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Published byAlberta Johns Modified over 9 years ago

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**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

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**Quadratic Inequalities in 2 variables…**

The graph contains all solutions (x,y) that make the inequality true

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**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.

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Example Graph y < 2x2 – 5x -3

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Graph y > - x2 +5

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Graph y>x2 – 4x +3

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**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.

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

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

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**Quadratic Inequalities in 1 variable**

ax2 + bx + c > 0 ( or >) ax2 + bx + c < 0 ( or <) Two ways to solve: Graphically Algebraically

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**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)

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**Solve by graphing: x2 – 5x + 6 > 0**

Where is your graph above the x-axis?

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**Solve by graphing: x2 – 11x + 5 < 0**

Where is your graph below the x – axis?

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**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:

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

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**Ticket out: Solve 16 – x2 > 0**

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

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