Transformation of graphs — translation, reflection, scaling, composition

Question

Given $f(x) = x^2$ , sketch the graphs of: (a) $f(x) + 3$ , (b) $f(x - 2)$ , (c) $-f(x)$ , (d) $f(-x)$ , (e) $2f(x)$ , (f) $f(2x)$ . Describe each transformation in words.

Solution — Step by Step

(a) $f(x) + 3 = x^2 + 3$ : Shifts the parabola up by 3 units. Every point moves 3 units upward.

(b) $f(x-2) = (x-2)^2$ : Shifts the parabola right by 2 units. The vertex moves from $(0,0)$ to $(2,0)$ .

Key insight: adding to $x$ inside the function moves the graph LEFT (counterintuitive). Subtracting from $x$ moves it RIGHT.

(c) $-f(x) = -x^2$ : Reflects across the x-axis. The parabola opens downward instead of upward.

(d) $f(-x) = (-x)^2 = x^2$ : Reflects across the y-axis. For $x^2$ , this happens to give the same graph (since $x^2$ is an even function), but for most functions the reflection is visible.

(e) $2f(x) = 2x^2$ : Vertical stretch by factor 2. The parabola becomes narrower (steeper). Every y-coordinate doubles.

(f) $f(2x) = (2x)^2 = 4x^2$ : Horizontal compression by factor 2. The graph squeezes toward the y-axis. At $x = 1$ , the function value is now $f(2) = 4$ instead of $f(1) = 1$ .

Why This Works

graph TD
    A["Graph Transformation Type"] --> B["Changes OUTSIDE f: affect y-values"]
    A --> C["Changes INSIDE f: affect x-values and reverse direction"]
    B --> D["f x + k: shift UP by k"]
    B --> E["-f x: reflect in x-axis"]
    B --> F["a times f x: vertical stretch by a"]
    C --> G["f x - h: shift RIGHT by h"]
    C --> H["f -x: reflect in y-axis"]
    C --> I["f bx: horizontal compress by b"]

The golden rule: transformations that happen outside $f$ (affecting the output) work as expected — add means up, multiply means stretch vertically. Transformations that happen inside $f$ (affecting the input) work in the opposite direction — adding to $x$ shifts left (not right), multiplying $x$ compresses (not stretches).

Why the reversal for horizontal transformations? Think about it: $f(x - 2) = 0$ when $x - 2 = 0$ , i.e., $x = 2$ . The root has moved to the right. You need a larger $x$ to get the same input to $f$ , hence the graph shifts right.

Alternative Method

For JEE, when you need to sketch a transformed graph quickly, use these 3 anchor points: find where the function equals 0, where it is maximum/minimum, and one additional point. Transform these anchor points and connect smoothly.

For composed transformations like $y = 2f(3x - 1) + 4$ , apply in this order:

Horizontal transformations first (inside-out): factor out the coefficient of $x$ , giving $f(3(x - 1/3))$ — shift right by $1/3$ , then compress horizontally by factor 3
Vertical transformations next (outside-in): stretch by 2, then shift up 4

Common Mistake

Getting the direction of horizontal shifts backwards. $f(x + 2)$ shifts the graph LEFT by 2, not right. $f(x - 3)$ shifts RIGHT by 3, not left. The sign is opposite to what you might expect. If you keep getting confused, substitute a test point: for $f(x - 3)$ , the original point at $x = 0$ now needs $x = 3$ to give $f(0)$ . So the point has moved right.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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