Section 1.1. Chapter 1. Quadratics. Parabolas. Example. Example. ( ) = ax 2 + bx + c -2-1

Transcription

1 Chapter 1 Quadratic Functions and Factoring Section 1.1 Graph Quadratic Functions in Standard Form Quadratics The polynomial form of a quadratic function is: f x The graph of a quadratic function is a parabola. = ax + bx + c Our book mistakenly calls this standard form, not polynomial form. Make sure your notes are correct. A parabola is a U shaped graph that either opens up or down. It is not a V shape, it does not have a sharp corner! The parent function for the family of parabolas is the function f(x) = x. Parabolas The vertex of the graph is the highest or lowest point of the curve The axis of symmetry divides the parabola into mirror images and passes through the vertex. -15 Graph f ( x) = x x y Our graph is narrower than the parent function, but has the same vertex The method of graphing used here is called point plotting, basically making a table. Graph f x =! 1 x + 3 x y Our graph is wider, opens down and the vertex is moved 3 units up The direction a graph opens is called its concavity. This one is concave down.

2 Properties = ax + bx + c f x The graph opens up if a > 0 and down if a < 0. The graph is narrower than the parent function if a > 1 and wider if a < 1. The axis of symmetry is x =!b x =!b a a and that is the x coordinate of the vertex. The y intercept is c, so the point (0, c). ( 0,c) Graphing in Polynomial Form Identify the axis of symmetry, sketch it. Find the vertex through evaluation. Identify the y intercept, if it visible on the graph. Evaluate the function at at least one other point. Use symmetry to get as many points as possible. Graph f x = x! 8x + 6 x =!b a x =!!8 x = 8 4 = f =! f = 8! f =! (,!) ( 0,6) vertex y - intercept x y Graph f x = x! 8x + 6 x y Maxima and Minima Parabolas are the first functions we come across in algebra that actually have maximum and minimum values. The vertex is a maximum value if a < 0. The vertex is a minimum value if a > 0. Maxima and minima are the plurals of maximum and minimum respectively. f x Find the max (or min) of the function. = 3x! 18x + 0 x =!b a x =!!18 3 x = 18 6 x = 3 f ( 3) = 3( 3)! 18( 3) + 0 f ( 3) = 7! f ( 3) =!7 ( 3,!7) The minimum is at the point (3, -7)

3 Section 1. Graph Quadratic Functions in Vertex and Intercept Form This is actually the standard form of a quadratic function. The vertex is the point (h, k). The graph is translated h units horizontally. The graph is translated k units vertically. Vertex Form f ( x) = a( x! h) + k The graph is concave up if a > 0 and concave down if a < 0. This is the best form for graphing parabolas. Identify the vertex! 5 y = x! y = 4( x + 6)! 7 y = 1 ( x! 3) + 4 y =! ( 5 x + 9) + 1 (,!5) (!6,!7) ( 3, 4) (!9,1) There are several different ways of looking at how to find h and make sure you have the correct sign. Graph y = x!! 5 (,!5) x y Transformation Transformation

4 Transformation Transformation Transformation Transformation Transformation Intercept Form p and q are the x intercepts of the graph. f ( x) = a( x! p) ( x! q) The axis of symmetry and vertex are found with: x = p + q The graph is concave up if a > 0 and concave down if a < 0. This is not great for graphing, but good for other things.

5 Graph f x =! 1 ( 5 x + 6) ( x! 4) p =!6 q = 4 x = p + q + ( 4) x =!6 f!1 x =! x =!1 =! 1 ( ( 5!1) + 6 )((!1)! 4) f (!1) =! 1 ( 5 5) (!5) f (!1) = 5 (!1, 5) (!1,5) x -11 y f ( x) =! 1 ( 5 x + 6) ( x! 4) A projectile is fired up from the ground at a speed of 80 ft/s. Its height is given by the function h( t) =!16t( t! 5), where t is time in seconds and h is the projectile s height in feet. How many seconds after it is fired will it hit the ground? What is the maximum height? The important thing to realize before we start is horizontal and vertical motion are described by different functions! h( t) =!16t( t! 5) h( t) =!16( t! 0) ( t! 5) The object is on the ground when h = 0, or at the t intercepts. t = 0 t = 5 The object hits the ground in 5 seconds. The object is at its maximum height when it is at the vertex. t = p + q t = 0 + ( 5) t =.5 =!16(.5) ((.5)! 5) = 100 h.5 h.5 The maximum height is 100 ft. F FOIL FOIL is used to convert from intercept form ( x + 3) ( x + 5) I O to polynomial form. L = x + 5x + 3x + 15 = x + 8x + 15 Remember, FOIL means First Outer Inner Last, it is only used to multiply a binomial by a binomial. Write in polynomial form. ( x + 6) y =! 1 x + 8 y = 3 x! 4 y = 3 x + x! 4 If you dare to say this is x We will have issues... Serious ones y = 3x + 6x! 7 y =! 1 x + 16x + 64 y =! 1 x! 8x! y =! 1 x! 8x + 3

