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Published byMerilyn Parker Modified over 9 years ago
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Warm Up Write as an inequality and interval notation.
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Graphs of Quadratic Inequalities
I can graph and find the solution set of quadratic inequalities.
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Remember having to graph
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Remember having to graph
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Steps for Graphing (quickly)
Graph the quadratic (use your calculator for points) For <> use DASHED for ≤≥ use SOLID line Locate the roots/solutions of the quadratic 4. Shade the appropriate region using a test point
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Graph: y ≤ x2 + 6x – 4 * Vertex: (-3,-13) * Solid Line
* Roots: X = {-6.6, 0.6} * Interval Notation:
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Graph: y > -x2 + 4x – 3 Vertex: Solid or Dotted: Roots: Test Point:
Interval Notation:
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Graph: y ≥ x2 – 8x + 12 Vertex: Solid or Dotted: Roots: Test Point:
Interval Notation:
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Graph: y > -x2 + 4x + 5 Vertex: Solid or Dotted: Roots: Test Point:
Interval Notation:
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Warm Up Pass HW forward Factor and solve: 1. x2 – 5x = – 4
2. -x2 + 7x = 12 (hint: use rooftop) Sketch the graph -(x – 1)2 – 3
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Solving a Quadratic Inequality
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Steps for solving Write the original inequality as an equation
Set equal to 0, factor, and solve. Plot the points on a number line and test points in each interval back into the original inequality. Write the answer (the TRUE part) as an inequality
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Solve: x2 – 5x ≤ – 4 x2 – 5x = -4 x2 – 5x + 4 = 0
False True False Answer: 1 ≤ x ≤ 4 Interval Notation:
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Solve: -x2 + 7x < 12 -x2 + 7x = 12 -x2 + 7x – 12 = 0
False True True Answer: x < 3 or x > 4 Interval:
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What if it can’t factor? Graph it!
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Solve: -(x – 1)2 – 3 < 0 -(x – 1)2 – 3 < y y > -(x – 1)2 – 3
Answer: all real numbers
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Solve: x2 + 4 ≤ 0 x2 + 4 ≤ y y ≥ x2 + 4 Answer: no solution
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