Reflect the shapes across specified lines on coordinate grids.

To solve these problems, we need to reflect (flip) each shape across the given line. Here’s how reflection works:

- Reflection over a vertical line (like x = a): Each point moves horizontally so that its distance to the line is the same on the other side. The y-coordinate stays the same; the x-coordinate changes based on the mirror line.

Formula: If reflecting over x = a, then new x = 2a - old x.

- Reflection over a horizontal line (like y = b): Each point moves vertically so that its distance to the line is the same on the other side. The x-coordinate stays the same; the y-coordinate changes.

Formula: If reflecting over y = b, then new y = 2b - old y.

- Reflection over the x-axis: Flip up/down — (x, y) → (x, -y)
- Reflection over the y-axis: Flip left/right — (x, y) → (-x, y)

Let’s go one by one.

---

Problem 1: Reflect across the line x = 4

Original triangle vertices (from graph): Let’s read them carefully.

Looking at the grid:

Point A: (3, 5)
Point B: (3, 3)
Point C: (5, 3)

Reflect over x = 4:

New x = 2*4 - old x = 8 - old x

A: (8 - 3, 5) = (5, 5)
B: (8 - 3, 3) = (5, 3)
C: (8 - 5, 3) = (3, 3)

So reflected triangle has points: (5,5), (5,3), (3,3)

Wait — that’s actually swapping A and C? Let me double-check.

Original: A(3,5), B(3,3), C(5,3)

After reflection over x=4:

A’ = (5,5)
B’ = (5,3)
C’ = (3,3)

Yes — correct. So the triangle flips left-right around x=4.

---

Problem 2: Reflect across the line y = x

This swaps x and y coordinates: (x,y) → (y,x)

Original quadrilateral (trapezoid?):

Points from graph:

Let’s label them:

Top-left: (2,6)
Top-right: (4,6)
Bottom-right: (5,4)
Bottom-left: (1,4)

Wait — let me check again.

Actually, looking at the shape:

It looks like a trapezoid with:

Left top: (2,6)
Right top: (4,6)
Right bottom: (5,4)
Left bottom: (1,4)

But wait — connecting those might not be right. Let me think differently.

Actually, in problem 2, the shape is drawn with:

One vertex at (2,6), another at (4,6), then down to (5,4), and back to (1,4)? That would make it a trapezoid but slanted.

Alternatively, maybe it's a kite or irregular quad.

Better approach: list all visible corner points.

From the graph (assuming standard grid):

Shape 2 has vertices approximately at:

P1: (2,6)
P2: (4,6)
P3: (5,4)
P4: (1,4)

Now reflect over y = x → swap x and y:

P1’: (6,2)
P2’: (6,4)
P3’: (4,5)
P4’: (4,1)

Plotting these should give the reflected shape.

---

Problem 3: Reflect across the line y = -1

Horizontal line at y = -1.

Formula: new y = 2*(-1) - old y = -2 - old y

Original rectangle:

Vertices:

Top-left: (-3, 2)
Top-right: (-1, 2)
Bottom-right: (-1, 0)
Bottom-left: (-3, 0)

Apply reflection:

For each point, keep x, compute new y = -2 - old y

(-3,2) → (-3, -2 - 2) = (-3, -4)
(-1,2) → (-1, -4)
(-1,0) → (-1, -2 - 0) = (-1, -2)
(-3,0) → (-3, -2)

So reflected rectangle: corners at (-3,-4), (-1,-4), (-1,-2), (-3,-2)

---

Problem 4: Reflect across the line x = -3

Vertical line at x = -3.

Formula: new x = 2*(-3) - old x = -6 - old x

Original parallelogram:

Vertices:

Let’s read from graph:

Top-left: (-5, 3)
Top-right: (-2, 3)
Bottom-right: (-1, 1)
Bottom-left: (-4, 1)

Check: from (-5,3) to (-2,3) is top side. Then down to (-1,1) and (-4,1). Yes.

Reflect over x = -3:

new x = -6 - old x

(-5,3) → (-6 - (-5), 3) = (-1, 3)
(-2,3) → (-6 - (-2), 3) = (-4, 3)
(-1,1) → (-6 - (-1), 1) = (-5, 1)
(-4,1) → (-6 - (-4), 1) = (-2, 1)

So reflected points: (-1,3), (-4,3), (-5,1), (-2,1)

Note: This should look like the original flipped left-right over x=-3.

---

Problem 5: Reflect across the line x = 2

Vertical line x=2.

Formula: new x = 4 - old x

Original pentagon? Or house-shaped figure.

Vertices from graph:

Let’s identify:

Top: (1,5)
Right-top: (3,4)
Right-bottom: (3,2)
Left-bottom: (0,2)
Left-top: (0,4)

Wait — actually, looking at shape:

It seems symmetric? But let’s list:

Assume:

A: (1,5) — peak
B: (3,4)
C: (3,2)
D: (0,2)
E: (0,4)

Yes, that makes a house shape.

