Math 30-1 · Transformations
Moving graphs around
Every transformation question is really about one thing: where does each point go? Learn the mapping rule and you can move any graph — no matter what function it is.
Translations and stretches
The idea: in y = a·f(b(x − h)) + k, anything outside the function changes y and does what it looks like. Anything inside, with x, changes x and does the opposite of what it looks like.
| In the equation | What happens |
|---|---|
| y = f(x) + k | up k (down if k is negative) |
| y = f(x − h) | right h (left if written x + h) |
| y = a·f(x) | vertical stretch by a factor of |a| |
| y = f(bx) | horizontal stretch by a factor of 1/|b| |
The mapping rule
For y = a·f(b(x − h)) + k, every point moves like this:
(x, y) → (x/b + h, a·y + k)
Stretch first, then translate. For y = 3f(x − 1) + 2, the point (4, −6) becomes (4 + 1, 3 × (−6) + 2) = (5, −16).
Watch out: y = f(2x − 6) is not a shift of 6. Factor out the 2 first: f(2(x − 3)), so the shift is 3 right.
Reflections and inverses
The idea: a negative outside flips y; a negative inside flips x; swapping x and y reflects the whole graph in the line y = x — that is the inverse.
| Equation | Reflection in | (3, 4) becomes |
|---|---|---|
| y = −f(x) | the x-axis | (3, −4) |
| y = f(−x) | the y-axis | (−3, 4) |
| x = f(y), the inverse | the line y = x | (4, 3) |
Finding an inverse algebraically
For f(x) = 2x − 6: write y = 2x − 6, swap to x = 2y − 6, and solve for y: f⁻¹(x) = (x + 6)/2. Notice it undoes the original steps in reverse — add 6, then divide by 2.
When the inverse is not a function
Reflect y = x² in y = x and you get a sideways parabola, which fails the vertical line test. Restricting the original domain to x ≥ 0 keeps only half, and then the inverse, y = √x, is a function.
Watch out: f⁻¹(x) does not mean 1/f(x). The −1 is a label for "inverse", not an exponent.
Combining functions
The idea: (f + g)(x), (f − g)(x), (f·g)(x) and (f/g)(x) just mean do that operation to the two outputs. The only trap is the domain.
With f(x) = x² − 4 and g(x) = x − 2:
- (f + g)(x) = x² + x − 6
- (f − g)(x) = x² − x − 2
- (f·g)(x) = (x² − 4)(x − 2) = x³ − 2x² − 4x + 8
- (f/g)(x) = (x − 2)(x + 2)/(x − 2) = x + 2, x ≠ 2
The domain of any combination is where both original functions are defined. For a quotient, you also remove every x that makes g(x) = 0 — and that restriction stays even after the factor cancels.
Watch out: simplifying f/g to x + 2 and forgetting x ≠ 2. The graph has a hole there.
Composition
The idea: f(g(x)) means put x into g, then put that answer into f. Always work from the inside out.
Worked example
f(x) = x + 1 and g(x) = 2x. Find f(g(3)) and g(f(3)).
- f(g(3)): g(3) = 6, then f(6) = 7.
- g(f(3)): f(3) = 4, then g(4) = 8.
Different orders, different answers. Composition is usually not the same both ways.
As expressions
With f(x) = x² and g(x) = x + 3: f(g(x)) = (x + 3)² but g(f(x)) = x² + 3.
Domain of a composition
f(g(x)) is only defined where g(x) is defined and its output is allowed into f. With f(x) = √x and g(x) = x − 5, f(g(x)) = √(x − 5), so x ≥ 5.
Working backwards
To write h(x) = (2x − 1)² as a composition, find the steps: first 2x − 1, then square. The first step is the inside function: g(x) = 2x − 1 and f(x) = x².
Watch out: reading f(g(x)) left to right and doing f first. The function closest to x always goes first.