Upcoming SlideShare
×

# 5.8 Graphing quadratic inequalities

602 views

Published on

Published in: Education, Technology
0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
Your message goes here
• Be the first to comment

• Be the first to like this

Views
Total views
602
On SlideShare
0
From Embeds
0
Number of Embeds
7
Actions
Shares
0
4
0
Likes
0
Embeds 0
No embeds

No notes for slide

### 5.8 Graphing quadratic inequalities

1. 1. 5.8 Graphs of Quadratic Inequalities p. 312
2. 2. Forms of Quadratic Inequalities y<ax2+bx+c y>ax2+bx+c y≤ax2+bx+c y≥ax2+bx+c   Graphs will look like a parabola with a solid or dotted line and a shaded section. The graph could be shaded inside the parabola or outside.
3. 3. Steps for graphing 1. Sketch the parabola y=ax2+bx+c (dotted line for < or >, solid line for ≤ or ≥) ** remember to use 5 points for the graph! 2. Choose a test point and see whether it is a solution of the inequality. 3. Shade the appropriate region. (if the point is a solution, shade where the point is, if it’s not a solution, shade the other region)
4. 4. Example: Graph y ≤ x2+6x- 4 * Opens up, solid line x * Vertex: (-3,-13) −b −6 x= = = −3 2a 2(1) y = (−3) 2 + 6(−3) − 4 = 9 − 18 − 4 = −13 •Test Point: (0,0) 0≤02+6(0)-4 0≤-4 -1 -2 -3 -4 -5 Test point y -9 - 12 - 13 - 12 -9 So, shade where the point is NOT!
5. 5. Graph: y>-x2+4x-3 * Opens down, dotted line. x * Vertex: (2,1) y −b −4 x= = =2 2a 2(−1) 0 -3 1 0 y = − 1(2) 2 + 4(2) − 3 y = −4 + 8 − 3 = 1 2 1 3 0 4 -3 * Test point (0,0) 0>-02+4(0)-3 0>-3 Test Point
6. 6. Last Example! Sketch the intersection of the given inequalities. 1 y≥x2 and 2 y≤-x2+2x+4 SOLUTION!  Graph both on the same coordinate plane. The place where the shadings overlap is the solution. Vertex of #1: (0,0) Other points: (-2,4), (-1,1), (1,1), (2,4)  Vertex of #2: (1,5) Other points: (-1,1), (0,4), (2,4), (3,1)  * Test point (1,0): doesn’t work in #1, works in #2.
7. 7. Last Example! Sketch the intersection of the given inequalities. 1 y≥x2 and 2 y≤-x2+2x+4 SOLUTION!  Graph both on the same coordinate plane. The place where the shadings overlap is the solution. Vertex of #1: (0,0) Other points: (-2,4), (-1,1), (1,1), (2,4)  Vertex of #2: (1,5) Other points: (-1,1), (0,4), (2,4), (3,1)  * Test point (1,0): doesn’t work in #1, works in #2.