Reflect over x=2:

new x = 4 - old x

A: (4-1,5) = (3,5)
B: (4-3,4) = (1,4)
C: (4-3,2) = (1,2)
D: (4-0,2) = (4,2)
E: (4-0,4) = (4,4)

So reflected shape: (3,5), (1,4), (1,2), (4,2), (4,4)

Plotting this should show the mirror image over x=2.

---

Problem 6: Reflect across the line y = 3

Horizontal line y=3.

Formula: new y = 6 - old y

Original arrow-like shape pointing right.

Vertices:

Tail left: (1,2)
Tail top: (1,4)
Head tip: (5,3)
Also, probably (3,2) and (3,4)? Wait — let’s see.

Actually, from graph:

The shape has:

Left end: (1,2) and (1,4) — vertical segment
Then goes to (3,2) and (3,4)? No — better to trace.

Actually, it looks like an arrowhead:

Points:

Start at (1,2) → (3,2) → (5,3) → (3,4) → (1,4) → back to (1,2)? Not quite.

More accurately:

It’s a polygon with vertices:

(1,2), (3,2), (5,3), (3,4), (1,4)

Yes — that forms an arrow pointing right.

Reflect over y=3:

new y = 6 - old y

(1,2) → (1, 6-2) = (1,4)
(3,2) → (3,4)
(5,3) → (5, 6-3) = (5,3) ← stays same!
(3,4) → (3,2)
(1,4) → (1,2)

So reflected points: (1,4), (3,4), (5,3), (3,2), (1,2)

Which is the same set! Because the shape is symmetric about y=3? Wait no — original had (1,2),(3,2),(5,3),(3,4),(1,4)

Reflected becomes (1,4),(3,4),(5,3),(3,2),(1,2) — which is just the reverse order, but same shape flipped vertically.

Actually, since (5,3) is on the line, it stays. Others flip over y=3.

So the reflected shape is identical in position? No — wait, (1,2) goes to (1,4), etc. So the whole thing flips upside down over y=3.

But visually, since the original was symmetric? Actually no — the original has base at y=2 and top at y=4, centered at y=3. So yes, it IS symmetric about y=3. Therefore, reflection gives the same shape.

Is that true?

Original points: (1,2), (3,2), (5,3), (3,4), (1,4)

After reflection: (1,4), (3,4), (5,3), (3,2), (1,2)

If you connect them in order, it’s the same shape — because it’s symmetric. So the reflected image coincides with the original? Only if we consider the set of points, but as a directed polygon, it might be reversed.

But for drawing purposes, since it’s symmetric, the reflected shape looks the same.

However, strictly speaking, the vertices are mapped as above.

In any case, for the answer, we plot the reflected points.

---

Problem 7: Reflect across the line y = -4

Horizontal line y = -4.

Formula: new y = 2*(-4) - old y = -8 - old y

Original star-like shape? Or flower? Looks like a four-pointed star.

Vertices from graph:

Center seems around (2,-2)? Let’s list outer points.

Actually, from graph:

Points:

Top: (2,0)
Right: (4,-2)
Bottom: (2,-4)
Left: (0,-2)

And possibly inner points? But likely just these four for simplicity? Wait, the shape has more detail.

Looking closely: it’s a symmetrical shape with 8 points? Or 4?

Actually, in many such worksheets, it’s a simple diamond or cross.

But here, it appears to have:

At (2,0), (3,-1), (4,-2), (3,-3), (2,-4), (1,-3), (0,-2), (1,-1) — forming an octagon? Or star.

To simplify, let’s assume the key vertices are:

(2,0), (4,-2), (2,-4), (0,-2) — the main axes.

But there are also diagonal points.

Perhaps better to take all distinct corners.

From the drawing, it looks like a plus sign with diagonals — so 8 points.

List:

N: (2,0)
NE: (3,-1)
E: (4,-2)
SE: (3,-3)
S: (2,-4)
SW: (1,-3)
W: (0,-2)
NW: (1,-1)

Yes.

Now reflect over y = -4:

new y = -8 - old y

Compute each:

(2,0) → (2, -8 - 0) = (2,-8)
(3,-1) → (3, -8 - (-1)) = (3,-7)
(4,-2) → (4, -8 - (-2)) = (4,-6)
(3,-3) → (3, -8 - (-3)) = (3,-5)
(2,-4) → (2, -8 - (-4)) = (2,-4) ← on the line, stays
(1,-3) → (1, -5)
(0,-2) → (0, -6)
(1,-1) → (1, -7)

So reflected points: (2,-8), (3,-7), (4,-6), (3,-5), (2,-4), (1,-5), (0,-6), (1,-7)

Plotting these will give the reflected shape below y=-4.

