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.
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.
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.