Worksheet on transforming functions with a table of transformations and a graph to sketch and analyze functions.

Let’s solve each part step by step.

We are given a graph of a function — let’s call it f(x) — which is drawn on the coordinate plane. From the graph, we can see:

- The graph has a “V” shape with a peak at (2, 3), and goes down to (0, 1) and (4, 1), then continues downward linearly on both sides.
- Actually, looking more carefully: from left to right, it goes through (-2, -1), (0, 1), (2, 3), (4, 1), (6, -1). So it's symmetric about x=2? Wait — actually, from (0,1) to (2,3) to (4,1) — that’s an upside-down V? No, wait — if you connect those points, it looks like a triangle pointing up? But then beyond x=4, it goes to (6,-1), so maybe it’s two lines: one from left going up to (2,3), then down after that.

Actually, let’s define f(x) based on key points visible in the graph:

From the grid:
- At x = 0 → y = 1
- At x = 2 → y = 3
- At x = 4 → y = 1
- At x = 6 → y = -1
- At x = -2 → y = -1

So between x=0 and x=4, it’s a triangle with vertex at (2,3). Then outside that, it continues as straight lines.

But for transformations, we don’t need the full formula — we just need to apply the transformation rules to the graph or to key points.

The worksheet gives us transformation rules:

For example:
- Vertical translation: f(x) + d → shifts graph up/down by d
- Horizontal translation: f(x - c) → shifts right by c; f(x + c) → shifts left by c
- Reflection over x-axis: -f(x)
- Reflection over y-axis: f(-x)
- Stretch/compression: af(x) → vertical stretch by |a|; f(bx) → horizontal compression by |b|

Also, order matters: do stretches/reflections first, then translations.

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Part a: 2f(x - 5)

This means:
1. First, shift the graph of f(x) right by 5 units → because of (x - 5)
2. Then, vertically stretch by factor of 2 → multiply all y-values by 2

Let’s take key points from original f(x):

Original key points (from graph):
- (-2, -1)
- (0, 1)
- (2, 3)
- (4, 1)
- (6, -1)

Apply transformation: (x, y) → (x + 5, 2y) [because f(x - 5) shifts right 5, then ×2 vertically]

Wait — careful:

If we have f(x - 5), that replaces x with x - 5 → so to get same output, input must be 5 more → so point (a,b) on f becomes (a+5, b) on f(x-5)

Then 2f(x-5) → multiplies y by 2 → so (a+5, 2b)

So transform each point:

Original → After f(x-5) → After 2f(x-5)

(-2, -1) → (3, -1) → (3, -2)
(0, 1) → (5, 1) → (5, 2)
(2, 3) → (7, 3) → (7, 6)
(4, 1) → (9, 1) → (9, 2)
(6, -1) → (11, -1) → (11, -2)

Now, what about domain and range?

Original domain: from graph, seems to go from x=-∞ to ? But since it’s drawn only from x=-2 to x=6, but likely defined everywhere — assuming it’s piecewise linear extending infinitely.

But for this problem, since no restriction is given, we assume domain is all real numbers unless specified.

After shifting right by 5, domain still all real numbers.

Vertical stretch doesn’t change domain.

So domain of 2f(x-5): all real numbers

Range: original range — from graph, lowest y is -1, highest is 3? But wait — at x=-2 and x=6, y=-1, and at x=2, y=3. And since it extends linearly beyond, does it go lower?

Looking at slope:

From (2,3) to (4,1): slope = (1-3)/(4-2) = -2/2 = -1

From (4,1) to (6,-1): slope = (-1-1)/(6-4) = -2/2 = -1 → consistent

From (0,1) to (2,3): slope = (3-1)/(2-0) = 2/2 = 1

From (-2,-1) to (0,1): slope = (1 - (-1))/(0 - (-2)) = 2/2 = 1

So actually, the function is:

For x ≤ 2: line with slope 1 → y = x + 1? Check: at x=0, y=1 → yes; x=2, y=3 → yes; x=-2, y=-1 → yes.

For x ≥ 2: line with slope -1 → y = -x + 5? At x=2, y=3 → -2+5=3; x=4, y=1 → -4+5=1; x=6, y=-1 → -6+5=-1 → yes.

So f(x) =
{
x + 1, if x ≤ 2
-x + 5, if x > 2
}

Actually, at x=2, both give 3, so continuous.

So domain: all real numbers

Range: as x→ -∞, y→ -∞; as x→ ∞, y→ -∞; maximum at x=2, y=3.

