Algebra Chapter 12 Chapter 13 Graphing quadratic functions Solving quadratic equations A quadratic function can be written as

"ax² + bx + c = y", as well as "ax² + bx + c = f(x)". In the equation "ax² + bx + c = y", the a, b, and c are real numbers, and a ≠ 0. When graphing a function, you first must find the axis of symmetry. When your quadratic function is graphed with a set of real numbers, it is called a parabola. -b

2a x = For our equation "y = 2x² + 4x - 1", the line of symmetry is -1. x = -4

2 (2) y = x² - 8x + 16 In the equation "y = 2x² + 4x - 1",

a = 2

b = 4

c = -1 Next, you need to solve for the points on either side of this axis of symmetry. x y -1 -2 0 ? ? ? Point on one side Point on other side Axis of symmetry Plug in each x coordinate into the equation and find y. y = 2(-2)² + 4(-2) - 1 y = 2(-1)² + 4(-1) - 1 y = 2(0)² + 4(0) - 1 y = -1 y = -3 y = -1 Now, our T-chart looks like this x y -2 -1 0 -1 -3 -1 With our ordered pairs, we can now make a parabola!! (-1, -3) (-2, -1) (0, -1) Vertex The vertex is the maximum or minimum point of a parabola. In this case, it is the maximum point. 1 Parabola Round 2 Round 1 Congratulations! You have survived round one. Excelente

!!!! You can do this, I believe in you! Got that? Ready for a real challenge now? Stay strong my friend. If I can do this math, you can too! A quadratic equation written in standard form is ax² + bx + c = 0 WHY? This is because y = o in the quadratic function ax² + bx + c = y Once you find the standard form, you can use the quadratic fomula to solve. It is an equation where a, b, and c are real numbers and a ≠ 0 The quadratic formula! x = -b ± ─ b² - ac 4 √ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ 2a When solving a quadratic equation, the first thing you need to do is put it into standard form.

Let's solve x² - 4x = 21 Remember that it needs to be in standard form ax² + bx + c = 0 x² - 4x = 21 -21 So, subtract 21 to make equation equal to 0 x² - 4x - 21 = 0 Standard

form! Are you

ready? Now, plug your numbers into the quadratic formula x = -b √ ¯¯¯¯¯¯¯¯¯¯¯¯¯ b² - ac 2a ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ± 4 x² - 4x - 21 = 0 a = 1

b = -4

c = -21 x = - ² - ± (-4) (-4) (1)(-21) 4 √ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ 2 (1) x = 4 ± √ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ 16 8 + 4 2 x = √ √ ± ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ 4 100 2 ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ 100 = 10 Multiply everything out... Add 16 and 84... √ √ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ x = x = 4 4 2 2 + - 10 10 x = -3 , 7 YEA! You defeated those sand-blasted grease monkeys! Round 3 The Final One x² = 6x - 9 Subtract x² to make equation equal to 0 x² = 6x - 9 -x² -x² -x² + 6x - 9 Just to make things easier, multiply -x² + 6x - 9 by a negative 1, so that "a" is a positive number (-x² + 6x - 9) -1 = x² - 6x + 9 a = 1

b = -6

c = 9 x = - (-6) ± √ (-6) ² - 4 (1)(9) ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ 2 (1) Use the quadratic formula... Solve... Multiply everything out... x = 6 ± √ ¯¯¯¯¯¯¯¯¯¯¯¯¯ 36 - 36 2 ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ Discriminant The discriminant is the number under the radical. This means, that if b² - 4ac (discriminant) is 0, it has only one real solution, one x-intercept. If the discriminant is positive, it has 2 real solutions, two points. If it is negative, there are no real solutions, and it never touches the x-axis. Because our discriminant is 0, we know that it has only one real solution. x = 6 2 x = 3 And there you go! You've just learned the basics of Chapters 12 and 13 in 8th grade Algebra!!! Oh please don't go yet... I will miss you! FINALLY DONE! :D

Made by Citlayi Villasenor ~ Thanks for watching :)

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