1. 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

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

3. 5.7 Graphs of Quadratic Inequalities

4. 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.

5. 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)

6. 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!

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

8. Last Example. Sketch the intersection of the given inequalities
Last Example! Sketch the intersection of the given inequalities. 1 y≥x2 and 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.