5.7 : Graphing and Solving Quadratic Inequalities

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Presentation transcript:

5.7 : Graphing and Solving Quadratic Inequalities Objectives: Students will be able to… Graph quadratic inequalities in 2 variables Solve quadratic inequalities in 1 variable

Quadratic Inequalities in 2 variables… The graph contains all solutions (x,y) that make the inequality true

Graphing a Quadratic Inequality in 2 variables: Graph the parabola with the equation y=ax2 + bx +c (dashed for > or < ; solid line for > or <) Choose a point (x,y) inside the parabola and check whether the point is a solution of the inequality If the point is a solution, shade the region inside the parabola. If it is not a solution, shade the region outside the parabola.

Example Graph y < 2x2 – 5x -3

Graph y > - x2 +5

Graph y>x2 – 4x +3

Systems of Quadratic Inequalities Graph each inequality one at a time on the same coordinate plane Where the shading overlaps is your solution region. Darken this area.

Example: y < - x2 + 9 y > x2 +5x -6

Example: y < - 3x2 y > (-1/2)x2 - 5

Quadratic Inequalities in 1 variable ax2 + bx + c > 0 ( or >) ax2 + bx + c < 0 ( or <) Two ways to solve: Graphically Algebraically

How to solve graphically Graph y = ax2 + bx +c (make sure to set=0 first) Be sure to identify the x – intercepts (may need to use Quadratic Formula…uh oh!!) If ax2 + bx + c < 0 ( or <), you are looking for the x values where the graph is below the x axis. ( or on or below) If ax2 + bx + c > 0 ( or >), you are looking for the x values where the graph is above the x axis. ( or on or above)

Solve by graphing: x2 – 5x + 6 > 0 Where is your graph above the x-axis?

Solve by graphing: x2 – 11x + 5 < 0 Where is your graph below the x – axis?

Solving 1 variable inequalities algebraically Solve 2x2 – x > 3 Write the corresponding equation: 2x2 – x = 3 2. Solve the equation: 3. Test these critical x values on a number line. Pick x values to the left and right of the critical values to see what numbers satisfy the inequality:

Solve algebraically. 1. 2x2 < 8 2. x2 – 10x + 24 > 0

Ticket out: Solve 16 – x2 > 0

Remember this material when you are finding domain and range in precalculus and calculus!!!!