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5.4 Prezi – By Anna

5.4 Prezi – By Anna. http://prezi.com/7celq_oomduu/algebra-two-section-54/. Chapter 5.5. By Corey Lorraine. Warm Up Question . Vocabulary. Example 1. Example 2. Example 3. Example 3. Quadratic Formula and The Discriminant. Chapter 5 Section 6 By, Tim Summers. Quadratic Formula.

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5.4 Prezi – By Anna

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  1. 5.4 Prezi – By Anna • http://prezi.com/7celq_oomduu/algebra-two-section-54/

  2. Chapter 5.5 By Corey Lorraine

  3. Warm Up Question

  4. Vocabulary

  5. Example 1

  6. Example 2

  7. Example 3

  8. Example 3

  9. Quadratic Formula and The Discriminant Chapter 5 Section 6 By, Tim Summers

  10. Quadratic Formula • X= (-b±√(b^2-4ac))/2a • This is used to find the solution(s) of the quadratic equation ax^2 + bx + c. • Before using the quadratic equation ax^2 + bx + c must be equal to zero. • To check your answer replace the 0 with a “y” and then insert the equation into a graphing calculator

  11. Discriminant • The discriminant is b^2 – 4ac, where the a, b, and c are coefficients. • The discriminant is used to tell how many solutions and what type of solutions there are for a quadratic equation.

  12. Discriminant • If b^2 – 4ac > 0, then there are two real solutions. • If b^2 – 4ac = 0, then there is one real solution. • If b^2 – 4ac < 0, then there are two imaginary solutions.

  13. Example Problem #1 5x^2 + 6x + 1 = 0 a = 5, b = 6, c = 1 x = (-6 ± √(6^2 - 4×5×1)) / (2×5) x = (-6 ± √(36 - 20)) / 10 x = (-6 ± √(16)) / 10 x = (-6 ± 4) / 10 x = (-6 – 4) / 10 or x = (-6 + 4) / 10 x = -1 or x = -0.2

  14. Example Problem #2 5x^2 + 2x + 1 = 0 a = 5, b = 2, c = 1 x = (-2 ±√(2^2 - 4×5×1)) / (2×5) x = (-2 ± √(4 - 20)) / 10 x = (-2 ± √(-16)) / 10 x = (-2 ± 4i) / 10 x = -0.2 ± 0.4i

  15. Example Problem #3 x^2 + 3x – 4 = 0 a = 1, b = 3, c = -4 x = (-3 ± √(3^2 - 4×1×(-4))) / (2×1) x = (-3 ± √(9 – (-16))) / 2 x = (-3 ± √(25)) / 2 x = (-3 ± 5) / 2 x = (-3 + 5) / 2 or x = (-3 - 5) / 2 x = 1 or x = -4

  16. Section: 5.7Graphing and Solving Quadratic Inequalities By: Andrew Fratoni

  17. Quadratic Inequalities Quadratic inequality in two variables Quadratic inequality in one variable

  18. Graphing a Quadratic InequalityStep: 1 • Make the necessary parabola using the equation y= ax2+ bx+ c

  19. Step: 2 • Make it with a dashed line for inequalities using < or > and a solid line for inequalities using ≤ or ≥. y > x2 + 3x – 4 y < 2x2 – 3x + 1

  20. Step: 3 • Choose a point either outside or inside of the parabola and check to see if the point is a solution to the inequality • If it is a solution, then shade the corresponding area. If not, then shade the other side of the parabola

  21. Graph y ≤ -2x2 Step 1: graph y = -2x2 Step 2: since the inequality symbol is ≤, make the line solid Step 3: test a point like (0,2) 2 ≤ -2(0) 2 2 ≤ 0 false. Shade the inside of the parabola

  22. Graph y < x2 – 4x + 1 Step 1: graph y = x2 – 4x + 1 Step 2: since the inequality symbol is <, make the line dotted Step 3: test a point like (2,0) 0 < (2)2 – 4(2) + 1 0 < -3 false Shade the area outside of the parabola.

