Math 30 - Day 11 - Graphing Linear Functions part 2

Overview

Christopher Lee reviews how to construct parallel and perpendicular lines, find their slopes, and solve linear systems by identifying the lines’ intersection. He then introduces quadratic functions, explaining parabola features, general and vertex forms, transformations, and how to derive a quadratic equation from its graph.

Key takeaways

Parallel lines have equal slopes, while nonvertical perpendicular lines have slopes that are negative reciprocals; for example, slope 3 pairs with -1/3.
To find the intersection of two nonparallel lines, set their equations equal, solve for the input, and substitute it back to find the output.
For a quadratic in general form ax² + bx + c, the axis of symmetry is x = -b/(2a); evaluating the function there gives the vertex’s y-coordinate.
Vertex form y = a(x - h)² + k directly identifies the vertex (h, k), opening direction, and transformations from y = x².
A quadratic equation can be reconstructed from a graph by reading its vertex, substituting one additional point to find a, and expanding if general form is needed.

Chapters

0:00 Recap: Writing a Parallel Line Through a Given Point

Find the slope of the given line, then use that same slope for any parallel line.
Substitute the new line’s slope and the given point into point-slope form or slope-intercept form.
Simplify to obtain the equation of the required line.

1:50 Example: Parallel to y = 3x + 6 Through (3, 0)

The given line y = 3x + 6 has slope 3, so a parallel line also has slope 3.
Using point-slope form with (3, 0) gives y - 0 = 3(x - 3).
Simplifying yields y = 3x - 9, which has the same slope and a different y-intercept.

9:02 Graphing Lines with Three Convenient Points

For a line in slope-intercept form, start at the y-intercept and use the slope to locate additional points.
When the slope is a whole number, treat its denominator as 1 and use horizontal steps of 1.
Three points can help check a graph, though two distinct points determine a line.

10:45 Perpendicular Lines Use Negative Reciprocal Slopes

To find a perpendicular slope, take the negative reciprocal of the original slope.
For example, slope 2 is 2/1, so its negative reciprocal is -1/2.
Lines with slopes that are negative reciprocals meet at right angles.

12:32 Why Horizontal and Vertical Lines Are a Special Case

A horizontal line has slope 0, while a vertical line has undefined slope.
Horizontal and vertical lines are perpendicular even though their slopes do not have a product of -1.
The negative-reciprocal product rule is stated for line functions; a vertical line is not a function of x.

13:16 Five Steps for Finding a Perpendicular Line

Find the original slope and calculate its negative reciprocal.
Use the new slope and given point in y = mx + b or point-slope form.
Solve for b when using slope-intercept form, then write the resulting line equation.

15:08 Example: Perpendicular to y = 3x + 3

The original slope is 3, so the perpendicular slope is -1/3.
For a line through (3, 0), substitute into y = -1/3x + b to obtain b = 1.
The resulting perpendicular line is y = -1/3x + 1.

21:49 Parallel and Perpendicular Lines Through the Origin

For h(x) = 2x - 4, the parallel slope is 2 and the perpendicular slope is -1/2.
A line through (0, 0) has y-intercept 0, so the parallel equation is y = 2x.
The perpendicular equation through the origin is y = -1/2x.

30:30 Finding a Perpendicular Slope from Two Points

Calculate the original line’s slope from two points using change in y divided by change in x.
For points (-2, 6) and (4, 5), the slope is (5 - 6) / (4 - (-2)) = -1/6.
The perpendicular slope is 6, which can be used with a third point to form the new line.

32:00 Example: Perpendicular Through (4, 5)

The line through (-2, 6) and (4, 5) has slope -1/6, so its perpendicular slope is 6.
Substituting (4, 5) into y = 6x + b gives 5 = 24 + b and b = -19.
The required perpendicular line is y = 6x - 19.

38:35 Practice: Perpendicular to the Line Through Two Points

For the points (-2, 15) and (2, 3), the original slope is (3 - 15) / (2 - (-2)) = -3.
The negative reciprocal is 1/3; using the required point (6, 4) gives b = 2.
The perpendicular line is y = 1/3x + 2.