6 Section 1.3 Solve x + bx + c = 0 by Factoring Some Vocabulary A monomial is an expression that is either a number, variable or product of numbers and variables. A polynomial is a monomial or sum of monomials. In a polynomial the monomials are called terms. A binomial is a polynomial with terms. A trinomial is a polynomial with 3 terms. Factoring Factoring is reversing multiplication. It takes a sum and rewrites it as a product. When you are asked to factor anything the question that is actually being asked is What do I have to multiply to get this? and you answer it by factoring. Polynomials of the form x + bx + c 1. Look at c, if it is positive, then the signs in both parenthesis are the same, if it is negative, they are different.. If the signs are the same, look at b, the sign of b will tell you what signs are in the parenthesis. 3. Find all the factor pairs of c. 4. Choose the factor pair of c that when added gives you b. Signs are taken into account. Factor, if possible. x + 5x + 6 ( x + 3) ( x + ) x + 6x! 16 ( x + 8) ( x! ) Special Cases special case formula example Difference of Squares a! b = ( a + b) ( a! b) x! 36 = ( x + 6) ( x! 6) x! 10x + 1 ( x! 7) ( x! 3) x + x + 6 Not Factorable Perfect Square Trinomial Perfect Square Trinomial a + ab + b = ( a + b) x + 6x + 9 = ( x + 3) a! ab + b = ( a! b) x! 14x + 49 = ( x! 7)

7 x! 81 x + 9 ( x! 9) x + x + 1 ( x + 1) x! 8x + 16 ( x! 4) s x! 11 x + 11 ( x! 11) x + 1x + 36 x + 6 x! 4x ( x! 1) The Zero Product Property This is the key to solving equations by factoring. The zero product property states that if the product of two expressions is equal to zero the either one, or both of the expressions must equal zero. If ab = 0 then either a = 0 or b = 0 ( x + ) ( x! 9) = 0 x + = 0 x =! x! 9 = 0 x = 9 x + 6x + 8 = 0 ( x + 4) x + x + 4 = 0 x =!4 = 0 x + = 0 x =! s u =!9u u + 9u = 0 = 0 u u + 9 Whenever we factor out a monomial, one of our answers will be zero! u = 0 u + 9 = 0 u =!9 When equations are not equal to zero, we must rewrite them. The outdoor dining area at Prusik s Chocolate Chip Cookie Company is a rectangle that is 100 feet by 50 feet. The brilliant, handsome, brilliant, generous, brilliant owner wants to triple the area by adding the same distance to each side of the rectangle. How much should this brilliant person add? 50 ft (50+x) ft continued 5000 sqft 100 ft 15,000 sqft (100+x) ft ( x + 50) ( x + 100) = x + 150x = x + 150x! = 0 ( x + 00) ( x! 50) = 0 x =!00 x = 50 Add 50 ft to each dimension, making the rectangle 150 by 100. Zeros In graphing, using intercept form, we can easily identify the x intercepts of the graph. The x intercepts of the graph are are also called the zeros of the function. The zeros of the function are also the solutions to the equation of the function set equal to zero. I am trying to illustrate that the factors we get from factoring, the zeros of the function, the x intercepts, the solutions to equation are all the same thing!

8 Find the zeros of: y = x! 4x! 1 ( x + ) y = x! 6 x = 6 x =! These are exactly the same solutions we would get to the equation x! 4x! 1 = 0 These are exactly the same values we would get if we labeled the x intercepts of the graph at (-, 0) and (6, 0). Section 1.4 Solving ax + bx + c = 0 by Factoring a is not 1 Factoring gets a lot more complicated if the coefficient of x is not 1. We have two choices here: The guess and check method (long and drawn out) The ac method (different, but works well) I don t really care what the book says, I am teaching the ac method. The ac Method 1. Multiply a and c.. Fill in the box (you will see) using the first term, last term and two terms that when multiplied give you the product from step 1 and added give you the linear (x) term. 3. Factor to the left and above, using the sign closest to the border. If you can t take anything out, you can always take 1. x! 3x! 10x + 8 3x!4 3x!6x!4x + 8 ( x! ) ( 3x! 4) 3! 8 = 4 10,1 60, 40, 3 30, 4 4, 5 0, 6 1,10 5x!4 15x! x! 8 3x + 15x +10x!1x! 8 ( 5x! 4) ( 3x + ) 15 (!8 ) =!10