So range of f(x): (-∞, 3]

Now, for 2f(x-5):

First, f(x-5) has same range as f(x), since just shifted horizontally → range still (-∞, 3]

Then multiplying by 2: range becomes (-∞, 6] (since 2*3=6, and negative values become more negative)

Domain: still all real numbers.

So for part a:

Graph: plot transformed points: (3,-2), (5,2), (7,6), (9,2), (11,-2) — and draw lines connecting them with slopes adjusted.

Since original had slope 1 for x≤2, now for f(x-5), the "kink" is at x=7 (since 2+5=7), and for x≤7, slope is still 1? Wait no:

Original f(x) for x≤2: slope 1 → f(x-5) for x-5 ≤2 → x≤7: slope 1

Then for x>7: slope -1

Then 2f(x-5): slopes become 2*1=2 for x≤7, and 2*(-1)= -2 for x>7

So graph: line with slope 2 until x=7, then slope -2 after.

Points: at x=7, y=6; x=5, y=2; x=3, y=-2; etc.

Domain: all real numbers

Range: (-∞, 6]

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Part b: -f(3x) + 3

Transformations:

Order: according to worksheet, do stretches/compressions and reflections first, then translations.

Here: f(3x) → horizontal compression by factor 3 (replace x with 3x → compress toward y-axis)

Then -f(3x) → reflect over x-axis

Then +3 → shift up by 3

So steps:

Start with f(x)

Step 1: f(3x) → horizontal compression by 3 → so x-coordinates divided by 3

Key points of f(x): (-2,-1), (0,1), (2,3), (4,1), (6,-1)

After f(3x): replace x with x/3? Wait — if g(x) = f(3x), then to get same value as f(a), set 3x = a → x = a/3

So point (a,b) on f → (a/3, b) on f(3x)

So:

(-2,-1) → (-2/3, -1)

(0,1) → (0,1)

(2,3) → (2/3, 3)

(4,1) → (4/3, 1)

(6,-1) → (2, -1)

Step 2: -f(3x) → reflect over x-axis → negate y-values

So:

(-2/3, -1) → (-2/3, 1)

(0,1) → (0,-1)

(2/3, 3) → (2/3, -3)

(4/3, 1) → (4/3, -1)

(2, -1) → (2, 1)

Step 3: +3 → add 3 to y-values

So:

(-2/3, 1) → (-2/3, 4)

(0,-1) → (0,2)

(2/3, -3) → (2/3, 0)

(4/3, -1) → (4/3, 2)

(2, 1) → (2, 4)

Now, domain and range.

Original domain: all real numbers

f(3x): still all real numbers (just compressed)

-f(3x)+3: still all real numbers → domain: all real numbers

Range: original range of f(x): (-∞, 3]

f(3x): same range, since horizontal compression doesn't affect range → (-∞, 3]

-f(3x): reflects over x-axis → range becomes [-3, ∞) (since max was 3, becomes min -3; min was -∞, becomes ∞)

Then +3: shift up → range becomes [0, )

Check with points: we have y-values: 4, 2, 0, 2, 4 → minimum is 0, and since original went to -∞, after reflection and shift, goes to ∞.

Yes.

So for part b:

Graph: plot points: (-2/3,4), (0,2), (2/3,0), (4/3,2), (2,4)

And since original had kink at x=2, now for f(3x), kink at x=2/3 (since 3x=2 → x=2/3)

Slopes: original for x<2: slope 1 → for f(3x), when x < 2/3, slope is 1 * 3? Wait no.

Recall: if f(x) has slope m, then f(kx) has slope m*k? Let's think.

Suppose f(x) = mx + b for some interval.

Then f(3x) = m*(3x) + b = 3m x + b → so slope is 3m.

In our case, for x ≤ 2, f(x) = x + 1 → slope 1

So f(3x) for 3x ≤ 2 → x ≤ 2/3: f(3x) = 3x + 1 → slope 3

Then for x > 2/3: f(3x) = -(3x) + 5 = -3x + 5 → slope -3

Then -f(3x): for x ≤ 2/3: - (3x + 1) = -3x -1 → slope -3

For x > 2/3: - (-3x + 5) = 3x -5 → slope 3

Then +3: for x ≤ 2/3: -3x -1 + 3 = -3x + 2 → slope -3

For x > 2/3: 3x -5 + 3 = 3x -2 → slope 3

So graph: line with slope -3 until x=2/3, then slope 3 after.

