Upload
merilyn-parker
View
228
Download
0
Tags:
Embed Size (px)
Citation preview
Warm UpWrite as an inequality and interval notation. 5 4x
5 or 6x x
5 or 0x x
7 4x
Graphs of Quadratic Inequalities
I can graph and find the solution set of quadratic
inequalities.
Remember having to graph
24
3y x
Remember having to graph
13
2y x
Steps for Graphing Steps for Graphing (quickly)(quickly)
1. Graph the quadratic (use your calculator for points)
2. For <> use DASHED for ≤≥ use SOLID line
3. Locate the roots/solutions of the quadratic
4. Shade the appropriate region using a test point
Graph: y ≤ x2 + 6x – 4* Vertex: (-3,-13)
* Solid Line
* Roots: X = {-6.6, 0.6}
* Interval Notation:
Graph: y > -x2 + 4x – 3
Vertex:
Solid or Dotted:
Roots:
Test Point:
Interval Notation:
Graph: y ≥ x2 – 8x + 12
Vertex:
Solid or Dotted:
Roots:
Test Point:
Interval Notation:
Graph: y > -x2 + 4x + 5Vertex:
Solid or Dotted:
Roots:
Test Point:
Interval Notation:
Warm UpPass HW forward
• Factor and solve:1. x2 – 5x = – 42. -x2 + 7x = 12 (hint: use rooftop)
• Sketch the graph-(x – 1)2 – 3
Solving a Solving a Quadratic Quadratic InequalityInequality
Steps for solvingSteps for solving
1. Write the original inequality as an equation
2. Set equal to 0, factor, and solve.3. Plot the points on a number line
and test points in each interval back into the original inequality.
4. Write the answer (the TRUE part) as an inequality
Solve: x2 – 5x ≤ – 4x2 – 5x = -4
Answer: 1 ≤ x ≤ 4
Interval Notation:
x2 – 5x + 4 = 0(x – 4) (x – 1) = 0
x = 1, 4
False True False
Solve: -x2 + 7x < 12-x2 + 7x = 12
Answer: x < 3 or x > 4
Interval:
-x2 + 7x – 12 = 0
(x – 4) (x – 3) = 0
x = 3, 4
True Fa
lse
True
What if it What if it can’t can’t
factor?factor?Graph it!Graph it!
Solve: -(x – 1)2 – 3 < 0-(x – 1)2 – 3 < y
y > -(x – 1)2 – 3
Answer: all real numbers
Solve: x2 + 4 ≤ 0x2 + 4 ≤ y
y ≥ x2 + 4
Answer: no solution