on the real number line and on cartesian coordinates presented by: vicki angel and jan whisonant...
TRANSCRIPT
GRAPHING
onThe Real Number Line
and onCartesian Coordinates
Presented by:Vicki Angel and Jan WhisonantTransitions Math ConferenceAustin, TXMay 6-7, 2011
OBJECTIVE
The participants will learn strategies for teaching graphing of equations and
inequalities in one and two variables.
Points, intervals, lines, and parabolas will be included.
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For each real number there corresponds exactly
one point on the real number line and for each
point on the real number line there corresponds
exactly one real number.
X = -2
Solution to an equation in one unknown
X < 2
Solution to an inequality in one unknown
x ≤ -2 or x ≥ 1
Solution of a compound inequality
-3 ≤ x ≤ 4
Solution of a compound inequality
Linear Equations
Standard Form
ax + by + c = 0
Slope- Intercept Form
y = mx + b
m = slope b = y intercept
Graphing Methodsfor Linear Equations
1. Find both the x and y intercepts2. Use the y intercept and the slope3. Use a T chart, selecting values for x and then finding the values of y for each x
Graphing using Intercepts
y = - 4x + 8
If x = 0, then y = 8y = -4(0) + 8y = 0 + 8 = 8
If y = 0, then x = 20 = - 4x + 84x = 8, x = 2
Graphing using the slope and y intercept
y = 5/4 x – 2
y intercept = - 2
slope = 5/4
Graphing using T Chart
y = 3/2 x + 3
x y
0 3
2 6
- 4 - 3
Linear Inequalities 1. Graph the inequality by replacing the
inequality with an = sign and graph the line just like it was an equation…using any of the three methods we discussed
2. If the inequity is < or > make the line a dotted line.
3. Use a test point to determine which side of the line to shade.
Using a Test Point
Select a point that is not on the line. Substitute the x and y values of the point
into the inequality. If those values make the inequality true,
then shade the side of the line that the point is on.
If those values make the inequality false, then shade the opposite side of the line from the selected point.
y ≤ - 8/3 x + 2
y intercept = 2
slope = - 8/3
y > 6x + 4
y intercept = 4
slope = 6/1
Graphing a quadratic equation
Find the vertex and at least two other points on the graph, then use the axis of symmetry to add two more points to the graph.
When the equation is in the form
The vertex is (h , k) and the line of symmetry is x = h
)0(a )( 2 khxay
y = x²
x y
0 0
2 4
- 3 9
Vertex (3,4) Line of symmetry: x=3
x y
4
1
2
- 4
4 )32(- 2 xy
Vertex: (1, -5) Line of symmetry: x=1
7
-2
3
0
x y
2 - 6 - 3 2 xxy
x
- 4 1
y
- 1
1 6
- 2
y ≥ (x+2)² - 3