Sketching Slope Fields

The derivative is the slope of the tangent line: rise over run is -2.

• Correct: falls 2 units for each unit to the right:
• rises 2 units for each unit to the right:
• is horizontal:
• falls 1 unit for each 2 units to the right:

The derivative of $x^2 + 4$ is $2x$.

• Correct: $y = x^2 + 4$:
• $y = 2x^2$:
• $y = 2$:
• $y = x^2 + x$:

When $\frac{dy}{dx}$ depends on $x$ alone, moving up or down a column does not change the slope.

• The segments in each horizontal row are parallel.: Along a row $x$ changes, so $\cos x$ changes.
• The segments share one slope everywhere.: $\cos x$ varies from $-1$ to $1$.
• The segments are horizontal on the x-axis.: On the $x$-axis the slope is $\cos x$, 0 at $x = \frac{\pi}{2}$ but 1 at $x = 0$.
• Correct: The segments in each vertical column are parallel.: The slope depends on $x$ alone, so it is the same for every $y$ in a column.

Zero numerator gives slope 0; zero denominator gives an undefined, vertical slope.

• Correct: Segments are vertical at points on the $x$-axis with $x \ne 0$.: There $y = 0$ and the slope $\frac{x}{0}$ is undefined.
• Correct: Segments are horizontal at points on the $y$-axis with $y \ne 0$.: There the slope is $\frac{0}{y} = 0$.
• Slopes are positive at points with $x < 0$ and $y > 0$.: A negative over a positive is negative.
• The slope at $(2, 4)$ is 2.: $\frac{2}{4} = \frac{1}{2}$; the fraction was flipped.

At $(1, 2)$: $1 - 2^2 = -3$. The student added instead of subtracting $y^2$.

• The slope there is $1 - 2 = -1$, so the segment should fall gently.: $y$ is squared: $2^2 = 4$.
• Correct: The slope there is $1 - 4 = -3$, so the segment should fall.: $x - y^2 = 1 - 2^2 = -3$.
• Nothing: $1 + 2 = 3$.: The equation subtracts $y^2$; it does not add $y$.
• The slope there is $2 - 1 = 1$.: The coordinates were swapped and $y$ was not squared.