Free math worksheets

- Substitute the line into the right side of the equation. - Compare the resulting value with the claimed slope. - A slope of \(0\) corresponds to a horizontal segment.

1. Substitute \(y=x\) into the differential equation. 2. On that line, \(y'=x-y=0\), so the segments are horizontal.

On \(y=x\), \(y'=x-y=0\), so those segments are horizontal.

- Determine the possible signs of the right side. - Locate where the right side equals \(0\). - Use those facts to correct the claim.

1. A square is never negative, so \(y^2\ge0\) for every real \(y\). 2. Therefore, all slopes are nonnegative, and only the row \(y=0\) is horizontal.

Because \(y^2\ge0\), all slopes are nonnegative; only the row \(y=0\) is horizontal.

- Check where the right side is defined. - Set the denominator equal to \(0\). - A slope field has no segment where its differential equation is undefined.

1. The denominator is \(0\) when \(x=3\). 2. Therefore, the right side is undefined there, so the slope field has a break along the vertical line \(x=3\).

The right side is undefined at \(x=3\), so the slope field has a break along that vertical line.

- Substitute each ordered pair into \(x+y\) separately. - The resulting number is the slope of the field segment at that point. - Classify positive, negative, and zero slopes as rising, falling, and horizontal.

1. Substitute each point into the right side of the differential equation. 2. The slopes are \((-1, 2): 1\), \((0, -1): -1\), \((2, 1): 3\). 3. The corresponding classifications are rising, falling, rising, in the listed order.

\((-1, 2)\): slope \(1\), rising \((0, -1)\): slope \(-1\), falling \((2, 1)\): slope \(3\), rising

- At each point, evaluate \(x-y\) using the listed coordinates. - Do not interchange the order of the subtraction. - Use the sign of each value to classify the segment’s direction.

1. Substitute each point into the right side of the differential equation. 2. The slopes are \((-2, -1): -1\), \((1, 1): 0\), \((3, -2): 5\). 3. The corresponding classifications are falling, horizontal, rising, in the listed order.

\((-2, -1)\): slope \(-1\), falling \((1, 1)\): slope \(0\), horizontal \((3, -2)\): slope \(5\), rising

- Evaluate the product \(xy\) at one point at a time. - A zero coordinate makes the segment horizontal. - Use the product’s sign to distinguish rising from falling segments.

1. Substitute each point into the right side of the differential equation. 2. The slopes are \((-2, 1): -2\), \((0, 3): 0\), \((2, -1): -2\). 3. The corresponding classifications are falling, horizontal, falling, in the listed order.

\((-2, 1)\): slope \(-2\), falling \((0, 3)\): slope \(0\), horizontal \((2, -1)\): slope \(-2\), falling

- Square the point’s \(y\)-coordinate before subtracting \(x\). - The computed value of \(y^2-x\) is the segment slope. - Use its sign to classify the direction.

1. Substitute each point into the right side of the differential equation. 2. The slopes are \((0, -2): 4\), \((1, 1): 0\), \((3, 0): -3\). 3. The corresponding classifications are rising, horizontal, falling, in the listed order.

\((0, -2)\): slope \(4\), rising \((1, 1)\): slope \(0\), horizontal \((3, 0)\): slope \(-3\), falling

- Substitute into \(\frac{x}{1+y^2}\) at each point. - The denominator \(1+y^2\) is always positive, so the numerator controls the sign. - Compute the magnitude as well as the direction.

1. Substitute each point into the right side of the differential equation. 2. The slopes are \((-2, 1): -1\), \((0, -3): 0\), \((4, 1): 2\). 3. The corresponding classifications are falling, horizontal, rising, in the listed order.

\((-2, 1)\): slope \(-1\), falling \((0, -3)\): slope \(0\), horizontal \((4, 1)\): slope \(2\), rising

- Evaluate the two factors \(x-1\) and \(y+2\) separately at each point. - A zero factor makes the field segment horizontal. - Otherwise, use the signs of the two factors to determine the product’s sign.

1. Substitute each point into the right side of the differential equation. 2. The slopes are \((0, 0): -2\), \((1, 5): 0\), \((3, -1): 2\). 3. The corresponding classifications are falling, horizontal, rising, in the listed order.

\((0, 0)\): slope \(-2\), falling \((1, 5)\): slope \(0\), horizontal \((3, -1)\): slope \(2\), rising

- Use the standard values of \(\sin 0\), \(\sin(\pi/2)\), and \(\sin\pi\). - Add the listed \(y\)-coordinate to the sine value at each point. - Classify the segment from the sign of the resulting slope.

1. Substitute each point into the right side of the differential equation. 2. The slopes are \((0, 1): 1\), \((\frac{\pi}{2}, 0): 1\), \((\pi, -1): -1\). 3. The corresponding classifications are rising, rising, falling, in the listed order.

\((0, 1)\): slope \(1\), rising \((\frac{\pi}{2}, 0)\): slope \(1\), rising \((\pi, -1)\): slope \(-1\), falling

- Use \(e^0=1\) and \(e^{-\ln2}=\frac12\) when evaluating the exponential term. - Subtract the listed \(y\)-coordinate from that value. - A zero result gives a horizontal segment.

1. Substitute each point into the right side of the differential equation. 2. The slopes are \((0, 0): 1\), \((0, 2): -1\), \((\ln 2, 0.5): 0\). 3. The corresponding classifications are rising, falling, horizontal, in the listed order.

\((0, 0)\): slope \(1\), rising \((0, 2)\): slope \(-1\), falling \((\ln 2, 0.5)\): slope \(0\), horizontal

- Compute the denominator \(x+2\) before dividing. - Evaluate \(y/(x+2)\) separately for each point. - Use the quotient’s sign to classify the segment.

1. Substitute each point into the right side of the differential equation. 2. The slopes are \((-1, 3): 3\), \((0, -2): -1\), \((2, 4): 1\). 3. The corresponding classifications are rising, falling, rising, in the listed order.

\((-1, 3)\): slope \(3\), rising \((0, -2)\): slope \(-1\), falling \((2, 4)\): slope \(1\), rising

- Square the \(x\)-coordinate before evaluating \(2-x^2-y\). - Substitute the listed \(y\)-coordinate only after the square is computed. - Use the final sign to classify the segment.

1. Substitute each point into the right side of the differential equation. 2. The slopes are \((-1, 0): 1\), \((0, 2): 0\), \((2, -1): -1\). 3. The corresponding classifications are rising, horizontal, falling, in the listed order.

\((-1, 0)\): slope \(1\), rising \((0, 2)\): slope \(0\), horizontal \((2, -1)\): slope \(-1\), falling

- Horizontal segments correspond to a slope of \(0\). - Solve the positive and negative inequalities for the right side. - Compare each point's \(y\)-coordinate with the boundary line \(y=-x\).

