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# Algebra slideshow

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### Algebra slideshow

1. 1. GRAPHING QUADRATIC EQUATIONSGraph y=2x2-8x+6Solution:Note: Coefficients for this function are a= 2, b= -8, and c= 6. Since a>0, the parabola opens up.1. Find and plot the vertex. The x-coordinate is: x= -b/ = --8/2(2) = 2 2a The y-coordinate is: y = 2(2)2 – 8(2) + 6 = -2 So, the vertex is (2, -2).2. Draw the axis of symmetry x = 2.3. Plot two points on one side of the axis of symmetry, such as (1,0) and (0,6). Use symmetry to plot two or more points, such as (3,0) and (4,6).4. Draw a parabola through the plotted points.-----------------------------------The quadratic function y = ax2 + bx + c is written in standard form.
2. 2. GRAPHING A QUADRATIC FUNCTION IN VERTEX FORMGraph y = -1/2 (x +3)2 + 4Solution:The function is in vertex form y = a(x – h)2 + k where a = -1/2, h = -3, and k = 4. Since a < 0, the parabola opens down. To graph the function, first plot the vertex (h, k) = (-3, 4). Draw the axis of symmetry x = -3 and plot two points on one side of it, such as (-1, 2) and (1, -4). Use symmetry to complete the graph.
3. 3. GRAPHING A QUADRATIC FUNCTION IN INTERCEPT FORMGraph y = -( x + 2)( x – 4).Solution:The quadratic function is in intercept form y = a( x – p)( x – q) where a = -1, p = -2. and q = 4. The x- intercepts occur at ( -2,0) and ( 4, 0). The axis of symmetry lies halfway between these points, at x = 1. So, the x-coordinate of the vertex is x = 1 and the y-coordinate of the vertex is: y = -( 1 + 2)( 1 - 4) = 9The graph of the function is shown.
4. 4. WRITING QUADRATIC FUNCTIONS IN STANDARD FORMWrite the quadratic function in standard form.a. Y = -( x+4)( x – 9) b. y = 3( x – 1)2 +8Solution: a. y = -( x +4)( x – 9) Write original function. = -( x2 – 9x +4x – 36) Multiply by using FOIL. = -( x2 – 5x – 36) Combine like terms. = -x2 + 5x +36 Use distributive property. b. y = 3( x – 1)2 +8 Write original function. = 3( x – 1)( x – 1) + 8 Rewrite ( x - 1)2. = 3(x2 - x – x + 1) + 8 Multiply using FOIL. = 3( x2 – 2x + 1) +8 Combine like terms. = 3x2 – 6x + 3 + 8 Use distributive property. = 3x2 – 6x + 11 Combine like terms.
5. 5. SOLVING QUADRATIC EQUATIONS WITH FACTORING( X + 3)( X + 5) can be written as x2 + 8x + 15. The expressions x + 3 and x + 5 are binominals because they have two terms. The expression x2 + 8x + 15 is a trinomial because it has three terms. You can use factoring to write a trinomial as a product of binominals. To factor x2 + bx + c, find integers m and n such that: x2 + bx + c = ( x + m)( x + n) = x2 + ( m + n)x + mnSo, the sum of m and n must equal b and the product of m and n must equal c.Factoring Shortcuts: Difference of Two Squares: a2 – b2 = (a + b)(a – b), Example: x2 – 9 = ( x + 3)( x – 3) Perfect Square Trinomial: a2 + 2ab + b2 = ( a + b)2, Example: x2 +12x + 36 = ( x +6)2 a2 - 2ab + b2 = ( a - b)2, Example: x2 – 8x + 16 = ( x – 4)2
6. 6. FACTORING A TRINOMIAL OF THE FORM X2 + BX + C Factor x2 - 12x – 28 Solution: You want x2 – 12x – 28 = ( x + m)(x + n) where mn = -28 and m + n = -12.Factors of -28 -1, 28 1, -28 -2, 14 2, -14 -4, 7 4, -7 (m, n)Sum of factors 27 -27 12 -12 3 -3 ( m + n) The table shows that m = 2 and n = -14. So, x2 – 12x – 28 = ( x + 2)( x – 14).
7. 7. FACTORING A TRINOMIAL OF THE FORM AX2 + BX + CFactor 3x2 – 17x + 10.Solution:You want 3x2 – 17x + 10 = ( kx + m)( lx + n ) where k and l are factors of 3 and m and n are (negative) factors of 10. Check possible factorization by multiplying. ( 3x – 10)( x – 1) = 3x2 – 11x + 10 ( 3x – 1)( x – 10) = 3x2 – 31x + 10 ( 3x – 5)( x – 2) = 3x2 – 11x + 10 ( 3x – 2)( x – 5) = 3x2 – 17x + 10  CorrectThe correct factorization is 3x2 – 17x + 10 = ( 3x – 2)( x – 5).
