The discriminant tells you how many

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Solving Systems of Inequalities by Graphing

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Presentation transcript:

The discriminant tells you how many The discriminant tells you how many solutions and what type you will have. If the discriminant:  Is positive – 2 real solutions  Is negative – 2 imaginary solutions  Is zero – 1 real solution

Find the discriminant and give the number and type of solutions. 9x2+6x+1=0 9x2+6x-4=0

5.7 Graphs of Quadratic Inequalities

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.

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)

Example: Graph y ≤ x2+6x- 4 * Opens up, solid line * Vertex: (-3,-13) Test point * Opens up, solid line * Vertex: (-3,-13) Test Point: (0,0) 0≤02+6(0)-4 0≤-4 So, shade where the point is NOT!

Graph: y>-x2+4x-3 Test Point * Opens down, dotted line. * Vertex: (2,1) x y 0 -3 1 0 1 -3 * Test point (0,0) 0>-02+4(0)-3 0>-3

Last Example. Sketch the intersection of the given inequalities 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.