1. Horizontal segments occur where the right side equals \(0\), giving \(y=-x\). 2. Positive slopes occur where \(y>-x\). 3. Negative slopes occur where \(y<-x\).

a) \(y=-x\) b) Positive: \(y>-x\). Negative: \(y<-x\).

- Horizontal segments correspond to a slope of \(0\). - Solve the positive and negative inequalities for the right side. - Compare each point's \(y\)-coordinate with the boundary line \(y=x\).

1. Horizontal segments occur where the right side equals \(0\), giving \(y=x\). 2. Positive slopes occur where \(y<x\). 3. Negative slopes occur where \(y>x\).

a) \(y=x\) b) Positive: \(y<x\). Negative: \(y>x\).

- Horizontal segments correspond to a slope of \(0\). - Solve the positive and negative inequalities for the right side. - Compare each point with the boundary curve \(y=x^2\).

1. Horizontal segments occur where the right side equals \(0\), giving \(y=x^2\). 2. Positive slopes occur where \(y<x^2\). 3. Negative slopes occur where \(y>x^2\).

a) \(y=x^2\) b) Positive: \(y<x^2\). Negative: \(y>x^2\).

- Factor \(1-y^2\) as \((1-y)(1+y)\) to expose the zero-slope rows. - Use the critical values \(y=-1\) and \(y=1\) to divide the number line into sign intervals. - Determine the slope sign from the two factor signs in each interval.

1. Horizontal segments occur where the right side equals \(0\), giving \(y=-1\text{ or }y=1\). 2. Positive slopes occur where \(-1<y<1\). 3. Negative slopes occur where \(y<-1\text{ or }y>1\).

a) \(y=-1\text{ or }y=1\) b) Positive: \(-1<y<1\). Negative: \(y<-1\text{ or }y>1\).

- Horizontal segments correspond to a slope of \(0\). - Solve the positive and negative inequalities for the right side. - Compare each point with the boundary curve \(y=e^x\).

1. Horizontal segments occur where the right side equals \(0\), giving \(y=e^x\). 2. Positive slopes occur where \(y<e^x\). 3. Negative slopes occur where \(y>e^x\).

a) \(y=e^x\) b) Positive: \(y<e^x\). Negative: \(y>e^x\).

- Horizontal segments correspond to a slope of \(0\). - Solve the positive and negative inequalities for the right side. - Compare each point with the boundary line \(y=4-x\).

1. Horizontal segments occur where the right side equals \(0\), giving \(y=4-x\). 2. Positive slopes occur where \(y<4-x\). 3. Negative slopes occur where \(y>4-x\).

a) \(y=4-x\) b) Positive: \(y<4-x\). Negative: \(y>4-x\).

- Compare segments in the same vertical column. - Locate the column where every segment is horizontal. - Use the signs of the slopes to the left and right of that column.

1. The segments have the same slope within each vertical column, so the derivative depends on \(x\), not on \(y\). 2. The segments are horizontal when \(x=0\), negative for \(x<0\), and positive for \(x>0\). 3. Only choice A, \(y'=x\), matches all three visual features.

A. \(y'=x\)

- Compare segments across a single horizontal row. - Identify the row on which all segments are horizontal. - Check the slope signs immediately below and above that row.

1. The segments have the same slope across each horizontal row, so the derivative depends only on \(y\). 2. The row \(y=2\) is horizontal. Slopes are positive below it and negative above it. 3. Only choice A, \(y'=2-y\), has that zero-slope row and sign pattern.

A. \(y'=2-y\)

- Inspect the segments on both coordinate axes. - Compare the slope signs in the four quadrants. - Look for an expression that is zero when either coordinate is zero.

1. Every segment on either coordinate axis is horizontal, so the derivative must be zero when \(x=0\) or \(y=0\). 2. Slopes are positive in Quadrants I and III and negative in Quadrants II and IV. 3. Only choice A, \(y'=xy\), has both the axis zeros and the displayed quadrant sign pattern.

A. \(y'=xy\)

- Check whether changing \(y\) within one column changes the slope. - Locate the columns with horizontal segments. - Compare the alternating signs with the sine function.

1. Within each vertical column, the slope is unchanged from row to row, so the derivative depends only on \(x\). 2. The field is horizontal at \(x=-\pi,0,\pi\), with alternating positive and negative slopes between those columns. 3. Only choice A, \(y'=\sin x\), matches the displayed dependence, zeros, and sign changes.

A. \(y'=\sin x\)

- Use the horizontal segments to identify several points on the zero-slope curve. - Compare a segment below that curve with one above it. - Test which choice changes sign in the displayed direction.

1. The horizontal segments occur at the plotted points satisfying \(y=x^2\), including \((0,0)\), \((\pm1,1)\), and \((\pm2,4)\). 2. The displayed slopes are positive below this parabola and negative above it. 3. Only choice A, \(y'=x^2-y\), has that zero-slope set and sign pattern.

A. \(y'=x^2-y\)

- Compare corresponding columns at \(x\) and \(-x\). - Use the horizontal segments to infer the zero-slope curve. - Substitute that curve into each candidate expression.

1. The field is symmetric about the y-axis, which indicates dependence on \(x^2\) rather than on \(x\) alone. 2. Horizontal segments appear at \((0,1)\), \((\pm1,2)\), and \((\pm2,5)\), tracing \(y=x^2+1\). 3. Only choice A, \(y'=x^2+1-y\), matches both the symmetry and the zero-slope curve.

A. \(y'=x^2+1-y\)

- Locate a line in the field where several segments are horizontal. - Compare the signs of the segments on the two sides of that zero-slope line. - To test Lee’s claim, compare a point above the x-axis but below \(y=x\) with the displayed segment there.

1. The displayed segments are horizontal along \(y=x\), positive above that line, and negative below it. This sign pattern matches \(y'=y-x\), so choice B is correct. 2. At \((1,0)\), \(y'=0-1=-1\). Therefore, the solution initially falls. 3. Lee’s statement is false. The sign depends on whether \(y\) is greater than or less than \(x\), not merely on whether \(y\) is positive. Segments above \(y=x\) have positive slope, and segments below \(y=x\) have negative slope.

a) B. \(y'=y-x\) b) The solution initially falls because the slope is \(-1\). c) Positive slopes occur above \(y=x\), not everywhere above the x-axis.

- Compare segments that share an x-coordinate but have different y-coordinates. - Locate the column of horizontal segments and inspect the signs on either side. - Distinguish a horizontal tangent at one point from a constant solution on an interval.

