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Quiz P-6 1 1. (-3 – 4i)(1 + 2i) = ? 2. 2. (2 – 3i) – (-4 – 5i) = ? 3.

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Presentation on theme: "Quiz P-6 1 1. (-3 – 4i)(1 + 2i) = ? 2. 2. (2 – 3i) – (-4 – 5i) = ? 3."— Presentation transcript:

1 Quiz P-6 1 1. (-3 – 4i)(1 + 2i) = ? 2. 2. (2 – 3i) – (-4 – 5i) = ? 3.

2 P-7: Solving Inequalities; Algebraically and Graphically

3 Vocabulary: (review) Linear Inequality: Ax + B < C or Ax + B < C or Ax + B > C or Ax + B > C or Ax + B ≤ C or Ax + B ≤ C or Ax + B ≥ C. Ax + B ≥ C. 3x – 2 < 4

4 What would you call this? x > 3 and x 3 and x < 5 What would it look like on the number line? Compound inequality: 2 inequalities joined together by either “and” or “or”. 1 2 3 4 5 What part is x > 3 and x 3 and x < 5 ??

5 What would you call this? x ≤ 3 or x > 5 What would it look like on the number line? Compound inequality: 2 inequalities joined together by the either “and” or “or”. 1 2 3 4 5 Which part is x ≤ 3 OR x > 5

6 Your turn: Solve and graph the inequality 1. -3 < 4 – x ≤ 3 2. -5 > x + 1 or x + 1 ≥ 6

7 Quadratic Inequality (background) What does the “solution” mean? Let y = 0 then solve for ‘x’ “for what values of ‘x’ will y = 0” “where does the graph cross the x-axis?” How do you solve it algebraically?

8 Quadratic Inequality (background) 0 = (x – 4)(x + 3) What are the 4 solution methods? x = -3, 4

9 Your Turn: Solve the Inequality 3.

10 Solving Quadratic Inequalities What does the “solution” mean? “for what values of ‘x’ will the graph be below y = 0” “where is the graph below the x-axis?” -3 < x < 4

11 Solving Quadratic Inequalities Graphically 1. Enter the equation into: ‘y = ‘ 2. Set the window 3. Determine ‘x-intercepts’ 4. Determine whether the inequality means ‘above’ or ‘below’ x-axis. means ‘above’ or ‘below’ x-axis. 5. The solution is the inequality describing the x-values where the graph is above (or below if applicable) where the graph is above (or below if applicable) the ‘x-axis’. the ‘x-axis’.

12 Example Solve Graphically x > 3.73x < 0.27

13 Your Turn: 4. Solve Graphically

14 Your Turn: 5. Solve Graphically

15 What if it’s in a non-standard form? Use properties to get “0” on one side of the inequality. Input into “y =“ What interval of x-values result in the graph being below the x-axis? below the x-axis?

16 Your Turn: 6. Solve Graphically

17 Absolute Value Equations (review) x – 5 = 1 x – 5 = -1 The distance between ‘x’ and 5 is 1. x 5 x x 5 x +1

18 Absolute Value INEQUALITIES x – 5 < 1 x – 5 > -1

19 Absolute Value INEQUALITIES x – 5 > 1 x – 5 < -1

20 Your turn: 7. Solve algebraically

21 Graphs of Absolute Value Equations y = a x – h + k SLOPE (of right side of “V”: Lead coefficient (outside of absolute value symbols) value symbols) (h, k): vertex of the graph.

22 Effect of ‘h’ and ‘k’ on the Absolute Value Function h = 0, k = 0 h = 1, k = 1 1 1 Slope = -1 Slope = +1

23 Graphing Absolute Value Inequalities (Single Inequality Method) 1. Set equal to ‘0’ 2. Replace ‘0’ with ‘y’ 3. Use TI-83 to graph “Solution” means: the range of ‘x’ values where the graph is above or below the ‘x’ axis.

24 Graphing Absolute Value Inequalities 4. For what values of ‘x’ will the graph be below the x-axis? the x-axis? 5. Write the inequality solution. solution. -4 < x < 12 “Solution” means: the range of ‘x’ values where the graph is above or below the ‘x’ axis.

25 Which Graphing Method do you like better? 1. One graph Using properties of equality (‘same thing right, same thing left’) to put “0” on one side of inequality and having one graph. 2. Two graphs Setting each side of the inequality equal “y” then having two graphs that overlap?

26 Your Turn: 8. Graph on your calculator and then write the interval of x-values that satisfies the interval of x-values that satisfies the inequality. the inequality.

27 Real World What path does the cannon ball take?

28 Projectile Motion A vertical time-distance problem in two dimensions. Height as a function of time. Height at time = 0 (initial height) Vertical component of velocity (speed) multiplied by time gives the change in vertical position resulting from the initial velocity. Vertical component of acceleration (of Gravity) multiplied by time squared gives the change in vertical position due to gravity.

29 Projectile Motion Problem An object is launched vertically upward from the ground at an initial velocity of 250 ft per second. a.When will the object be above 1000 feet? b.When will the object be at least 500 feet high? 1. Write the projectile motion equation.

30 Projectile Motion Problem 2. Plug values from the problem into the equation. 1. Write the projectile motion equation. Initial velocity = ? Initial Height = ? Min. / max. height = ? Choose your method: 1.Algebraically 2. Single graph 3. Double graph

31 I like the double graph method

32 Your turn An object is launched vertically upward from the ground at an initial velocity of 125 ft per second. 9. When will the object be below 100 feet? 10. When will the object be at least 200 feet high?

33 Your turn: An object is kicked on the roof of a 500 foot building. It slides to the edge and falls off. 11. What is its vertical velocity immediately after it crosses off the edge of the building? crosses off the edge of the building? 12. What is the object’s initial height as it begins to fall? 13. When will the object be above 100 feet above ground level? 14. When is the object at least 200 feet high?


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