8. 8. SOLVING QUADRATIC EQUATIONS BY FINDING SQUARE ROOTSA number r is a square root of a number s if r2 = . A positive number s has two square roots denoted by √s and - √s. The symbol √ is a radical sign, the number s beneath the radical sign is the radicand, and the expression √s is a radical. For example, since 32 = 9 and ( -3)2 = 9, the two square roots of 9 are √9 = 3 and - √9 = - 3. You can use a calculator to approximate √s when s is not a perfect square. For instance, √2 ≈ 1,414.Example: Using Properties of Square Roots Simplify the expression: a. √24 = √4 * √6 = 2 √6 b. √6 * √15 = √90 = √9 * √10 = 3 √16 c. √7/16 = √7/√16 = √7/4 d. √7/2 = √7/√2 * √2/√2 = √14/2In part (d) of this example, the square root in the denominator of 7/2 was eliminated by multiplying both the numerator and denominator by √2. This process is called rationalizing the denominator. You can use square roots to solve some types of quadratic equations.
9. 9. SOLVING A QUADRATIC EQUATION (√)Solve 2x2 + 1 = 17Solution: Begin by writing the equation in the form x2 = s. 2x2 + 1 = 17 Write original equation. 2x2 = 16 Subtract 1 from each side. x2 = 8 Divide each side by 2. x = +/- √8 Take square roots of each side. x = +/- 2 √2 Simplify.The solutions are 2√2 and -2√2.
10. 10. SOLVING A QUADRATIC EQUATION (√)Solve 1/3( x + 5)2 = 7Solution: 1/ x + 5)2 = 7 3( Write original equation. ( x + 5)2 = 21 Multiply each side by 3 x + 5 = +/- √21 Take square roots of each side. x = -5 +/- √21 Subtract 5 from each side.The solutions are -5 + √21 and -5 - √21.
11. 11. COMPLETING THE SQUARECompleting the square is a process that allows you to write an expression of the form x2 + bx as the square of a binomial. This process can be illustrated by using an area model too.Example: Completing The SquareFind the value of c that makes x2 – 7x + c, note that b = -7, therefore: c = (b/2)2 = (-7/2)2 = 49/4Use this value of c to write x2 – 7x + c as a perfect square trinomial, and then as the square of a binomial. x2 – 7x + c = x2 – 7x + 49/4 Perfect square trinomial. = ( x – 7/2)2 Square of a binomial: ( x + b/2)2
12. 12. SOLVING A QUADRATIC EQUATION IF THE COEFFICIENT OF X2 IS NOT 1Solve 3x2 – 6x + 12 = 0 by completing the square.Solution: 3x2 – 6x + 12 = 0 Write original equation. x2 – 2x + 4 = 0 Divide each side by the coefficient of x2. x2 – 2x = -4 Write the left side in the form of x2 + bx. x2 – 2x + (-1)2 = -4 + 1 Add (-2/2)2 = (-1)2 = 1 to each side. ( x – 1)2 = -3 Write the left side as a binomial squared. x – 1 = +/- √-3 Take square roots of each side. x = 1 +/- √-3 Solve for x x = 1 +/- i√3 Write in terms of imaginary unit i.The solutions are 1 + i√3 and 1 - i√3.Because these solutions are imaginary, you cannot check them graphically. However, you can check the solutions algebraically by substituting them back into the original equation.
13. 13. SOLVING A QUADRATIC EQUATION IF THE COEFFICIENT OF X2 IS 1Solve x2 + 10x – 3 = 0 by completing the square.Solution: x2 + 10x – 3 = 0 Write original equation. x2 + 10x = 3 Write the left side in the form x2 + bx. x2 + 10x +52 = 3 + 25 Add (10/2)2 = 52 = 25 to each side. ( x + 5)2 = 28 Write the left side as a binomial squared. x + 5 = +/- √28 Take square roots of each side. x = -5 +/- √28 Solve for x. x = -5 +/- 2 √7 Simplify.The solutions are -5 + 2 √7 and -5 - 2 √7.
14. 14. QUADRATIC FORMULA AND DISCRIMINANTThe Quadratic Formula: Let a, b, and c be real numbers such that a ≠ 0. The solutions of the quadratic equation ax2 + bx + c = 0 are: x = -b +/- √b2 – 4ac/ 2aRemember that before you apply the quadratic formula to a quadratic equation, you must write the equation in standard form, ax2 + bx + c = 0.
15. 15. SOLVING A QUADRATIC EQUATION WITH TWO REAL SOLUTIONSSolve 2x2 + x = 5Solution: 2x2 + x = 5 Write original equation. 2x2 + x – 5 = 0 Write in standard form. 2 – 4ac x = -b +/- √b / 2a Quadratic formula x = -1 +/- √12 – 4(2)(-5)/ 2(2) a = 2, b = 1, c = -5 x = -1 +/- √41/4 Simplify. The solutions are x = -1 +/- √41/4 ≈ 1.35, and x = -1 +/- √41/4 ≈ 1.85