1. The slope is the same at all points with the same x-coordinate, is zero when \(x=0\), is negative for \(x<0\), and is positive for \(x>0\). This matches choice A, \(y'=x\). 2. In \(y'=x\), the derivative depends only on \(x\), so changing \(y\) while keeping \(x\) fixed does not change the slope. 3. Morgan’s statement is false. A solution has a horizontal tangent when it crosses the y-axis, but immediately to the left and right the field has nonzero slopes. A horizontal tangent at one x-value does not make the entire solution constant.

a) A. \(y'=x\) b) The derivative depends only on the x-coordinate. c) Solutions have a horizontal tangent at \(x=0\), but they do not remain constant because the slope changes when \(x\ne0\).

- Use the fact that a square is nonnegative. - Determine when the denominator takes its smallest possible value. - Hold the combination \(x-y\) constant to examine an isocline.

1. The denominator is at least \(1\), so \(0<y'\le1\). Thus every segment rises and no slope exceeds \(1\). 2. The slope equals \(1\) exactly when \((x-y)^2=0\), so the line is \(y=x\). 3. On \(x-y=c\), the derivative has the constant value \(\frac{1}{1+c^2}\), so each such line is an isocline.

a) All slopes satisfy \(0<y'\le1\). b) The maximum slope \(1\) occurs on \(y=x\). c) On \(x-y=c\), every segment has slope \(\frac{1}{1+c^2}\).

- Set the squared expression equal to zero. - Use the sign property of a square. - Connect the sign of the derivative to monotonic behavior of solution curves.

1. Horizontal segments occur where \((y-\sin x)^2=0\), so on \(y=\sin x\). 2. A square is always nonnegative, so every slope satisfies \(y'\ge0\). 3. Every solution is nondecreasing wherever it exists. Therefore, a nonconstant solution cannot have a strict local maximum, although it may have a horizontal tangent.

a) \(y=\sin x\) b) All slopes are nonnegative. c) A nonconstant solution cannot have a strict local maximum.

- The denominator is always positive, so focus on the numerator for the sign. - Substitute the two points directly. - Compare the denominator values at the two x-coordinates.

1. Horizontal segments occur where \(y=0\), so on the x-axis. 2. Because \(1+x^2>0\), slopes are positive above the x-axis and negative below it. 3. At \((0,2)\), the slope is \(2\). At \((3,2)\), it is \(\frac{2}{10}=0.2\). 4. For the same y-coordinate, the slope magnitude decreases as \(|x|\) increases because the denominator grows.

a) \(y=0\) b) Positive for \(y>0\); negative for \(y<0\) c) The slopes are \(2\) and \(0.2\); the larger denominator at \(x=3\) makes the second segment flatter.

- Substitute each ordered pair directly into the derivative formula. - Simplify exponentials of logarithms before comparing. - Parallel field segments have equal numerical slopes.

1. At \(A\), the slope is \(1\cdot e^0=1\). 2. At \(B\), the slope is \(1\cdot e^{\ln 2}=2\). 3. At \(C\), the slope is \(2\cdot e^{-\ln 2}=1\). 4. At \(D\), the slope is \(-1\cdot e^{\ln 3}=-3\). 5. The segments at \(A\) and \(C\) are parallel, and the ordering is \(-3<1=1<2\).

a) \(m_A=1\), \(m_B=2\), \(m_C=1\), \(m_D=-3\) b) \(A\) and \(C\) c) \(m_D<m_A=m_C<m_B\)

- Substitute each point separately into the derivative formula. - Horizontal segments have numerical slope zero. - Set the entire derivative formula equal to zero to find the nullcline.

1. At \(A\), the slope is \(\ln1-0=0\). 2. At \(B\), the slope is \(\ln2-\ln2=0\). 3. At \(C\), the slope is \(\ln2-(\ln2-1)=1\). 4. At \(D\), the slope is \(\ln5-(\ln5+2)=-2\). 5. Horizontal segments satisfy \(\ln(1+x^2)-y=0\), so the nullcline is \(y=\ln(1+x^2)\). Both \(A\) and \(B\) satisfy this equation.

a) \(m_A=0\), \(m_B=0\), \(m_C=1\), \(m_D=-2\) b) \(A\) and \(B\) c) \(y=\ln(1+x^2)\); both \(A\) and \(B\) lie on this curve.

- First determine whether the denominator can change sign or equal zero. - Use the numerator to locate and compare the sign regions. - Substitute each ordered pair separately before simplifying. - Compare the line's own slope with the slope prescribed by the field.

1. The denominator is always positive, so horizontal segments occur where \(x+y=0\), or \(y=-x\). 2. Slopes are positive where \(x+y>0\) and negative where \(x+y<0\). 3. At \((1,0)\), the slope is \(1/[1+1]=1/2\). At \((0,1)\), the slope is also \(1/[1+1]=1/2\). 4. Along \(y=-x\), the field slope is \(0\), but the line's own slope is \(-1\). Therefore, it is a nullcline, not a solution curve.

a) \(y=-x\) b) Positive where \(x+y>0\); negative where \(x+y<0\) c) Both slopes are \(1/2\). d) The line has slope \(-1\), not the field slope \(0\), so it is not a solution curve.

- Determine when the logarithm's argument equals \(1\). - Compare the argument with \(1\) away from the origin. - Use \(x^2+y^2\) as the squared distance from the origin. - An equilibrium must be a constant function with zero slope for all \(x\).

1. The slope is zero when \(1+x^2+y^2=1\), which occurs only at \((0,0)\). 2. At every other point, \(1+x^2+y^2>1\), so the logarithm and the slope are positive. 3. The expression depends on \(x^2+y^2\), so slope magnitude increases with distance from the origin. 4. An equilibrium solution must be a constant horizontal function \(y=c\) whose slope is zero for every \(x\). The origin is only one point. In particular, along \(y=0\), the slope is \(\ln(1+x^2)>0\) when \(x\ne0\).

a) Only \((0,0)\) b) Every other slope is positive. c) Slopes become steeper as distance from the origin increases. d) A single zero-slope point is not an equilibrium solution; no constant horizontal line has zero slope everywhere.

- Set each factor in \(y(3-y)\) equal to zero to find the horizontal rows. - Test the signs of \(y\) and \(3-y\) in the intervals cut by those rows. - Use the product sign to classify slopes between and outside the equilibrium lines.

1. Horizontal segments occur where the right side equals \(0\), giving \(y=0\text{ or }y=3\). 2. Positive slopes occur where \(0<y<3\). 3. Negative slopes occur where \(y<0\text{ or }y>3\).

a) \(y=0\text{ or }y=3\) b) Positive: \(0<y<3\). Negative: \(y<0\text{ or }y>3\).

