Set of six graphs displaying slope fields for various first-order differential equations.

Let’s go through each differential equation one by one and figure out what the slope field should look like. A slope field shows little line segments at different points (x, y) that have a slope equal to dy/dx at that point.

We’ll think about how the slope changes depending on x and/or y for each equation.

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1. dy/dx = x + 1

This means the slope only depends on x — not on y. So along any vertical line (same x), all slopes are the same.

- When x = -1 → slope = 0 → horizontal lines
- When x > -1 → slope positive → lines tilt up to the right
- When x < -1 → slope negative → lines tilt down to the right

Also, as x increases, slope gets steeper positively; as x decreases, slope gets steeper negatively.

So in the slope field:
- Left of x = -1: downward sloping lines
- At x = -1: flat lines
- Right of x = -1: upward sloping lines, getting steeper as you go right

✔ This matches the first graph shown (black lines).

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2. dy/dx = 2y

Slope depends only on y — not on x. So along any horizontal line (same y), all slopes are the same.

- When y = 0 → slope = 0 → horizontal lines (along x-axis)
- When y > 0 → slope positive → lines tilt up to the right
- When y < 0 → slope negative → lines tilt down to the right

Also, magnitude doubles with |y| — so farther from x-axis, steeper slopes.

In the slope field:
- On x-axis: flat
- Above x-axis: upward tilting, steeper as you go up
- Below x-axis: downward tilting, steeper as you go down

✔ This matches the second graph (red lines).

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3. dy/dx = x + y

Now slope depends on both x and y. Let’s test some key lines:

- Along y = -x → slope = 0 → horizontal lines
- Above that line (y > -x) → slope positive
- Below that line (y < -x) → slope negative

For example:
- At (0,0): slope = 0
- At (1,0): slope = 1
- At (0,1): slope = 1
- At (-1,-1): slope = -2 → steep down
- At (1,1): slope = 2 → steep up

The zero-slope line is diagonal: y = -x.

Looking at the third graph (red lines), we see:
- Horizontal segments along the line y = -x (diagonal from top-left to bottom-right)
- Above it: upward slopes
- Below it: downward slopes

✔ Matches perfectly.

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4. dy/dx = 2x

Again, slope depends only on x.

- When x = 0 → slope = 0 → vertical line at x=0 has flat segments
- When x > 0 → slope positive → right side: upward tilting
- When x < 0 → slope negative → left side: downward tilting

Magnitude doubles with |x| — so farther from y-axis, steeper.

In the slope field:
- On y-axis: flat
- Right of y-axis: upward, steeper as you go right
- Left of y-axis: downward, steeper as you go left

✔ Fourth graph (black lines) shows exactly this.

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5. dy/dx = y - 1

Slope depends only on y.

- When y = 1 → slope = 0 → horizontal line at y=1
- When y > 1 → slope positive → above y=1: upward tilting
- When y < 1 → slope negative → below y=1: downward tilting

Magnitude grows as you move away from y=1.

In the slope field:
- At y=1: flat
- Above: upward, steeper higher up
- Below: downward, steeper lower down

✔ Fifth graph (red lines) matches — note the flat row at y=1.

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6. dy/dx = -y/x

This one is special — undefined when x=0 (vertical axis). Also, slope depends on ratio y/x.

Let’s think:

- If y = 0 (and x ≠ 0) → slope = 0 → horizontal along x-axis
- If x = y → slope = -1 → diagonal down to right
- If x = -y → slope = 1 → diagonal up to right
- In quadrant I (x>0, y>0): slope negative → down to right
- Quadrant II (x<0, y>0): slope positive → up to right
- Quadrant III (x<0, y<0): slope negative → down to right? Wait:

Wait: dy/dx = -y/x

Quadrant III: x negative, y negative → - (negative)/(negative) = - (positive) = negative → still down to right?

Actually let’s compute:

At (1,1): slope = -1/1 = -1 → ↘
At (-1,1): slope = -1/(-1) = 1 → ↗
At (-1,-1): slope = -(-1)/(-1) = -1 → ↘
At (1,-1): slope = -(-1)/1 = 1 → ↗

So pattern:
- QI: ↘
- QII: ↗
- QIII: ↘
- QIV: ↗

And along axes:
- x-axis (y=0): slope 0 → →
- y-axis (x=0): undefined → no segments or vertical? But usually we skip x=0.

Also, notice symmetry: if you rotate 180°, same pattern? Or reflect?

Actually, this looks like circles centered at origin — because solutions to dy/dx = -y/x are circles: x² + y² = C.

So slope field should be tangent to circles around origin.

Looking at sixth graph (red lines):

- Around origin, arrows form circular patterns
- On x-axis: horizontal
- On y-axis: missing or vertical? Actually in the image, they show segments near but not on y-axis, and they curve appropriately.

✔ Yes — sixth graph matches.

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All six graphs provided match their respective equations.

Final Answer:
Each given slope field correctly corresponds to its differential equation as labeled:
1. dy/dx = x+1 → first graph
2. dy/dx = 2y → second graph
3. dy/dx = x+y → third graph
4. dy/dx = 2x → fourth graph
5. dy/dx = y−1 → fifth graph
6. dy/dx = −y/x → sixth graph