A dynamic practice lesson focused on finding slope directly from graphs by counting vertical rise and horizontal run. Students rotate through interactive challenge stations and consolidate their skills with a structured practice worksheet.
Four lines have slopes: Line W has \( m = -\frac{3}{4} \), Line X has \( m = 4 \), Line Y has \( m = 0 \), and Line Z has \( m = -5 \).
a) Which line is the steepest overall?
b) Which line is completely flat?
Graph G (Drone Launch): Change in height = \(+20\text{ m}\), time = \(+4\text{ s}\). Rate: \( \frac{20\text{ m}}{4\text{ s}} = \mathbf{+5\text{ m/s}} \).
Graph H (Tank Draining): Change in volume = \(-40\text{ gal}\), time = \(+5\text{ min}\). Rate: \( \frac{-40\text{ gal}}{5\text{ min}} = \mathbf{-8\text{ gal/min}} \).
Challenge Answers:
1. The drone climbs 5 meters in elevation every second.
2. Rate is negative because volume decreases; exactly 8 gallons drain out each minute.
Station Rotation Tip: Give a 1-minute warning before rotations. Ensure students record group consensus slopes. See Page 2 for Practice Worksheet Solutions →
Practice Worksheet Solutions
Complete Problem Key & Common Misconception Guide
Answer Key • Worksheet
Problems 1–8 Step-by-Step Calculation Master
#
Coordinate Points
Rise (\(\Delta y\))
Run (\(\Delta x\))
Unsimplified
Final Slope (\(m\))
Behavior
1
\((0, -1)\) and \((2, 3)\)
+4
+2
\( \frac{4}{2} \)
\( m = 2 \)
Rises left-to-right (steep)
2
\((-3, 2)\) and \((1, -2)\)
-4
+4
\( \frac{-4}{4} \)
\( m = -1 \)
Falls left-to-right (45°)
3
\((-2, -3)\) and \((2, -1)\)
+2
+4
\( \frac{2}{4} \)
\( m = \frac{1}{2} \)
Gentle uphill incline
4
\((-1, 3)\) and \((2, -3)\)
-6
+3
\( \frac{-6}{3} \)
\( m = -2 \)
Steep downhill fall
5
\((-3, 1)\) and \((3, 3)\)
+2
+6
\( \frac{2}{6} \)
\( m = \frac{1}{3} \)
Gradual uphill climb
6
\((-2, 2)\) and \((2, -1)\)
-3
+4
\( \frac{-3}{4} \)
\( m = -\frac{3}{4} \)
Downhill slope
7
\((-3, -2)\) and \((3, -2)\)
0
+6
\( \frac{0}{6} \)
\( m = 0 \)
Horizontal line (\( y = -2 \))
8
\((2, -3)\) and \((2, 2)\)
+5
0
\( \frac{5}{0} \)
Undefined
Vertical line (\( x = 2 \))
Problem 9: Error Analysis Key
Mistake 1 (Inverted Fraction): Marcus calculated \(\frac{\text{run}}{\text{rise}} = \frac{3}{2}\) instead of \(\frac{\text{rise}}{\text{run}}\).
Mistake 2 (Missing Sign): The line falls left-to-right, so the vertical rise is negative (\(-2\)).
Correct Slope: \( m = \frac{-2}{3} = \mathbf{-\frac{2}{3}} \).
Problem 10: Steepness Key
Given Slopes: \( W = -\frac{3}{4} \), \( X = 4 \), \( Y = 0 \), \( Z = -5 \).
a) Steepest Line:Line Z. Absolute value determines steepness: \( |-5| = 5 \), which is greater than \( |4| = 4 \) and \( |-\frac{3}{4}| = 0.75 \).
b) Completely Flat Line:Line Y (\( m = 0 \)).
Quick Intervention Guide for Common Student Stumbling Blocks
Flipping Rise & Run: Students put horizontal change on top because coordinate pairs write \( x \) first \((x, y)\). Reinforce: "Rise to the sky before you run to the right!"
Zero vs Undefined: Use the "NO vs OK" mnemonic: \( \frac{0}{k} = 0 \) is OK (a flat road you can walk on). \( \frac{n}{0} = \) NO! (an impossible vertical wall).
Sign Confusion: Train students to trace the line with their finger from left to right like reading. Moving downhill immediately means the slope must be negative.