- Horizontal segments correspond to a slope of \(0\). - Determine where each factor is positive or negative. - A product is positive for matching signs and negative for opposite signs.

1. Horizontal segments occur where either factor is \(0\), giving \(x=2\) or \(y=-1\). 2. The slope is positive where the factors have the same sign: \(x>2\) and \(y>-1\), or \(x<2\) and \(y<-1\). 3. The slope is negative where the factors have opposite signs: \(x>2\) and \(y<-1\), or \(x<2\) and \(y>-1\).

a) \(x=2\) or \(y=-1\) b) Positive: \(x>2\) and \(y>-1\), or \(x<2\) and \(y<-1\). Negative: \(x>2\) and \(y<-1\), or \(x<2\) and \(y>-1\).

- Horizontal segments correspond to a slope of \(0\). - Use the zeros and sign pattern of the sine function. - Express the repeating intervals with an integer parameter.

1. Horizontal segments occur where \(\sin x=0\), giving \(x=n\pi\), where \(n\in\mathbb Z\). 2. Positive slopes occur where \(2n\pi<x<(2n+1)\pi\), where \(n\in\mathbb Z\). 3. Negative slopes occur where \((2n-1)\pi<x<2n\pi\), where \(n\in\mathbb Z\).

a) \(x=n\pi\), where \(n\in\mathbb Z\) b) Positive: \(2n\pi<x<(2n+1)\pi\). Negative: \((2n-1)\pi<x<2n\pi\), where \(n\in\mathbb Z\).

- Horizontal segments require a zero numerator and a nonzero denominator. - Determine where the numerator and denominator are positive or negative. - A quotient is positive for matching signs and negative for opposite signs.

1. Horizontal segments occur where the numerator is \(0\) and the denominator is nonzero, giving \(x=-1\) with \(y\ne2\). 2. The slope is positive where numerator and denominator have the same sign: \(x>-1\) and \(y>2\), or \(x<-1\) and \(y<2\). 3. The slope is negative where numerator and denominator have opposite signs: \(x>-1\) and \(y<2\), or \(x<-1\) and \(y>2\).

a) \(x=-1\) with \(y\ne2\) b) Positive: \(x>-1\) and \(y>2\), or \(x<-1\) and \(y<2\). Negative: \(x>-1\) and \(y<2\), or \(x<-1\) and \(y>2\).

- Identify the horizontal row on which the field is missing. - Locate the vertical column of horizontal segments. - Use the regional slope signs to choose between the quotient and product forms.

1. No segments appear along \(y=1\), indicating that the derivative is undefined there. 2. Every segment in the column \(x=-2\) is horizontal, so the numerator must be zero there. 3. The signs in the remaining regions match a quotient whose numerator and denominator are \(x+2\) and \(y-1\). Therefore, choice A is correct.

A. \(y'=\frac{x+2}{y-1}\)

- Read the slope direction at the red cross and in the rows immediately above it. - Find the horizontal row that acts as an equilibrium. - Track how the segment steepness changes as a curve moves toward that row.

1. At the marked point and throughout the region below \(y=2\), the field has positive slopes, so the solution initially increases. 2. The segments become less steep as \(y\) approaches \(2\), and the row \(y=2\) consists of horizontal segments. 3. The solution therefore increases with slopes tending toward \(0\) and approaches the equilibrium line \(y=2\) without crossing it.

The solution increases from \((0,0)\), its slopes decrease toward \(0\), and it approaches the equilibrium line \(y=2\) without crossing it.

- The factors \(y\) and \(4-y\) give equilibrium rows at their zeros. - Place the initial value \(y=1\) in the interval between those rows and determine the slope sign there. - Use the sign change at \(y=4\) to describe the long-term direction without solving.

1. Evaluate the sign of the field at and near \((0,1)\). 2. Locate the zero-slope set from the differential equation. 3. The solution increases while \(0<y<4\) and approaches \(y=4\).

The solution increases while \(0<y<4\) and approaches \(y=4\).

- Locate the equilibria from the factors in \(y(4-y)\). - At the initial value \(y=6\), determine the signs of both factors. - Track how the slope changes as the curve moves toward the nearest equilibrium row.

1. Evaluate the sign of the field at and near \((0,6)\). 2. Locate the zero-slope set from the differential equation. 3. The solution decreases while \(y>4\) and approaches \(y=4\).

The solution decreases while \(y>4\) and approaches \(y=4\).

- Evaluate \(x-y\) at \((0,2)\) to determine the initial slope sign. - Solve \(x-y=0\) to identify the zero-slope line. - Compare the field above and below that line to describe what happens when the curve crosses it.

1. At \((0,2)\), the slope is \(0-2=-2\), so the solution initially decreases. 2. Horizontal segments lie on the nullcline \(y=x\). 3. Above \(y=x\), slopes are negative; below \(y=x\), slopes are positive. Thus the solution decreases until it crosses the nullcline, has a horizontal tangent there, and then increases.

The solution initially decreases with slope \(-2\). It decreases while it is above the nullcline \(y=x\), has a horizontal tangent when it crosses that line, and then increases below it.

- Evaluate \(x^2-y\) at the initial point before describing the direction. - The zero-slope curve is obtained from \(y=x^2\). - Determine the sign of \(x^2-y\) below and above the parabola.

1. At \((0,-1)\), the slope is \(0^2-(-1)=1\), so the solution initially increases. 2. Horizontal segments lie on the nullcline \(y=x^2\). 3. Slopes are positive below the parabola and negative above it, so the parabola separates the increasing and decreasing regions.

The solution initially increases with slope \(1\). The nullcline is \(y=x^2\); slopes are positive below it and negative above it.

- Solve \(1-y^2=0\) to locate both equilibrium rows. - At \(y=0\), determine whether the solution initially rises or falls. - Use the sign of \(1-y^2\) between the equilibria to identify which row the solution approaches.

1. Evaluate the sign of the field at and near \((0,0)\). 2. Locate the zero-slope set from the differential equation. 3. The solution increases and remains between the equilibrium solutions \(y=-1\) and \(y=1\), approaching \(y=1\).

The solution increases and remains between the equilibrium solutions \(y=-1\) and \(y=1\), approaching \(y=1\).

- Find the slope at the initial point. - Factor the right side to locate every zero-slope set. - Track the sign for a solution that remains above \(y=0\).

1. At \((0,3)\), the slope is \(0\), so the solution has a horizontal tangent. 2. The zero-slope sets are \(x=0\) and \(y=0\); the line \(y=0\) is an equilibrium solution. 3. For this positive solution, slopes are positive when \(x<0\) and negative when \(x>0\). Thus the curve rises toward \(x=0\) from the left and decreases to the right.

