MISSION · Algebra
Shifting and Stretching Graphs
One graph, and every version of it you'll ever need
LEVEL 1 · EXPLORE
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LEVEL 2 · LEARN
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The idea in one sentence
Change a function’s equation in one of four standard ways and its graph moves in a predictable way — so you only ever have to know the basic shape.
The four moves
Starting from any \(f(x)\):
| Change | Effect |
|---|---|
| \(f(x) + k\) | Shifts up by \(k\) |
| \(f(x + h)\) | Shifts left by \(h\) |
| \(a \cdot f(x)\) | Stretches vertically by \(a\) |
| \(f(bx)\) | Squashes horizontally by \(b\) |
Negative values reverse each: \(-f(x)\) flips the graph upside down, and \(f(-x)\) flips it left-to-right.
Outside is obvious, inside is backwards
Changes outside the function affect the output, which is vertical, and they do what you expect. \(f(x) + 3\) adds 3 to every height, so the graph rises by 3.
Changes inside the brackets affect the input, which is horizontal, and they do the opposite of what you expect. \(f(x + 3)\) shifts the graph left by 3, not right.
That trips up everyone once, and there’s a clean reason. To get the output the graph used to give at \(x = 0\), you now need \(x = -3\), because \(-3 + 3 = 0\). Every feature arrives three units earlier, so the whole picture slides left.
The same inversion applies to stretches. \(f(2x)\) reaches each value at half the input, so the graph is squashed to half its width — not doubled.
Reading a function in transformed form
Most curriculum equations are written to make this easy. Vertex form of a quadratic:
\[y = a(x - h)^2 + k\]is the basic parabola \(y = x^2\), shifted \(h\) right and \(k\) up, and stretched by \(a\). The vertex is at \((h, k)\), which you can read off without any algebra. The minus sign in \((x - h)\) is there precisely so that \(h\) reads as a rightward shift rather than backwards.
The same reading works everywhere. \(y = 2\sin(x) + 1\) is a sine wave twice as tall, raised by 1. \(y = \sqrt{x - 4}\) is the square root curve starting at 4 instead of 0.
Order matters
Applying a stretch and then a shift is not the same as shifting and then stretching. \(2f(x) + 3\) stretches first, then raises; \(2(f(x) + 3)\) raises first, then stretches the raised graph — which moves it further.
Work outward from the function, in the order the operations would be evaluated.
Why it’s worth it
Because it collapses an enormous amount of memorisation. There is one parabola, one sine wave, one exponential curve, one absolute-value V — and everything else on the syllabus is one of those with a few transformations applied. Learn four shapes and four moves instead of forty graphs.
Common mix-ups
- Reading \(f(x + 3)\) as a shift right. Inside the brackets runs backwards.
- Doubling the width for \(f(2x)\). It halves.
- Forgetting that \(-f(x)\) and \(f(-x)\) flip about different axes.
- Applying transformations in the wrong order when several are combined.
Check yourself
Describe how \(y = -(x - 2)^2 + 5\) relates to \(y = x^2\), and give its vertex.
Shifted 2 right, flipped upside down, and raised 5. The vertex is \((2, 5)\), and it’s a maximum rather than a minimum.
LEVEL 3 · REFLECT
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