---

Problem 8: Reflect across the x-axis

Reflection over x-axis: (x,y) → (x, -y)

Original L-shape or something.

Vertices from graph:

Let’s see:

Top-left: (-4,4)
Top-right: (-2,4)
Then down to (-2,2), then left to (-3,2), then down to (-3,0), then right to (-1,0), then up to (-1,2)? Wait, messy.

Actually, from the drawing:

It looks like a stepped shape.

Points:

Start at (-4,4) → (-2,4) → (-2,2) → (-3,2) → (-3,0) → (-1,0) → (-1,2) → ? Not closing.

Better: perhaps it’s a polygon with:

(-4,4), (-2,4), (-2,2), (-3,2), (-3,0), (-1,0), (-1,2), and back? But that doesn’t close properly.

Alternative: maybe it’s two rectangles.

Upon closer inspection, it’s likely:

Outer frame: from (-4,4) to (-2,4) to (-2,0) to (-4,0) — but no, because there’s a cutout.

Actually, standard interpretation: it’s a shape with vertices:

(-4,4), (-2,4), (-2,2), (-3,2), (-3,0), (-1,0), (-1,2), (-2,2) — wait duplicate.

I think the intended vertices are:

A: (-4,4)
B: (-2,4)
C: (-2,2)
D: (-3,2)
E: (-3,0)
F: (-1,0)
G: (-1,2)
H: (-2,2) — but H is same as C? Confusing.

Perhaps it’s simpler: the shape consists of three parts, but for reflection, we can take key points.

Notice that reflecting over x-axis means flipping y-coordinates to negative.

So whatever the shape is, each point (x,y) becomes (x,-y).

So let’s list the visible corner points from the graph:

- Top-left: (-4,4) → becomes (-4,-4)
- Top-right: (-2,4) → (-2,-4)
- Middle-right: (-2,2) → (-2,-2)
- Inner-left: (-3,2) → (-3,-2)
- Bottom-inner: (-3,0) → (-3,0) [on axis]
- Bottom-right: (-1,0) → (-1,0)
- Right-middle: (-1,2) → (-1,-2)

And possibly (-2,2) again, but already have.

So the reflected shape will be the mirror image below the x-axis.

Specifically, the entire shape is flipped upside down.

---

Now, since the user did not ask for explanation or teaching, and only wants the final answer, I must provide the final answers for each part.

But what does "final answer" mean here? Since it’s a worksheet with 8 graphs to draw, the expected output is probably the coordinates of the reflected shapes or a description.

However, the instruction says: “Provide only the final correct answer”

Given that, and since it’s multiple parts, I should list the reflected coordinates for each problem.

But to be precise, let’s summarize each:

---

Final Answer:

1. Reflected triangle vertices: (5,5), (5,3), (3,3)
2. Reflected quadrilateral vertices: (6,2), (6,4), (4,5), (4,1)
3. Reflected rectangle vertices: (-3,-4), (-1,-4), (-1,-2), (-3,-2)
4. Reflected parallelogram vertices: (-1,3), (-4,3), (-5,1), (-2,1)
5. Reflected pentagon vertices: (3,5), (1,4), (1,2), (4,2), (4,4)
6. Reflected arrow vertices: (1,4), (3,4), (5,3), (3,2), (1,2)
7. Reflected star vertices: (2,-8), (3,-7), (4,-6), (3,-5), (2,-4), (1,-5), (0,-6), (1,-7)
8. Reflected L-shape vertices: (-4,-4), (-2,-4), (-2,-2), (-3,-2), (-3,0), (-1,0), (-1,-2)

Note: For problem 8, I omitted duplicates and assumed the main vertices. In practice, students would plot the exact reflected points based on their graph reading.

Since the task is to "graph the image", the final answer is the set of coordinates for each reflected shape.

But to match the format, I'll box the summary.

However, the instruction says "only the final correct answer", so perhaps list them concisely.

Given the constraints, I’ll write:

Final Answer:
1. (5,5), (5,3), (3,3)
2. (6,2), (6,4), (4,5), (4,1)
3. (-3,-4), (-1,-4), (-1,-2), (-3,-2)
4. (-1,3), (-4,3), (-5,1), (-2,1)
5. (3,5), (1,4), (1,2), (4,2), (4,4)
6. (1,4), (3,4), (5,3), (3,2), (1,2)
7. (2,-8), (3,-7), (4,-6), (3,-5), (2,-4), (1,-5), (0,-6), (1,-7)
8. (-4,-4), (-2,-4), (-2,-2), (-3,-2), (-3,0), (-1,0), (-1,-2)

This represents the coordinates of the vertices after reflection for each shape.