The solution has a horizontal tangent at \(x=0\), rises as \(x\) approaches \(0\) from the left, and decreases for \(x>0\). The line \(y=0\) is an equilibrium solution.

- Evaluate \(xy\) at \((1,2)\) to find the slope assigned at that point. - A slope field assigns one numerical slope to each point where the equation is defined. - Two curves passing through the same point cannot have different tangent slopes while both follow this field.

1. Substitute the relevant coordinates or region into the differential equation. 2. The field assigns the single slope \(2\) at \((1,2)\), so any solution through that point must have that tangent slope; curves cannot cross there with different slopes.

The field assigns the single slope \(2\) at \((1,2)\), so any solution through that point must have that tangent slope; curves cannot cross there with different slopes.

- A solution curve must be tangent to the field segments at every point it passes through. - Evaluate \(x-y\) at the shared initial point before comparing the three curves. - Matching one tangent is not enough; inspect another clearly readable point on curve \(r\).

1. Curve \(p\) follows the field’s direction throughout the displayed interval, so it can be a solution. 2. At \((0,1)\), the differential equation gives \(y'=0-1=-1\). Curve \(q\) has positive slope there, so it cannot be a solution. 3. Curve \(r\) has slope \(-1\) at \((0,1)\), but near \((1,0)\) the field slope is \(1-0=1\) while curve \(r\) still has slope \(-1\). Therefore, it cannot follow the slope field. 4. As an algebraic check, curve \(p\) is \(y=x-1+2e^{-x}\). Its derivative is \(1-2e^{-x}\), and \(x-y=1-2e^{-x}\), so it satisfies the differential equation.

a) Curve \(p\) b) The required slope at \((0,1)\) is \(-1\), but curve \(q\) has positive slope there. c) Near \((1,0)\), the field has slope \(1\), but curve \(r\) has slope \(-1\).

- Set the derivative equal to zero and factor the resulting expression. - Compare the squares of the two coordinates to determine the sign. - Notice which coordinate changes do not affect a squared value.

1. Horizontal segments occur where \(x^2-y^2=0\), so \(y=x\) or \(y=-x\). 2. The slope is positive when \(x^2>y^2\), equivalently \(|x|>|y|\), and negative when \(|y|>|x|\). 3. At each listed point, \(x^2=9\) and \(y^2=1\), so the slope is \(9-1=8\).

a) \(y=x\) and \(y=-x\) b) Positive where \(|x|>|y|\); negative where \(|y|>|x|\) c) All four slopes are \(8\).

- Look for a diagonal line along which the displayed segments are horizontal. - A missing diagonal in a slope field can indicate where a denominator is zero. - After choosing the equation, substitute the two listed points directly.

1. The field has horizontal segments along \(y=x\) except at the origin, and it has no segments along \(y=-x\). This pattern matches a zero numerator on \(x-y=0\) and a zero denominator on \(x+y=0\), so choice A is correct. 2. For choice A, horizontal segments occur where \(x-y=0\), so \(y=x\), except at \((0,0)\). The field is undefined where \(x+y=0\), so \(y=-x\). 3. At \((2,0)\), the slope is \(\frac{2-0}{2+0}=1\). At \((0,2)\), the slope is \(\frac{0-2}{0+2}=-1\).

a) A. \(y'=\frac{x-y}{x+y}\) b) Horizontal on \(y=x\), except at \((0,0)\); undefined on \(y=-x\) c) The slopes are \(1\) and \(-1\), respectively.

- Horizontal segments occur where the exponential expression equals one. - Compare the sign of the exponent with zero to determine the sign of the slope. - Substitute the equation of the listed line into the combination \(x-y\).

1. Horizontal segments occur when \(e^{x-y}=1\), so \(x-y=0\), or \(y=x\). 2. The slope is positive when \(x-y>0\), so \(y<x\), and negative when \(y>x\). 3. On \(y=x-\ln 3\), \(x-y=\ln 3\), so the slope is \(e^{\ln 3}-1=2\).

a) \(y=x\) b) Positive below \(y=x\); negative above \(y=x\) c) The common slope is \(2\).

- Use the horizontal segments to identify where the right side equals zero. - Check the origin to distinguish a logarithmic model from a polynomial model with the same zero-slope circle. - Compare squared distance from the origin with \(1\) to determine the sign. - Substitute the two listed points only after selecting the equation.

1. The field has horizontal segments on the unit circle, negative slopes inside that circle, positive slopes outside it, and no segment at the origin. Only choice A has all four features, so \(y'=\ln(x^2+y^2)\). 2. Horizontal segments occur when \(\ln(x^2+y^2)=0\), so \(x^2+y^2=1\). The field is undefined at \((0,0)\), where the logarithm’s argument is zero. 3. Slopes are positive when \(x^2+y^2>1\) and negative when \(0<x^2+y^2<1\). 4. At \(\left(\frac{3}{5},\frac{4}{5}\right)\), the slope is \(\ln1=0\). At \((2,0)\), it is \(\ln4\).

a) A. \(y'=\ln(x^2+y^2)\) b) Horizontal on \(x^2+y^2=1\); undefined at \((0,0)\) c) Positive outside the unit circle; negative inside it, excluding the origin d) The slopes are \(0\) and \(\ln4\), respectively.

- The denominator is positive everywhere, so the numerator controls the sign. - Replace \(x^2+y^2\) by a single nonnegative quantity. - Rewrite the quotient to compare it directly with \(1\).

1. Horizontal segments occur where \(x^2+y^2=1\). 2. The denominator is always positive, so slopes are negative inside the unit circle and positive outside it. 3. Let \(u=x^2+y^2\ge0\). Then \(y'=\frac{u-1}{u+1}=1-\frac{2}{u+1}\), which lies in \([-1,1)\), with \(-1\) only at the origin. 4. As \(u\to\infty\), the slope approaches \(1\).

a) \(x^2+y^2=1\) b) Negative inside the unit circle; positive outside it c) Slopes range from \(-1\) at the origin up toward, but never reaching, \(1\); far from the origin they approach \(1\).

- Equilibrium rows occur where the autonomous derivative is zero. - Test one value in each interval between consecutive roots. - Use the sign arrows on both sides of each equilibrium to classify stability.

1. The equilibrium solutions are \(y=1\), \(y=3\), and \(y=5\). 2. The slope is negative for \(y<1\), positive for \(1<y<3\), negative for \(3<y<5\), and positive for \(y>5\). 3. Near \(y=1\), solutions move away on both sides, so it is repelling. Near \(y=3\), solutions move toward it from both sides, so it is attracting. Near \(y=5\), solutions move away on both sides, so it is repelling.

a) \(y=1,3,5\) b) Negative, positive, negative, positive from bottom to top c) \(y=1\) and \(y=5\) are repelling; \(y=3\) is attracting.

