Transformation of graphs — translation, reflection, scaling, composition

Question

Starting from the graph of $y = x^2$ , describe the transformations needed to get $y = -2(x-3)^2 + 5$ . Sketch the final graph.

(CBSE 11 & JEE Main — functions chapter)

Solution — Step by Step

Compare $y = -2(x-3)^2 + 5$ with $y = x^2$ :

$(x-3)$ : horizontal shift right by 3 units
Factor of $2$ : vertical stretch by factor 2
Negative sign: reflection about the x-axis
$+5$ : vertical shift up by 5 units

The standard order is: horizontal shift → stretch/reflect → vertical shift.

Start with $y = x^2$ (standard parabola, vertex at origin)
Replace $x$ with $(x-3)$ : shift right 3 → $y = (x-3)^2$ , vertex at $(3, 0)$
Multiply by $-2$ : stretch vertically by 2 and reflect → $y = -2(x-3)^2$ , vertex at $(3, 0)$ , opens downward
Add $5$ : shift up 5 → $y = -2(x-3)^2 + 5$ , vertex at $(3, 5)$

Vertex: $(3, 5)$ — this is the maximum point (parabola opens downward)
Axis of symmetry: $x = 3$
Direction: Opens downward (coefficient is negative)
Narrower than $y = x^2$ (stretched by factor 2)
y-intercept: Set $x = 0$ : $y = -2(9) + 5 = -13$

Why This Works

Each algebraic modification to a function corresponds to a geometric transformation of its graph. The rules are consistent across all functions — not just parabolas.

graph TD
    A["Graph Transformation"] --> B{"What changed in equation?"}
    B -->|"f(x) → f(x-h)"| C["Shift RIGHT by h<br/>(opposite sign!)"]
    B -->|"f(x) → f(x+h)"| D["Shift LEFT by h"]
    B -->|"f(x) → f(x) + k"| E["Shift UP by k"]
    B -->|"f(x) → f(x) - k"| F["Shift DOWN by k"]
    B -->|"f(x) → af(x), a>1"| G["Vertical STRETCH<br/>by factor a"]
    B -->|"f(x) → af(x), 0<a<1"| H["Vertical COMPRESS"]
    B -->|"f(x) → -f(x)"| I["Reflect about<br/>x-axis"]
    B -->|"f(x) → f(-x)"| J["Reflect about<br/>y-axis"]

The tricky part is that horizontal transformations are “opposite” to what you’d expect: $f(x-3)$ shifts RIGHT (not left), and $f(2x)$ compresses horizontally (not stretches). Vertical transformations are intuitive: $2f(x)$ stretches up, $f(x) + 5$ shifts up.

Alternative Method — Plot Key Points

Take 3-4 points on $y = x^2$ : $(-1, 1)$ , $(0, 0)$ , $(1, 1)$ , $(2, 4)$ . Apply each transformation to the coordinates:

Original $(0, 0)$ → shift right 3 → $(3, 0)$ → multiply y by $-2$ → $(3, 0)$ → shift up 5 → $(3, 5)$

Original $(1, 1)$ → $(4, 1)$ → $(4, -2)$ → $(4, 3)$

Plot the transformed points and connect smoothly.

For JEE: the order matters when combining transformations. For $y = af(bx + c) + d$ , apply in this order: (1) horizontal shift by $-c/b$ , (2) horizontal scale by $1/b$ , (3) vertical scale by $a$ , (4) vertical shift by $d$ . Getting the order wrong is the number one source of errors.

Common Mistake

The most common error: $f(x-3)$ shifts left instead of right. Students think “minus means left.” But substituting $x = 3$ into $f(x-3)$ gives $f(0)$ — the original y-intercept now appears at $x = 3$ , which is a shift to the RIGHT. Always test with a specific point if you’re unsure.

Question

Solution — Step by Step

Why This Works

Alternative Method — Plot Key Points

Common Mistake

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