(3\), a minimum of \(-1\), it reaches its lowest point at \(x = 0\), and one full cycle again spans \(2\pi\).
- Midline: \(D = \dfrac{3 + (-1)}{2} = 1\)
- Amplitude: \(A = \dfrac{3 - (-1)}{2} = 2\)
- Period is \(2\pi\), so \(B = 1\)
- An ordinary cosine curve starts at a maximum at \(x = 0\); this graph starts at a minimum instead, which means it is reflected, so \(A\) is negative rather than positive.
Combining these gives \(y = -2\cos(x) + 1\). Notice that the same wave could also be described using a shifted sine equation; a phase shift can turn a cosine curve into an equivalent sine curve and vice versa, so there is often more than one correct answer.
Common pitfalls
A few mistakes come up again and again when students determine a trig equation from a graph:
- Forgetting to subtract the midline before reading the amplitude, which produces a value that is too large.
- Measuring the period from a peak to the next peak but reading the x-axis in degrees when the intended answer is in radians (or vice versa).
- Assuming every graph must be written with sine; cosine, and for other shapes, tangent, secant, cosecant, or cotangent, are equally valid starting functions.
- Missing a reflection because only the amplitude was checked and the direction of the curve at the midline crossing was ignored.
If the curve you are given has vertical asymptotes rather than smooth peaks and valleys, it is not a sine or cosine graph at all. In that case check whether it matches a tangent graph, a secant, cosecant, or cotangent shape instead, since those functions are read differently and are covered in their own lessons.