At x=2/3, y=0 (as calculated)

Points match: at x=0, y=2; x=2/3, y=0; x=4/3≈1.333, y=3*(4/3)-2=4-2=2; x=2, y=3*2-2=4 → yes.

Domain: all real numbers

Range: [0, ∞)

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Part c: ½f(x + 4) - 2

Note: it says “remember to factor first!” — probably meaning write it as ½ [f(x+4)] - 2, so transformations are clear.

Order: stretches/reflections first, then translations.

Here: f(x+4) → horizontal shift left by 4

Then ½ f(x+4) → vertical compression by 1/2

Then -2 → shift down by 2

But according to order, we should do the vertical scaling before the vertical translation, which we are doing.

Horizontal shift can be done anytime, but typically we do horizontal shifts after horizontal scalings, but here no horizontal scaling, so fine.

Steps:

Start with f(x)

Step 1: f(x+4) → shift left by 4 → so point (a,b) → (a-4, b)

Key points:

(-2,-1) → (-6, -1)

(0,1) → (-4, 1)

(2,3) → (-2, 3)

(4,1) → (0, 1)

(6,-1) → (2, -1)

Step 2: ½ f(x+4) → multiply y by 1/2

So:

(-6, -1) → (-6, -0.5)

(-4, 1) → (-4, 0.5)

(-2, 3) → (-2, 1.5)

(0, 1) → (0, 0.5)

(2, -1) → (2, -0.5)

Step 3: -2 → subtract 2 from y

So:

(-6, -0.5) → (-6, -2.5)

(-4, 0.5) → (-4, -1.5)

(-2, 1.5) → (-2, -0.5)

(0, 0.5) → (0, -1.5)

(2, -0.5) → (2, -2.5)

Now, domain and range.

Domain: original all real numbers → after shift and scale, still all real numbers.

Range: original range (-∞, 3]

f(x+4): same range → (-, 3]

½ f(x+4): multiply by 1/2 → range (-∞, 1.5]

Then -2: shift down → range (-∞, -0.5]

Check with points: max y is -0.5 (at x=-2), and goes to -∞.

Yes.

Graph: plot points: (-6,-2.5), (-4,-1.5), (-2,-0.5), (0,-1.5), (2,-2.5)

Kink at x=-2 (since original kink at x=2, shifted left by 4 → x=-2)

Slopes: original for x≤2: slope 1 → for f(x+4), when x+4 ≤2 → x≤-2: slope 1

Then ½ f(x+4): slope ½ * 1 = 0.5 for x≤-2

For x>-2: original slope -1 → f(x+4) slope -1 → ½ f(x+4) slope -0.5

Then -2: doesn't change slope.

So graph: line with slope 0.5 until x=-2, then slope -0.5 after.

At x=-2, y=-0.5

Points: at x=-4, y=0.5*(-4+4)? Wait, better use the expression.

f(x+4) for x≤-2: since x+4 ≤2, f(x+4) = (x+4) + 1 = x + 5

So ½ f(x+4) = ½(x + 5)

Then minus 2: ½(x + 5) - 2 = ½x + 2.5 - 2 = ½x + 0.5

At x=-4: ½*(-4) + 0.5 = -2 + 0.5 = -1.5 → matches

At x=-6: ½*(-6) + 0.5 = -3 + 0.5 = -2.5 → matches

For x > -2: f(x+4) = -(x+4) + 5 = -x -4 + 5 = -x +1

So ½ f(x+4) = ½(-x +1) = -½x + 0.5

Minus 2: -½x + 0.5 - 2 = -½x - 1.5

At x=0: -0 -1.5 = -1.5 → matches

At x=2: -½*2 -1.5 = -1 -1.5 = -2.5 → matches

Good.

Domain: all real numbers

Range: (-, -0.5]

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Now, summarize answers.

Final Answer:

a. Graph: Shift original graph right by 5, then vertically stretch by 2. Key points: (3,-2), (5,2), (7,6), (9,2), (11,-2). Domain: all real numbers. Range: (-, 6]

b. Graph: Horizontally compress by 3, reflect over x-axis, shift up by 3. Key points: (-2/3,4), (0,2), (2/3,0), (4/3,2), (2,4). Domain: all real numbers. Range: [0, ∞)

c. Graph: Shift left by 4, vertically compress by 1/2, shift down by 2. Key points: (-6,-2.5), (-4,-1.5), (-2,-0.5), (0,-1.5), (2,-2.5). Domain: all real numbers. Range: (-∞, -0.5]