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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. 9x 2 +6x+1=0 9x 2 +6x-4=0.
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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
Forms of Quadratic Inequalitiesy<ax2+bx+c y>ax2+bx+cy≤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 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 • 0 • -3 * Test point (0,0) 0>-02+4(0)-3 0>-3
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.