Multiply outside → stretch the y's: y = a·f(x) stretches the graph vertically by factor a — every y-coordinate is multiplied by a.
(x-intercepts stay put, since 0 × a = 0.)
IB-style question — vertical stretch
y = f(x) passes through (2, 3).
Where does y = 4f(x) pass through (at x = 2)?
Step by step
- Multiply the y-coordinate by 4.
Final answer
(2, 12).
x-intercepts don't move: A vertical stretch leaves the x-intercepts fixed (their y is 0) but moves every other point away from the x-axis.
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Multiply inside → stretch the x's by 1/b: y = f(bx) stretches the graph horizontally by factor 1/b — the reciprocal.
So f(2x) squashes by ½ (factor 1/2), and f(x/2) stretches by 2.
Inside changes are counterintuitive again.
IB-style question — horizontal stretch
Describe the transformation y = f(x) → y = f(2x).
Step by step
- Inside factor 2 → horizontal stretch by 1/2.
Final answer
Horizontal stretch with scale factor ½ (the graph is squashed toward the y-axis).
Reciprocal factor: f(bx) stretches by 1/b, not b. f(3x) → factor 1/3 (squash); f(x/3) → factor 3 (stretch).
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Minus outside / inside flips an axis: y = −f(x) reflects the graph in the x-axis (flip the y-coordinates).
y = f(−x) reflects it in the y-axis (flip the x-coordinates).
y = −f(x)
- Reflect in the x-axis
- (x, y) → (x, −y)
- minus is OUTSIDE
y = f(−x)
- Reflect in the y-axis
- (x, y) → (−x, y)
- minus is INSIDE
Outside flips y, inside flips x: Same theme: outside the function changes y; inside changes x.
a·f(x) stretches vertically (and if a < 0, also flips in the x-axis); f(−x) flips in the y-axis. The base here is deliberately lop-sided so the two reflections look clearly different.
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Track one point through the change: Apply the rule to a point's coordinates: a·f(x) multiplies y by a; f(bx) divides x by b; −f(x) negates y; f(−x) negates x.
IB-style question — image under a reflection
The point (2, 5) lies on y = f(x).
Find its image on y = −f(x).
Step by step
- −f(x) negates the y-coordinate.
Final answer
(2, −5).
Combine carefully: For 2f(x) the point (2, 5) → (2, 10); for f(−x) it → (−2, 5).
Apply each rule to the correct coordinate.
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Reflect in the x-axis → the range flips: g(x) = −f(x) reflects f in the x-axis: every output y becomes −y.
So take the range of f and negate both ends — which swaps ≤ and ≥.
An open bound (<) stays open and a closed bound (≤) stays closed; the symbol just moves to the other end of the range.
g = −f flips f over the x-axis — the highest point becomes the lowest, and vice versa.
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IB-style question — reflect, then find the range
The graph of y = f(x), for x ≥ 0, has range −2 < f(x) ≤ 4.
The function g is defined by g(x) = −f(x) for x ≥ 0.
Find the range of g.
Step by step
- g = −f reflects f in the x-axis, so every output changes sign.
- The closed end: f's maximum 4 becomes g's minimum −4 (still reached).
- The open end: f's lower bound −2 becomes g's upper bound 2 (still not reached).
- Combine, smaller bound first.
Final answer
Range of g is −4 ≤ g(x) < 2.
Don't lose the open/closed end: The easy mark to drop is the inequality type.
Reflecting keeps each bound open or closed, but the direction flips (≤ ↔ ≥) because the values change sign: f's closed top (≤ 4) must become g's closed bottom (≥ −4).