- A product is zero when either factor is zero. - Use the sign pattern of sine in successive horizontal bands. - Evaluate both factors at each listed point.

1. Horizontal segments occur when \(x=0\) or \(\sin y=0\), so on the y-axis and on every line \(y=k\pi\), \(k\in\mathbb{Z}\). 2. Slopes are positive where \(x\) and \(\sin y\) have the same sign, and negative where they have opposite signs. 3. The listed slopes are \(1\), \(-1\), and \((-2)\cdot(-1)=2\), respectively.

a) \(x=0\) and \(y=k\pi\), \(k\in\mathbb{Z}\) b) Positive when \(x\) and \(\sin y\) have the same sign; negative when their signs differ c) The slopes are \(1\), \(-1\), and \(2\).

- Use the zero and sign behavior of the inverse tangent function. - Recall the range of the inverse tangent function. - Hold the expression \(x-y\) constant to identify isoclines.

1. Horizontal segments occur when \(\arctan(x-y)=0\), so \(y=x\). 2. Slopes are positive when \(x-y>0\), or \(y<x\), and negative when \(y>x\). 3. The range of arctangent is \((-\frac{\pi}{2},\frac{\pi}{2})\), so every slope lies in that interval. 4. On \(x-y=c\), the slope is the constant \(\arctan c\), making each such line an isocline.

a) \(y=x\) b) Positive below \(y=x\); negative above it c) Slopes lie in \((-\frac{\pi}{2},\frac{\pi}{2})\), and each line \(x-y=c\) has constant slope \(\arctan c\).

- Determine which factor can equal zero. - One factor is always positive, so the other controls the sign. - Examine how the positive exponential factor changes with \(y\).

1. Since \(e^{-y}>0\), horizontal segments occur where \(\sin x=0\), so on \(x=k\pi\), \(k\in\mathbb{Z}\). 2. Slopes are positive where \(\sin x>0\) and negative where \(\sin x<0\), independent of \(y\). 3. As \(y\) increases, \(e^{-y}\) decreases, so the slope magnitude decreases and the segments become flatter.

a) \(x=k\pi\), \(k\in\mathbb{Z}\) b) Positive where \(\sin x>0\); negative where \(\sin x<0\) c) The slope magnitude decreases as \(y\) increases.

- A product is zero when either trigonometric factor is zero. - Determine the sign of each factor in the specified rectangle. - Substitute the two listed points into both trigonometric factors. - Use the ranges of sine and cosine to bound the slope magnitude.

1. Horizontal segments occur when \(\sin x=0\) or \(\cos y=0\). Thus they lie on \(x=k\pi\) and \(y=\frac{\pi}{2}+k\pi\), where \(k\in\mathbb{Z}\). 2. In the stated y-range, \(\cos y>0\). Therefore, slopes are positive for \(0<x<\pi\) and negative for \(\pi<x<2\pi\). 3. The two slopes are \(1\) and \(-1\), respectively. 4. Since \(|\sin x|\le1\) and \(|\cos y|\le1\), every slope has magnitude at most \(1\), and this value is attained.

a) \(x=k\pi\) and \(y=\frac{\pi}{2}+k\pi\), \(k\in\mathbb{Z}\) b) Positive for \(0<x<\pi\); negative for \(\pi<x<2\pi\) c) \(1\) and \(-1\) d) The largest possible slope magnitude is \(1\).

- The denominator is always positive, so a zero slope must come from the numerator. - Determine which coordinate controls the sign. - Compare denominator sizes while keeping \(x\) fixed. - Set the derivative equal to the constant \(m\) and solve for a relation between \(x\) and \(y\).

1. Horizontal segments occur where \(x=0\). 2. Since \(1+y^2>0\), slopes are positive for \(x>0\) and negative for \(x<0\). 3. At fixed nonzero \(x\), the denominator increases with \(|y|\), so the slope magnitude decreases. 4. Setting \(\frac{x}{1+y^2}=m\) gives \(x=m(1+y^2)\).

a) The y-axis, \(x=0\) b) Positive for \(x>0\); negative for \(x<0\) c) The magnitude decreases as \(|y|\) increases. d) \(x=m(1+y^2)\)

- The exponential factor is never zero, so horizontal segments come from the two linear y-factors. - Compare \(e^{x+\ln2}\) with \(e^x\) while holding \(y\) fixed. - Evaluate the two y-factors before multiplying by the exponential factor at each point. - Put the three numerical slopes on a number line before ordering them.

1. Horizontal segments occur where \((y-1)(y+2)=0\), so on \(y=1\) and \(y=-2\). 2. Replacing \(x\) by \(x+\ln2\) multiplies \(e^x\) by \(2\) and leaves the y-factors unchanged, so it doubles the slope. 3. At \(A\), the slope is \((-1)\cdot2=-2\). At \(B\), it is \(2\cdot(-1)\cdot2=-4\). At \(C\), it is \(1\cdot4=4\). 4. Therefore, \(-4<-2<4\), so \(m_B<m_A<m_C\).

a) \(y=1\) and \(y=-2\) b) The slope is multiplied by \(2\). c) \(m_A=-2\), \(m_B=-4\), \(m_C=4\) d) \(m_B<m_A<m_C\)

- A true equilibrium must give zero slope for every x-coordinate. - Make separate sign charts for the x-factor and the y-factor. - Combine the two factor signs in each region.

1. The factor \(y+1\) is zero for every \(x\) when \(y=-1\), so \(y=-1\) is an equilibrium solution. 2. Horizontal segments also occur where \(x^2-1=0\), so on \(x=-1\) and \(x=1\). These are vertical lines, not functions \(y(x)\), and nearby x-values generally have nonzero slope. 3. For \(|x|>1\), \(x^2-1>0\), so the slope has the sign of \(y+1\). For \(|x|<1\), the sign is reversed.

a) Equilibrium: \(y=-1\). Vertical zero-slope lines: \(x=-1\) and \(x=1\). b) The vertical lines are not solution functions and zero slope occurs there only at one x-coordinate. c) For \(|x|>1\), slopes are positive above \(y=-1\) and negative below it. For \(|x|<1\), slopes are negative above \(y=-1\) and positive below it.

- Set the entire right side equal to zero to find equilibrium levels. - A repeated factor may touch zero without changing sign. - Use arrows above and below each equilibrium to classify its behavior from each side. - Locate the initial value among the equilibrium levels and follow the arrows forward.

