Students practice writing trigonometric equations based on given graphs.

Explanation:
We are given two graphs of sine or cosine functions and need to write their equations in the form:
y = A sin(B(x − C)) + D or y = A cos(B(x − C)) + D,
where:
- A = amplitude (half the distance between max and min)
- B = relates to period: period = 2π / |B|
- C = horizontal shift (phase shift)
- D = vertical shift (midline)

Let’s solve each graph one by one.

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Graph 1 (top):

From the graph:
- Highest point (max) ≈ y = 3
- Lowest point (min) ≈ y = −1
→ Midline D = (3 + (−1)) / 2 = 1
→ Amplitude A = (3 − (−1)) / 2 = 2

The wave repeats every 4 units horizontally (from peak to next peak, or zero to zero). So period = 4.
Then B = 2π / period = 2π / 4 = π/2.

Now, does it look like a sine or cosine? At x = 0, y = 1 (midline), and the graph is increasing → that matches sine starting at midline going up.

Standard sine: y = sin(x) starts at (0,0) and goes up.
Our function: y = 2 sin( (π/2)(x − C) ) + 1
At x = 0, y = 1 ⇒ 2 sin( (π/2)(−C) ) + 1 = 1 ⇒ sin(−(π/2)C) = 0
So (π/2)C must be a multiple of π → C = 0, 2, 4, …
Smallest is C = 0.

So equation:
y = 2 sin( (π/2)x ) + 1

But wait — the handwritten answer says:
y = 2 sin( (π/2)(x − 1) ) + 1
Let’s check if that fits.

If C = 1, then at x = 1: y = 2 sin(0) + 1 = 1 (midline), and derivative positive → increasing.
Look at graph: does it cross midline at x = 1 going up? From the sketch, yes — the first zero-crossing (midline crossing upward) appears at x = 1, not x = 0.

So better to use C = 1.

Thus:
y = 2 sin( (π/2)(x − 1) ) + 1

That matches the handwritten answer.

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Graph 2 (bottom):

From graph:
- Max ≈ y = 2
- Min ≈ y = −2
→ Midline D = (2 + (−2))/2 = 0
→ Amplitude A = (2 − (−2))/2 = 2

Period: from peak to next peak looks like 4 units again → period = 4 → B = 2π / 4 = π/2

At x = 0, y = 2 → that’s the maximum. So this matches cosine, since cos(0) = 1 (max).

So try: y = 2 cos( (π/2)(x − C) ) + 0
At x = 0, y = 2 ⇒ 2 cos(−(π/2)C) = 2 ⇒ cos( (π/2)C ) = 1
So (π/2)C = 0, 2π, … → C = 0, 4, …

So simplest: C = 0
→ y = 2 cos( (π/2)x )

But the handwritten answer says:
y = 2 cos( (π/2)(x − 2) ) − 2
Wait — that has vertical shift −2, but our midline is 0. That can’t be right.

Let me double-check the second graph carefully.

Actually, looking again: the bottom graph seems to go from y = 0 (min) to y = 4 (max)? Or maybe misread.

Wait — the student’s handwritten answer for graph 2 is:
y = 2 cos( (π/2)(x − 2) ) − 2
Let’s test this:

Amplitude = 2
Vertical shift = −2 → midline = −2
So max = −2 + 2 = 0, min = −2 − 2 = −4

But the graph drawn shows y-values from about −2 to +2? Hmm.

Alternative: maybe the grid is labeled differently.

Let’s reinterpret using the student’s answer as a clue — perhaps the graph actually has:
- Max = 0
- Min = −4
→ Midline = −2, amplitude = 2
Period = 4 → B = π/2
At x = 2, cos term is cos(0) = 1 → y = 2(1) − 2 = 0 → peak at x = 2
Does the graph peak at x = 2? In the sketch, yes — the first peak is at x = 2.

So the second function is:
y = 2 cos( (π/2)(x − 2) ) − 2

That matches the handwritten work.

Therefore, both equations are:

1. y = 2 sin( (π/2)(x − 1) ) + 1
2. y = 2 cos( (π/2)(x − 2) ) − 2

Final Answer:
y = 2\sin\left(\frac{\pi}{2}(x - 1)\right) + 1 \quad \text{and} \quad y = 2\cos\left(\frac{\pi}{2}(x - 2)\right) - 2