51:01 Systems of Linear Equations and Intersection Points

A system contains two or more linear equations; nonparallel lines intersect at one point.
Coincident lines are the same line and share every point, while distinct parallel lines do not intersect.
The intersection point satisfies both equations, so it can be found by setting their formulas equal.

52:11 Solve a Linear System by Setting Equations Equal

For h(t) = 3t - 4 and j(t) = 5 - t, set 3t - 4 = 5 - t.
Solving gives t = 9/4; substituting into either equation gives an output of 11/4.
The intersection is (9/4, 11/4), using the input-output order.

1:02:18 Recognizing No Solution for Distinct Parallel Lines

Setting the equations of distinct parallel lines equal ultimately produces a contradiction.
An impossible statement such as 0 = 6 signals that the system has no solution.
A graph can confirm the result because the two distinct parallel lines never intersect.

1:03:21 Read a Line’s Intercepts, Slope, and Behavior from Its Graph

For the orange line J(t), the graph crosses the y-axis at (0, 5) and the x-axis at (5, 0).
Using those intercepts gives slope (0 - 5) / (5 - 0) = -1.
Because the line descends from left to right, J(t) is decreasing.

1:11:21 Compare Line Slopes and Describe a Transformation

The comparison line H(t) has slope 3, so a line with slope -1 is neither parallel nor perpendicular to it.
Parallel lines have equal slopes; a line perpendicular to slope 3 would have slope -1/3.
Relative to the identity function f(t) = t, J(t) = -t + 5 reflects the slope and shifts the graph up 5 units.

1:19:40 Quadratic Functions Model Parabolas and Real-World Shapes

A quadratic function graphs as a U-shaped or upside-down U-shaped curve called a parabola.
Parabolas can model the cross-section of a curved satellite antenna, as well as area and projectile motion.
The lesson shifts from linear functions to quadratic graphs and their features.

1:26:32 Identify a Parabola’s Vertex and Axis of Symmetry

An upward-opening parabola has a minimum at its vertex; a downward-opening parabola has a maximum.
The vertex is the turning point, and the vertical line through it is the axis of symmetry.
The y-intercept occurs where the graph crosses the y-axis; x-intercepts, when present, are zeros of the function.

1:30:11 Read Vertex, Symmetry Axis, Zeros, and Y-Intercept

For the illustrated graph, the vertex is (3, 1), so the axis of symmetry is x = 3.
The graph stays above the x-axis, so it has no real zeros or x-intercepts.
Its y-intercept is (0, 7).

1:35:45 General Form Gives the Axis of Symmetry

In general form, f(x) = ax² + bx + c, with a not equal to zero; positive a opens upward and negative a opens downward.
The axis of symmetry is x = -b/(2a), and the vertex lies on that line.
For y = x² + 4x + 3, the axis is x = -2; setting the function to zero gives x-intercepts at (-3, 0) and (-1, 0).

1:40:22 Vertex Form Reveals a Quadratic’s Turning Point

Vertex form is y = a(x - h)² + k, where the vertex is (h, k).
If a is positive, the parabola opens upward and the vertex is a minimum; if a is negative, it opens downward and the vertex is a maximum.
For y = -3(x + 2)² + 4, the vertex is (-2, 4) and the parabola opens downward.

1:44:33 Quadratic Transformations and Converting Between Forms

In y = a(x - h)² + k, h controls the horizontal shift and k controls the vertical shift.
The sign of a determines opening direction; |a| greater than 1 makes the parabola narrower, while 0 < |a| < 1 makes it wider.
Expanding vertex form connects it to ax² + bx + c; the vertex’s x-coordinate is -b/(2a), and its y-coordinate is the function value there.

1:49:34 Procedure: Build a Quadratic Equation from Its Graph

Read the vertex from the graph to identify h and k in y = a(x - h)² + k.
Substitute another point on the graph to solve for the stretch factor a.
Use the parabola’s opening to check the sign of a, then expand if general form is required.

1:52:21 Example: Derive and Expand a Quadratic from Its Graph

The graph has vertex (-2, -3), so its equation begins y = a(x + 2)² - 3.
Substituting the point (0, -1) gives -1 = 4a - 3, so a = 1/2.
Vertex form is y = 1/2(x + 2)² - 3; expanding gives y = 1/2x² + 2x - 1.

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Christopher Lee.