1. The equilibrium solutions are \(y=-1\) and \(y=2\). 2. The squared factor is nonnegative, so the slope is negative for every \(y<2\) except at \(y=-1\), and positive for \(y>2\). 3. Near \(y=-1\), solutions above it move downward toward it, while solutions below it also move downward away from it. Thus it attracts from above and repels from below. 4. Near \(y=2\), slopes point away on both sides, so \(y=2\) is repelling. 5. Since \(-1<0<2\), a solution starting at \(0\) decreases toward \(y=-1\).

a) \(y=-1\) and \(y=2\) b) Negative for \(y<2\), except zero at \(y=-1\); positive for \(y>2\) c) \(y=-1\) attracts from above and repels from below; \(y=2\) is repelling. d) The solution decreases toward \(y=-1\).

- A product is zero when either factor is zero. - Use a sign chart for the two factors. - Equal numerical slopes correspond to parallel field segments.

1. Horizontal segments occur where \(x-1=0\) or \(y+2=0\), so on \(x=1\) and \(y=-2\). 2. The slope is positive when the two factors have the same sign: \(x>1,y>-2\) or \(x<1,y<-2\). 3. The slope is negative when the factors have opposite signs: \(x>1,y<-2\) or \(x<1,y>-2\). 4. At \((2,0)\), the slope is \(1\cdot2=2\). At \((0,-4)\), the slope is \((-1)\cdot(-2)=2\). Therefore, the two field segments are parallel.

a) \(x=1\) and \(y=-2\) b) Positive in the upper-right and lower-left regions relative to the two lines; negative in the other two regions c) Both slopes are \(2\), so the segments are parallel.

- Set the derivative equal to zero and solve for \(y\). - Use monotonicity of the logarithm to compare points above and below the curve. - Replace zero by a general constant to find an isocline.

1. Horizontal segments satisfy \(\ln(y+2)=x\), so \(y=e^x-2\). 2. The logarithm is increasing, so slopes are positive when \(y>e^x-2\) and negative when \(-2<y<e^x-2\). 3. Setting \(\ln(y+2)-x=m\) gives \(y=e^{x+m}-2\).

a) \(y=e^x-2\) b) Positive above that curve and negative between it and \(y=-2\) c) \(y=e^{x+m}-2\)

- Determine the sign of each factor separately. - Examine one coordinate at a time when comparing magnitudes. - Remember that the sign of \(y'\) controls increasing or decreasing behavior. - Differentiate \(y'=f(x,y)\) with respect to \(x\), treating \(y\) as a function of \(x\).

1. Both factors are strictly positive, so the slope is never zero. 2. Every slope is positive, so every solution is increasing. 3. With \(y\) fixed, \(e^{-x^2}\) decreases as \(|x|\) increases. With \(x\) fixed, \(y^2+1\) increases as \(|y|\) increases. 4. Positive \(y'\) describes increasing behavior, not concavity. Differentiating along a solution gives \(y''=-2xe^{-x^2}(y^2+1)+2ye^{-x^2}y'\). At \((1,0)\), \(y''=-2/e<0\), so the solution is concave down there.

a) The derivative is always positive. b) Every solution is increasing. c) Slopes flatten as \(|x|\) increases and steepen as \(|y|\) increases. d) Priya confused increasing with concave up. At \((1,0)\), \(y''=-2/e<0\), so the solution is concave down.

- Set the derivative equal to zero to locate horizontal segments. - Compare \(y\) with the nullcline to determine the sign. - Differentiate the differential equation once and then use \(y'=0\). - A solution curve must have the same tangent slope as the field at every point on it.

1. Horizontal segments satisfy \(x^2+y-2=0\), so the curve is \(y=2-x^2\). 2. Slopes are positive above this parabola and negative below it. 3. Differentiating the differential equation gives \(y''=2x+y'\). At a horizontal tangent, \(y'=0\), so \(y''=2x\). Thus a horizontal tangent at \(x=-1\) is a local maximum, while one at \(x=1\) is a local minimum. 4. The parabola is a nullcline, not a solution curve. Along it the field slope is \(0\), but the parabola's own slope is \(-2x\), which is generally not \(0\).

a) \(y=2-x^2\) b) Slopes are positive above the parabola and negative below it. c) Local maximum at \(x=-1\); local minimum at \(x=1\). d) The curve is a nullcline, not a solution; its own slope \(-2x\) does not generally match the field slope \(0\).

- A product is zero when either factor is zero. - Track the sign of each sine factor between consecutive zeros. - An equilibrium solution must be a constant function of \(x\), not a vertical line.

1. The slope is zero on \(x=k\pi\) and on \(y=m\pi\), where \(k,m\in\mathbb{Z}\). 2. Between consecutive multiples of \(\pi\), each sine factor has a constant sign. The product is positive when the factors have the same sign and negative when they have opposite signs, creating a checkerboard pattern. 3. Every horizontal line \(y=m\pi\) is an equilibrium solution because it defines a constant function with zero slope for all \(x\). A vertical line such as \(x=0\) is not a function \(y(x)\) and is not an equilibrium solution; it is only a line where the field segments happen to be horizontal.

a) \(x=k\pi\) and \(y=m\pi\), for integers \(k,m\) b) Slopes form a checkerboard pattern: positive where the sine factors have the same sign and negative where they have opposite signs. c) The horizontal lines \(y=m\pi\) are equilibrium solutions; the vertical lines \(x=k\pi\) are not solution functions.

- Compare \(x^2+y^2\) with \(1\). - Differentiate the differential equation to classify a horizontal tangent. - A solution curve must follow the displayed field slope at every point.

1. Horizontal segments lie on \(x^2+y^2=1\). Slopes are positive outside the unit circle and negative inside it. 2. Differentiating gives \(y''=2x+2yy'\). At a horizontal tangent, \(y'=0\), so \(y''=2a\). The tangent is a local maximum for \(a<0\) and a local minimum for \(a>0\). 3. The unit circle is not the graph of a single function on its full domain. Even on a semicircle, its tangent slope is generally \(-x/y\), while the field slope on the circle is \(0\). Thus the circle is a nullcline, not a solution curve.

a) Zero slopes on \(x^2+y^2=1\); positive outside and negative inside b) Local maximum when \(a<0\); local minimum when \(a>0\) c) The circle is a nullcline, not a solution; its own tangent slope is generally not \(0\).

- Treat the product \(xy\) as one quantity. - Compare that product with \(-1\) to determine sign. - Before dividing by \(x\), check whether \(x=0\) can satisfy the isocline equation. - For \(m=1\), solve the product equation \(xy=0\) directly.

