Transforming Functions Worksheet - Fill Online, Printable ... - Free Printable

Explanation:
We are given three functions to graph and analyze:
a. $ f(x) = -2|x - 5| $
b. $ f(x) = -x^2 + 7 $
c. $ f(x) = \frac{3}{4}|x + 4| - 2 $

For each, we need to:
- Sketch the graph (describe key features),
- State the domain and range,
- Identify the vertex (for absolute value) or turning point (for quadratic),
- Note any transformations from the parent function.

Let’s go one by one.

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a. $ f(x) = -2|x - 5| $

Parent function: $ y = |x| $ — V-shaped, vertex at (0, 0), opens upward.

Transformations:
- Horizontal shift right 5 → $ |x - 5| $ → vertex moves to $ x = 5 $
- Vertical stretch by factor 2 → steepens the V
- Reflection over x-axis (negative sign) → opens downward

So vertex is at $ (5, 0) $, and since it opens downward and is stretched, the maximum value is 0.

Domain: all real numbers → $ (-\infty, \infty) $
Range: since max is 0 and goes down forever → $ (-\infty, 0] $

Sketch: upside-down V with tip at (5, 0), slopes of ±2 (but negative because reflected): left side slope = +2? Wait—let’s check:

For $ x < 5 $: $ |x - 5| = -(x - 5) = 5 - x $, so
$ f(x) = -2(5 - x) = -10 + 2x $ → slope = +2
For $ x > 5 $: $ |x - 5| = x - 5 $, so
$ f(x) = -2(x - 5) = -2x + 10 $ → slope = -2
So yes: left side rises with slope +2, right side falls with slope -2 — classic upside-down V.

✔ Vertex: $ (5, 0) $
✔ Domain: $ (-\infty, \infty) $
✔ Range: $ (-\infty, 0] $

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b. $ f(x) = -x^2 + 7 $

Parent: $ y = x^2 $ — parabola opening up, vertex at (0, 0)

Transformations:
- Reflection over x-axis (negative sign) → opens down
- Vertical shift up 7 → vertex moves to (0, 7)

So vertex: $ (0, 7) $, opens downward.

Domain: all real numbers → $ (-\infty, \infty) $
Range: max is 7, goes down forever → $ (-\infty, 7] $

Sketch: upside-down parabola, symmetric about y-axis, highest point at (0, 7)

✔ Vertex: $ (0, 7) $
✔ Domain: $ (-\infty, \infty) $
✔ Range: $ (-\infty, 7] $

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c. $ f(x) = \frac{3}{4}|x + 4| - 2 $

Parent: $ y = |x| $

Transformations:
- Horizontal shift left 4 → $ |x + 4| $ → vertex at $ x = -4 $
- Vertical compression by factor $ \frac{3}{4} $ (makes it flatter)
- Vertical shift down 2 → subtract 2

So vertex: $ (-4, -2) $

Since coefficient $ \frac{3}{4} > 0 $, opens upward.

Domain: all real numbers → $ (-\infty, \infty) $
Range: minimum value is -2 (at vertex), goes up forever → $ [-2, \infty) $

Check:
For $ x < -4 $: $ |x + 4| = -(x + 4) = -x - 4 $, so
$ f(x) = \frac{3}{4}(-x - 4) - 2 = -\frac{3}{4}x - 3 - 2 = -\frac{3}{4}x - 5 $ → slope = $ -\frac{3}{4} $
For $ x > -4 $: $ |x + 4| = x + 4 $, so
$ f(x) = \frac{3}{4}(x + 4) - 2 = \frac{3}{4}x + 3 - 2 = \frac{3}{4}x + 1 $ → slope = $ +\frac{3}{4} $

So V-shape, vertex at (-4, -2), gentle slopes.

✔ Vertex: $ (-4, -2) $
✔ Domain: $ (-\infty, \infty) $
✔ Range: $ [-2, \infty) $

Now, the worksheet asks for:
- Sketch (we described)
- Domain & Range
- Vertex/turning point
- Also in part 1, “Know the form” column: match each to standard form.

Standard forms:
- Absolute value: $ y = a|x - h| + k $, vertex at $ (h, k) $
- Quadratic: $ y = a(x - h)^2 + k $, vertex at $ (h, k) $

Let’s write each in that form:

a. $ f(x) = -2|x - 5| = -2|x - 5| + 0 $ → $ a = -2, h = 5, k = 0 $

b. $ f(x) = -x^2 + 7 = -1(x - 0)^2 + 7 $ → $ a = -1, h = 0, k = 7 $

c. $ f(x) = \frac{3}{4}|x + 4| - 2 = \frac{3}{4}|x - (-4)| + (-2) $ → $ a = \frac{3}{4}, h = -4, k = -2 $

All correct.

Now, the problem says:
“Given the graph of $ f(x) $, sketch the graph of the following functions, and state the domain and range for each.”
Then lists a, b, c — which we just did.

Also, there’s a note: “Note: A: Domain; r: Range; 1–3, 4–6, 7–9” — likely referencing rows, but not needed for answer.

So final answers per part:

a. Domain: $ (-\infty, \infty) $, Range: $ (-\infty, 0] $, Vertex: $ (5, 0) $
b. Domain: $ (-\infty, \infty) $, Range: $ (-\infty, 7] $, Vertex: $ (0, 7) $
c. Domain: $ (-\infty, \infty) $, Range: $ [-2, \infty) $, Vertex: $ (-4, -2) $

Make sure ranges are written correctly with brackets.

Double-check:
- a: max at 0 → includes 0 → $ (-\infty, 0] $ ✔
- b: max at 7 → includes 7 → $ (-\infty, 7] $ ✔
- c: min at -2 → includes -2 → $ [-2, \infty) $ ✔

All good.

Final Answer:
a. Domain: $ (-\infty, \infty) $, Range: $ (-\infty, 0] $, Vertex: $ (5, 0) $
b. Domain: $ (-\infty, \infty) $, Range: $ (-\infty, 7] $, Vertex: $ (0, 7) $
c. Domain: $ (-\infty, \infty) $, Range: $ [-2, \infty) $, Vertex: $ (-4, -2) $