This question requires using slope field reasoning to analyze solution behavior. For dy/dx = y - x at point (0,1), the slope is 1 - 0 = 1, so the curve is initially increasing. Moving slightly right from (0,1) to small positive x values, the slopes remain positive and even increase since y > x in that region, indicating the curve is concave up. Nearby points show slopes that increase as we move right and up, confirming concave up behavior. Choice A fails because the solution increases rather than decreases initially. To read slope fields effectively, first evaluate the slope at the given point, then examine how slopes change in the immediate neighborhood to determine concavity.

This question involves slope field reasoning to find points with horizontal tangents. For dy/dx = x - y, horizontal tangents occur when dy/dx = 0, so x - y = 0, giving y = x. We need to find which point lies on the line y = x: (1,0) gives 0 ≠ 1; (2,2) gives 2 = 2 ✓; (0,1) gives 1 ≠ 0; (-1,1) gives 1 ≠ -1; (1,2) gives 2 ≠ 1. Only (2,2) satisfies y = x, and we can verify: dy/dx = 2 - 2 = 0. Choice A fails because at (1,0), dy/dx = 1 - 0 = 1 ≠ 0. To locate points with horizontal tangents in slope fields, solve the differential equation equal to zero and identify which given points satisfy this condition.

This problem tests slope field evaluation at specific points. For dy/dx = x/(1+y²), we evaluate the slope at the point (0,2). Substituting x = 0 into the differential equation gives dy/dx = 0/(1+2²) = 0/5 = 0. A slope of 0 means the solution has a horizontal tangent at this point. The answer "vertical tangent" might seem plausible if one confuses the roles of numerator and denominator, but the denominator 1+y² is never zero. When evaluating slopes from differential equations, substitute the exact coordinates and simplify carefully to determine the tangent line's behavior.

This question involves analyzing a separable equation's behavior using slope field reasoning. For dy/dx = y/x with x > 0, at point (2,1) the slope is 1/2 > 0, indicating the solution increases. This differential equation has the property that along any ray y = kx (k > 0), the slope is k, which is constant and positive. Since we start with y/x = 1/2 and slopes are always positive for y > 0, the ratio y/x and hence y itself increases as x increases. The answer "decreases for all x > 0" fails because slopes are positive when y > 0. For equations of the form dy/dx = f(y/x), analyze how the ratio y/x evolves to understand solution behavior.

This question tests direct slope calculation from a differential equation. For dy/dx = x + y, we substitute the point (-1, 2) to find the slope. This gives dy/dx = -1 + 2 = 1 > 0, so the slope is positive. The solution curve passes through (-1, 2) with an upward (positive) slope. The answer "zero" might be tempting if one incorrectly thinks the negative x-value cancels the positive y-value exactly, but the calculation shows otherwise. When finding slopes at specific points in a slope field, always substitute both coordinates into the differential equation and evaluate carefully.

This question uses slope field reasoning to find horizontal slope segments. For dy/dx = y - 2x, horizontal slopes occur when dy/dx = 0, so y - 2x = 0, which gives y = 2x. Along this line, all slope segments are horizontal since the derivative equals zero there. We can verify: at any point (x, 2x) on this line, dy/dx = 2x - 2x = 0. Choice D fails because y = 2 gives dy/dx = 2 - 2x, which is horizontal only when x = 1, not along the entire line. To find horizontal slope segments in slope fields, solve the differential equation equal to zero and identify the resulting curve or line where dy/dx = 0.

This question requires slope field reasoning for cubic differential equations. For dy/dx = y³ with y(0) > 0, since y starts positive and dy/dx = y³ > 0 when y > 0, the solution increases. As y increases, y³ increases even more rapidly, making dy/dx larger, which accelerates the growth. This creates a positive feedback loop where increasing y makes dy/dx larger, causing y to increase faster. Therefore, solutions with positive initial conditions increase for all x where defined (though they may blow up in finite time). Choice A fails because positive y gives positive dy/dx, so solutions increase, not decrease. When analyzing slope fields for power functions, consider how the sign and magnitude of the solution affects the growth rate through the differential equation.

This question uses slope field reasoning to analyze slopes along horizontal lines. For dy/dx = cos y along the line y = π, we have dy/dx = cos(π) = -1 at every point on this line. Since cosine of π equals -1, all slope segments along y = π have slope -1, which is negative. The slope is constant along this horizontal line but definitely negative. Choice B fails because the slope is -1, not -1, and choice A fails because slopes are not horizontal (zero). When analyzing slope fields along specific lines, substitute the line's equation into the differential equation to find the slope value throughout that line.

This question uses slope field reasoning about exponential-type behavior. For dy/dx = y - 1 at point (0,0), we have dy/dx = 0 - 1 = -1 < 0. A negative slope indicates the solution is decreasing at x = 0. Moving slightly right from (0,0), y becomes more negative while x remains small, so dy/dx = y - 1 becomes more negative, maintaining the decreasing trend. This creates exponential decay behavior moving away from the equilibrium y = 1. Choice A fails because the slope is negative, indicating decreasing rather than increasing behavior. When analyzing solution behavior at specific points in slope fields, evaluate the differential equation there and consider how the slope changes as the solution evolves.

This question requires slope field reasoning for solutions in specific regions. For dy/dx = y/x in Quadrant I, both y > 0 and x > 0, so dy/dx = (positive)/(positive) > 0. Therefore, all solutions in Quadrant I have positive slope everywhere, meaning they are always increasing. The magnitude of the slope equals y/x, which varies depending on the specific values but is always positive. Choice A fails because slopes are positive, not negative, throughout Quadrant I. To analyze solution behavior in different regions using slope fields, determine the sign of the differential equation in each quadrant based on the signs of the variables involved.