  23. Graph the system of inequalities: y ≤ - x2 + 3 y ≥ x2 + 2x - 4 y ≤ - x2 + 3 Step 1: graph y = - x2 + 3 Step 2: since the inequality symbol is ≤, make the line solid Step 3: test a point like (0,0) 0 ≤ - (0)2 + 3 0 ≤ 3 True y ≥ x2 + 2x – 4 Step 1: graph y = x2 + 2x – 4 Step 2: since the inequality symbol is ≥, make the line solid Step 3: test a point like (0,0) 0 ≥ (0)2 + 2(0) – 4 0 ≥ -4 True The shaded region is the area where both parabolas would be shaded and that satisfy both inequalities.

  24. Solving Quadratic Inequalities Algebraically First, replace the inequality symbol with an equals sign. Then, solve for x using one of the methods (quad. formula, completing the square, factoring, etc.) Last, test an x-value in-between the two x values found above. Figure out if it satisfies the inequality. Write the solution based on the known information. Ex. X2 + 3x – 18 ≥ 0 X2 + 3x – 18 = 0 replace w/ equals sign (x + 6)(x -3) = 0 factor x = -6 x = 3Set both equal to zero Test x = 0 (0)2 + 3(0) – 18 ≥ 0 -18 ≥ 0 false So, because a number between the two values does not satisfy the inequality Solution: x ≤ -6 or x ≥ 3

  25. Real life Application Finding the weight of theater equipment that a rope can support. Ex. Weight that the rope can safely support (W) with diameter (d) W≤ 1480d2

  26. Using Properties of Exponents 6.1

  27. Properties of Exponents PRODUCT OF POWERS PROPERTY am. an = am+n POWER OF A POWER PROPERTY (am ) n = amn POWER OF A PRODUCT PROPERTY (ab) m = am bm NEGATIVE EXPONENT PROPERTY a-m= 1 / (am1) ZERO EXPONENT PROPERTY a0 = 1, a 0 0 QUOTIENT OF POWERS PROPERTY am/ an = am-n, 0 POWER OF A QUOTIENT PROPERTY (a/b)m= am /bm 0

  28. Evaluating Numerical Expressions (25)3 = ? 215= 32,768 (3/4)2 =? (32/42) = 9/16 (4)3. (4)-6 =? (4) -3= 1/4 3 = 1/64

  29. Student Practice 1.) 52 . 52 2.) (3-3)4 3.) (2/4)3 4.) 50.63 • 1.) 625 • 2.) 1/531,441 • 3.) 1/8 • 4.) 108

  30. Properties of Rational Exponents (7.2)

  31. Properties of Rational Exponents PRODUCT OF POWERS PROPERTY am. an = am+n POWER OF A POWER PROPERTY (am ) n = amn POWER OF A PRODUCT PROPERTY (ab) m = am bm NEGATIVE EXPONENT PROPERTY a-m= 1 / (am1) ZERO EXPONENT PROPERTY a0 = 1, a 0 0 QUOTIENT OF POWERS PROPERTY am/ an = am-n, 0 POWER OF A QUOTIENT PROPERTY (a/b)m= am /bm 0

  32. Student Practice 1.) n√a. b = ? 2.) n(√a/b) = ? 1.) n√a. n√b 2.) n√a/n√b

  33. 6.2evaluating and graphing polynomial function ALGEBRA 2

  34. a polynomial function is a function of the form when a0 is not zero, the exponents will be all whole numbers, and the coefficients will be all real numbers. for this polynomial function, the leading coefficient will be an, constant will be a0, and the degree will be na standard form polynomial function is when the terms of exponents are written in descending order and it also from left to right.

  35. Here is a summary of common types of polynomial function

  36. x-value 4 3 0 -8 5 -7 12 48 160 660 3 12 40 165 653 f(4)=653 Example1 use synthetic substitution to evaluate f(x)=38+5x-7 when x=4 Solution polynomial in standard form coefficients

  37. 0.000025 -0.004 0.33 5.3 100 0.0025 -0.21 12 0.000025 -0.0021 0.12 17.3 the recharge time is about 17 seconds Example 2 find the recharge time  of battery after 100 flashes, if the modeled is  where t(in sec) is  t=0.000025+0.004+0.33n+5.3.the time of a battery to recharge after flashing n times of uses.

