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Graphing linear functions using a single point and slope
- Lesson: 1a1:01
- Lesson: 1b1:48
- Lesson: 1c0:45
- Lesson: 1d0:42
- Lesson: 1e1:15
- Lesson: 1f0:57
- Lesson: 2a1:21
- Lesson: 2b1:08
- Lesson: 2c1:07
- Lesson: 2d0:46
Graphing linear functions using a single point and slope
If we are given a single point and the slope of a straight line, we will be able to graph it. By the same token, we can determine the slope and intercepts if we are given from its graph.
Basic Concepts: Solving linear equations by graphing, Slope intercept form: y = mx + b, Point-slope form: y−y1=m(x−x1)
Related Concepts: Special case of linear equations: Horizontal lines, Special case of linear equations: Vertical lines, Parallel line equation, Perpendicular line equation
Lessons
- 1.Graph the following line that passes through the given point and the slopea)(−3,2);m =32b)(4,−2);m = -53c)(1,2);m = undefinedd)(2,5);m = 0e)x int = 2; m = -1f)y int = -3; m = 23
- 2.Find the slope, x and y intercepts of each line.a)
b)
c)
d)
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8.
Linear Functions
8.1
Distance formula: d=(x2−x1)2+(y2−y1)2
8.2
Midpoint formula: M=(2x1+x2,2y1+y2)
8.3
Gradient equation: m=x2−x1y2−y1
8.4
Gradient intercept form: y = mx + b
8.5
General form: Ax + By + C = 0
8.6
Gradient-point form: y−y1=m(x−x1)
8.7
Rate of change
8.8
Graphing linear functions using table of values
8.9
Graphing linear functions using x- and y-intercepts
8.10
Graphing from slope-intercept form y=mx+b
8.11
Graphing linear functions using a single point and gradient
8.12
Word problems of graphing linear functions
8.13
Parallel and perpendicular lines in linear functions
8.14
Applications of linear relations
8.15
Perpendicular line equation