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1
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2
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- x and y are most commonly used as variables in an equation
- Variables are placeholders for values
- Variable y is dependent on the variable x
- The independent variable, x, is plotted along the horizontal axis (x
axis)
- The dependent variable, y, is plotted along the vertical axis (y axis)
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3
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- The graph of an equation consists of all the points which satisfy or
solve the equation.
- Solve the equation to find the points,then plot the points.
- These points are called coordinates.
- Each point (coordinate) consists of one value for y and one value for x,
and is written as the ordered pair
(x, y).
- The values of (x, y) solve the equation y = x + c
- where c represents a constant
value
- The coordinates are ordered pairs of real numbers.
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4
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- A linear equation is a first degree equation (the highest power exponent
is 1)
- Examples: y = x + 2, 3y = 4x + 2, y = mx + b,
- and Ax + By + C = 0
- where A an B
are not equal to 0
- The graph of a linear equation is a non vertical straight line
- Example: y = 2x + 3 is the graph of a
straight line
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5
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- Linear graphs are drawn by solving the equations and plotting the points
- All the points on the graph solve the equation
- Only two points are needed to graph a straight line
- Find a third point as a check for accuracy
- Draw the graph through the points
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6
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- Consider the linear equation y
= x + 2
- The y value is dependent on the value of x
- Find ordered pairs by assigning any value to x
- Let x = 0 and solve for y to find (0, y)
- y = 0 + 2 or y = 2 The ordered pair is (0, 2)
- Let x = 1 and solve again for y to find (1, y)
- y = 1 + 2 or y = 3 The ordered pair is (1, 3)
- Let x = -2 and solve again for y to find (-2, y)
- y = -2 + 2 or y = 0 The ordered pair is (-2, 0)
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7
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- Solutions for y = x + 2 put in a table form
- x y
- 0 2
- 1 3
- -2 0
- -1 1
- Graph of y = x + 2
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8
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- Set up a chart and substitute values for x, solving for y
- Let x = 0
- Let x = 1
- Let x = -1
- y is dependent on x
- Chart of Ordered Pairs
- x
y pair
- 0 3 (0, 3)
- 1 5 (1, 5)
- -1 1 (-1, 1)
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9
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10
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- The slope-intercept formula for a straight line is y = mx + b
- m and b are constant real numbers
- m is the slope of the line
- if m is positive, the line slopes up
- if m is negative, the line slopes down
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11
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- The y intercept of a line y = mx + b is the point where it crosses
the y axis (the point where x =
0).
- Let x = 0 and solve for y
- the x coordinate of this point is, by definition, 0
- since the y axis represents the zero value for x
- The x intercept of a line y = mx + b is the point where it crosses
the x axis (the point where y =
0).
- Let y = 0 and solve for x
- the y coordinate of this point is, by definition, 0
- since the x axis represents the zero value for y
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12
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- If the slope of a line is 3 and the y intercept is -2, then
- the equation of the line becomes
y = 3x - 2
- The ordered pair representing the y intercept is (0, -2)
- The y intercept is found by solving for y when x = 0.
- So when x = 0, y = 3(0)
-2 or y = -2
- The slope, m, equals 3 means that there is a change in x of 1 unit for each change in y of 3 units.
- x goes from 0 to 1 (add 1 to the x value of the y
intercept)
- y goes from -2 to 1
(add 3 to the y value of the y intercept)
- Another ordered pair on the graph is
(1, 1).
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13
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- Given any two points on a line, (x1, y1) and (x2,
y2)
- m is the difference between the y values divided by the difference
between the x values or
- m is the ratio of the change in y coordinates to the change in x
coordinates
- The slope of a vertical line is undefined
- The denominator, xi -
xj, would have the value of zero
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14
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- For a straight line we can make
- a table from a graph or an equation
- a graph from an equation or a table
- an equation from a graph or a table
- Use the difference in y values divided by the difference in x values
(point-slope formula)
- Use the slope-intercept formula
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15
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- Graph a line from
- an equation
- Solve for the y intercept
- Solve for the x intercept
- Draw a line through both
- Find a third point to check the accuracy
- Write the equation of a line
- find two points on the line and substitute the x and y values into the
point-slope formula or
- find two points on the line and substitute the x and y values into
the slope-intercept formula
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16
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- For the following exercises, you will need paper to sketch the
- axes as a vertical and a horizontal line.
- Squared paper is ideal.
- Lined paper is OK, too. Count the printed lines as increments of one
unit up and down, and sketch in the same increments side to side.
- Even plain paper works: a ruler is useful and you could use 1 cm
increments to mark off each axis.
- These are only sketches, DO NOT take measurements from the paper as a
draftsman might.
- You will need at least 7 increments in each direction of both axes.
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17
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- 1. Plot the points (0, 0) “the origin”, and (4, 3) i.e., x = 4 and y = 3.
- Draw a line segment between them.
- Draw another line segment from (4, 3)
- vertically to (4, 0).
- What are the lengths of each side of the resulting
- triangle? Hint: The figure is a right triangle and
- you can apply the Pythagorean Theorem.
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18
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- 2. On the same set of axes as Exercise 1, plot the points (2, 1), (-2,
-2), and (2, -2).
- Connect these three points with line segments.
- What is different?
- 3. On a new set of axes, plot the points (0, 0) and (3, 4).
- Connect these points. Is this
the same line segment as in
- Exercise 1?
- Plot the point (-3, 4).
- The slope of a line from (0, 0) to (3, 4) is , that is .
- What would be the slope of a line from (0, 0) to (-3, 4)?
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19
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- 4. On a new set of axes, plot the points (6, -6), (-6, -1), and (6, -1).
- 5. Find the distances between the points.
- 6. Find the slope of the line between the first two points.
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