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Math I UNIT QUESTION: What is a quadratic function? Standard: MM2A3, MM2A4 Today’s Question: How do you solve quadratic inequalities by algebra or a graph? Standard: MM2A4.d.
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3.9 Graphs of Quadratic Inequalities Day 1 p. 96-101
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Warm Up Write the expression as a complex number in standard form Determine if the following are true or false: 4. 4 -2 Graph the quadratic.
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Forms of Quadratic Inequalities y<ax 2 +bx+cy>ax 2 +bx+c y≤ax 2 +bx+cy≥ax 2 +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.
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Steps for graphing 1. Sketch the parabola y=ax 2 +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)
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Example: Graph y ≤ x 2 +6x- 4 * Vertex: (-3,-13) * Opens up, solid line Test Point: (0,0) 0≤0 2 +6(0)-4 0≤-4 So, shade where the point is NOT! Test point
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Graph: y>-x 2 +4x-3 * Opens down, dotted line. * Vertex: (2,1) * Test point (0,0) 0>-0 2 +4(0)-3 0>-3 x y 0 -3 1 0 2 1 3 0 4 -3 Test Point
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Day 2
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Last Example! Sketch the intersection of the given inequalities. 1 y≥x 2 and 2 y≤-x 2 +2x+4 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. SOLUTION!
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Steps for using algebra 1. Write the original inequality as an equation. 2. Write the equation in standard form by setting it equal to 0. 3. Factor. 4. Use the zero product property to find the critical x-values. 5. Plot the points on a number line and test points in each interval. 6. State the inequality.
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Example 1: Solve: x 2 - 5x ≤ - 4 x 2 – 5x = 4 x 2 – 5x + 4 = 0 (x – 4) (x – 1) = 0 x = 1, 4 14 XX√ Answer: 1≤x≤4
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Example 2: Solve: -x 2 + 7x < 12 -x 2 + 7x = 12 -x 2 + 7x - 12 = 0 x 2 - 7x + 12 = 0 (x – 4) (x – 3) = 0 x = 3, 4 34 √√X Answer: x 4
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Practice Problems From “Solving Quadratic Inequalities” Worksheet
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Day 1 - p. 98 #5-10, 14; p. 100 #7, 11 Day 2 - p. 101 #22-26 Homework
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