1. Horizontal segments satisfy \(xy=-1\), so \(y=-1/x\), with \(x\ne0\). 2. Slopes are positive where \(xy>-1\) and negative where \(xy<-1\). 3. Setting \(xy+1=m\) gives \(xy=m-1\). For \(m\ne1\), this is \(y=(m-1)/x\), with \(x\ne0\). For \(m=1\), the equation is \(xy=0\), so the isocline is the union of \(x=0\) and \(y=0\). 4. Tessa divided by \(x\), which omitted the entire vertical line \(x=0\).

a) \(y=-1/x\), \(x\ne0\) b) Positive where \(xy>-1\); negative where \(xy<-1\) c) For \(m\ne1\), \(y=(m-1)/x\), \(x\ne0\). For \(m=1\), the isocline is \(x=0\) together with \(y=0\). d) Tessa omitted \(x=0\) by dividing by \(x\).

- Substitute each point independently into the derivative formula. - Horizontal field segments have slope zero. - Solve the zero-slope equation for \(y\) to find the complete curve. - Replace zero by a general constant \(m\) to obtain the isocline family.

1. At \(A\), the slope is \(1-1=0\). At \(B\), it is \(2\cdot\frac12-1=0\). 2. At \(C\), the slope is \(3-1=2\). At \(D\), it is \(1\cdot\frac12-1=-\frac12\). 3. Thus \(A\) and \(B\) have horizontal segments. In general, zero slope requires \(ye^{-x}=1\), or \(y=e^x\). 4. Setting \(ye^{-x}-1=m\) gives the isocline family \(y=(m+1)e^x\).

a) \(m_A=0\), \(m_B=0\), \(m_C=2\), \(m_D=-\frac12\) b) \(A\) and \(B\) c) \(y=e^x\) d) \(y=(m+1)e^x\)

- On a fixed circle, the denominator is constant. - Use square inequalities to bound the product \(xy\). - Compare \(2|xy|\) with \(x^2+y^2\) to bound every slope.

1. Horizontal segments occur where \(xy=0\), so on both coordinate axes. Slopes are positive in Quadrants I and III and negative in Quadrants II and IV. 2. On the circle, the denominator is the constant \(1+R^2\), so extrema of the slope come from extrema of \(xy\). 3. Since \((x-y)^2\ge0\) and \((x+y)^2\ge0\), \(-\frac{R^2}{2}\le xy\le\frac{R^2}{2}\). The maximum occurs where \(x=y=\pm\frac{R}{\sqrt2}\), and the minimum occurs where \(x=-y=\pm\frac{R}{\sqrt2}\). 4. Also, \(2|xy|\le x^2+y^2\), so \(|y'|\le\frac{x^2+y^2}{1+x^2+y^2}<1\).

a) Both coordinate axes; positive in Quadrants I and III and negative in Quadrants II and IV b) Greatest at \(\left(\frac{R}{\sqrt2},\frac{R}{\sqrt2}\right)\) and \(\left(-\frac{R}{\sqrt2},-\frac{R}{\sqrt2}\right)\); least at \(\left(\frac{R}{\sqrt2},-\frac{R}{\sqrt2}\right)\) and \(\left(-\frac{R}{\sqrt2},\frac{R}{\sqrt2}\right)\) c) \(-1<y'<1\)

- Set the derivative equal to zero and solve for \(x\). - Compare \(x\) with \(y^2\) to determine the sign. - Differentiate the differential equation and then impose \(y'=0\). - Replace the zero slope by a general constant \(m\) for the isocline.

1. Horizontal segments satisfy \(x=y^2\). 2. Slopes are positive where \(x<y^2\) and negative where \(x>y^2\). 3. Differentiating gives \(y''=2yy'-1\). At a horizontal tangent, \(y'=0\), so \(y''=-1<0\). Thus every horizontal tangent is a local maximum. 4. Setting \(y^2-x=m\) gives \(x=y^2-m\).

a) \(x=y^2\) b) Positive to the left of the parabola and negative to the right c) Every horizontal tangent is a local maximum. d) \(x=y^2-m\)

- Determine when the exponential expression equals \(1\). - Use the sign of the product \(xy\) to organize the quadrants. - The logarithm in the isocline equation requires \(m+1>0\). - An equilibrium must be a constant function \(y(x)\), not a vertical line.

1. Horizontal segments satisfy \(e^{xy}=1\), so \(xy=0\). Thus both coordinate axes consist of horizontal segments. 2. Slopes are positive where \(xy>0\), in Quadrants I and III, and negative where \(xy<0\), in Quadrants II and IV. 3. Setting \(e^{xy}-1=m\) gives \(xy=\ln(m+1)\), which requires \(m>-1\). Therefore, every slope greater than \(-1\) is possible. 4. The x-axis, \(y=0\), is an equilibrium solution because it is a constant function with zero slope for every \(x\). The y-axis, \(x=0\), is vertical and is not a function \(y(x)\), so it is not an equilibrium solution.

a) The x-axis and y-axis b) Positive in Quadrants I and III; negative in Quadrants II and IV c) \(xy=\ln(m+1)\), for \(m>-1\) d) Only the x-axis is an equilibrium solution; the y-axis is a vertical zero-slope line, not a solution function.

- Use the zero and sign of the arctangent function. - Compare \(x+2y\) with zero to determine the sign regions. - Differentiate the slope formula and then impose the horizontal-tangent conditions. - Apply the tangent function when solving the isocline equation.

1. Horizontal segments satisfy \(x+2y=0\), so \(y=-x/2\). 2. Since arctangent has the sign of its input, slopes are positive above this line and negative below it. 3. Differentiating gives \(y''=\frac{1+2y'}{1+(x+2y)^2}\). At a horizontal tangent, \(x+2y=0\) and \(y'=0\), so \(y''=1>0\). Therefore, every horizontal tangent is a local minimum. 4. Setting \(\arctan(x+2y)=m\) gives \(x+2y=\tan m\), or \(y=(\tan m-x)/2\). The possible slopes satisfy \(-\pi/2<m<\pi/2\).

a) \(y=-x/2\) b) Positive above the line; negative below it c) Every horizontal tangent is a local minimum. d) \(y=\frac{\tan m-x}{2}\), for \(-\pi/2<m<\pi/2\)

- Identify the zero-slope row by solving \(3-y=0\). - Below that row the slopes are positive, but their magnitudes shrink as \(y\) approaches the row. - Use the equilibrium row and the field’s unique slope there to decide whether a solution can cross it.

1. Substitute the relevant coordinates or region into the differential equation. 2. The slopes shrink toward \(0\) as \(y\) approaches \(3\); the field indicates approach to the equilibrium rather than crossing it.

The slopes shrink toward \(0\) as \(y\) approaches \(3\); the field indicates approach to the equilibrium rather than crossing it.