  38. f (x)→+∞ as x →-∞ f (x)→-∞ as x →+∞ Graph function f (x)→+∞ as x →+∞ f (x)= x f (x)=-x f (x)→-∞ as x →-∞

  39. f (x)→ -∞ As x→ +∞ f (x)→ -∞ As x→ -∞ f (x)→ +∞ As x→ -∞ f (x)→ +∞ As x→ -∞ f (x)=x f (x)=-x

  40. X X X -3 -3 -3 -2 -2 -2 -1 -1 -1 0 0 0 1 1 1 2 2 2 3 3 3 f(x) -16 -3 0 -1 9 32 0 Solution a. To graph this function, make a table of values and plot with corresponding points. Connect the points with a smooth curve and check the end behavior. The degree is odd and the leading coefficient is positive so f(x) → -∞ as x →-∞ and f(x) → +∞ as x →+∞

  41. x -3 -2 -1 0 1 2 3 1 f(x) -15 8 1 0 -32 -147 solution b. To graph this function, make a table of values and plot with corresponding points. Connect the points with a smooth curve and check the end behavior The end The degree is even and the leading coefficient is negative so f(x) → -∞ as x →-∞ and f(x) → -∞ as x →+∞

  42. Algebra IILesson 6.3 Adding,Subtracting, Multiplying PolynomialsJ.W.Tang

  43. ~Adding Polynomials~ • To add polynomials • Combine like terms(make sure all degrees are account for!) • You can either do this vertically or horizontally • Example (pg 341 #25) • (10X-3+7X²)+(X³-2X+17)→=( X³+7X²+8X+14) • →___7X²+10X-3 • +X³ ___-2X+17 • ------------------------------- • X³+7X²+8X+14

  44. ~Subtracting Polynomials~ • To subtract polynomials • Subtract like terms • To make it easier flip signs : • multiply the equation you are subtracting by -1 and then add the equations • Example(pg341 #23) • (10X³-4X²+3X)-(X³-X²+1) • →10X³-4X²+3X__ 10X³-4X²+3X__ • -X³ - X² __ +1 →+-X³ + X² __ -1 • ------------------- ------------------------ • 9X³-3X²+3X-1 9X³-3X²+3X+1

  45. ~Multiplying Polynomials~ • To multiply polynomials set the equation like a normal multiplication equation then multiply one by one like a normal multiplication problem • See Next 2 Slides for Examples on how to multiply polynomials vertically and horizontally

  46. -To multiply Vertically- Example (pg.341 #41) (3X²-2) (X²+4X+3) X²+4X+3 * 3X²-2 ------------------ -2x²-8X-6 + 3X412x³+9x²__ ________________ 3X412x³+7X²-8X-6

  47. -To Multiply horizontally- -Distribute Example (pg.341 #41) (3X²-2) (X²+4X+3)→3X²(X²+4X+3)+-2 (X²+4X+3) =3X4+12X³+9X² +-2X²-8X-6 =3X4+12X³+7X²-8x-6

  48. ~Special Patterns~ ~Sum and Difference~ (a+b)(a-b)=a²+b² →(X+7)(X-7)=X²+49(pg342#53) ~Square of a Binomial~ (a+b)²=a²+2ab+b²→(X+4)²=X²+2(X4)+4²= X²+8X+16(pg.342 #54) (a-b)²=a²-2ab+b²→(6-X)²= 6²-2(6)(-X)+X²=36+12X+X²=X²+12X+36 ~Cube of a Binomial~ (a+b)³=a³+3a²b+3ab²+b³ →(X+4)³=X³+3(X²)(4)+3(X)(4²)+4³=X³+12X²+48x+64 (a-b)³=a³-3a²b+3ab²-b³ →(X-2)³=X³-3(X²)(2)+3(X)(2²)-2³= X³-6X²+12X-8

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