9 Other Factoring Tools Some things you may have to do other than using the ac method: Factor out a common monomial first Use special case factors Rewrite in standard form first Same rules for solving. Factor and use the zero product property! Rewrite equations to solve by factoring 81x! 5 9x + 5 ( 9x! 5) 49z + 11z + 64 ( 7z + 8) 9r! 66r + 11 ( 3r! 11) Know your perfect squares and you will be able to factor these without a second thought! s 3x! x! 100 ( x! 10) 3 x + 10!7t! 63t!7t t + 9 8m + 8m! 10 4 m + 7m! 30 ( m + 6) 4 m! 5!5y + 60y! 35!5 5y! 1y + 7 ( y! 1)!5 5y! 7 4x! 17x! 15 = 0 ( 4x + 3) ( x! 5) = 0 x =! 3 4 x = 5 3y + y + 60 =!14y! 48 3y + 36y = 0 3( y + 1y + 36) = 0 3( y + 6) = 0 y =!6 (y+6) is just (y+6)(y+6) so you get the same answer twice. This is called a repeated root. You have been granted the amazing task, by your awesome leader, of designing a garden to go outside the offices of Prusik Pool Professionals. You want the garden to be made up of a rectangular flower bed to be surrounded by a border of uniform width. You want a x 15 ft flower bed and your budged allows you to buy 10 sqft of stone for the border. How thick should the stone be? x continued x ft x +x ft 15 ft x 15+x ft ( x + ) ( x + 15)! ( 15 " ) = 10 4x + 74x + 330! 330 = 10 4x + 74x = 10 4x + 74x! 10 = 0 = 0 ( x + 0) = 0 x + 37x! 60 x! 3 x = 3 x =!0 The stone should be 1.5 ft, or 18 in wide.

10 Section 1.5 Solve Quadratic Equations by Finding Square Roots Square Roots r is a square root of s if r = s Remember, we only take square roots of positive numbers! A positive number s has two square roots, the principal square root (positive) and the negative square root. Radical Sign s Radicand Properties of Square Roots ab = a! b a b = a b Simplifying Radicals In order to simplify radicals, you must break the radicand up into a product of the largest perfect square possible, and whatever is left over. Use the properties on the previous slide to simplify from there. s Rationalizing = = One of the little rules of math that goes back a long long time is that radicals are not allowed in the denominator of fractions. In order to rewrite these expressions we use a process called rationalizing. There are two cases, and it is easier to show them than explain them.

11 7 5 75! s 3! ! 8 4 Notice what happened, what was outside the radical worked with what was outside. What was inside worked with what was inside ( 3) ( 5 + )! 5 " 5 " 15! ! 5! 4 s If the denominator involves addition or subtraction, you have to multiply by the conjugate of the denominator. A conjugate has the same integer value, but opposite radical value = 4 15! 3 5! = 15! 3 3 x! 15 = 65 x = 80 x = 40 x = ± 40 When we take a square root, we have to account for both the positive and negative, so we use the plus minus. x = ± 4 10 x = ± 10 Some books don t show the work between the steps, you are expected to know that we added 15 to both sides, divided by, etc. The key to solving equations with square roots, isolate the squared item, take the square root, solve as normal. 1 3 x! 4 = 11! 3$ " # 1% & 1 ( 3 x ' 4) = 11! 3 $ " # 1 % & ( x! 4) = 33 ( x! 4) = ± 33 x! 4 = ± x = 4 ± 33 What happens when you get something like 4±7? You do 4+7 and 4-7 and write down 11 and -3...that s it... A cookie is dropped off the 30 meter high roof of Prusik s Chocolate Chip Cookie factory...waste of a good cookie... The formula for something free falling is h( t) =!4.9t + h 0, where h0 is initial height in meters and t is time in seconds. How long before the cookie hits the ground? h( t) =!4.9t + h 0 h( t) =!4.9t =!4.9t t = 30 t = t = ± t = ± t! ±.474 In a truly wasteful display of total disregard for amazing product produced at Prusik s Chocolate Chip Cookie Company, the employee who threw the cookie will be fired before it hits the ground in approximately.474 seconds.

12 Section 1.6 Perform Operations with Complex Numbers Imaginary Numbers Sometimes, we end up taking the square root of a negative. We have always been told that taking a square root of a negative isn t allowed. Technically that is true, its not allowed in the real number system, because taking the square root of a negative produces an imaginary number. The Imaginary Unit The book likes to put i before the radical sign, I like to put it!1 after, = it makes i no difference. Property!r = i r!5 = 5i ( i r ) =!r 3i =!3 x =!13 x = ± 13i x =!38 x = ± 38i x + 11 = 3 x =!8 x = ± 8i x = ± i s x! 8 =!36 x =!8 x = ± 8i x = ± 7i 3x! 7 =!31 3x =!4 x =!8 x = ± 8i x = ± i 5x + 33 = 3 5x =!30 x =!6 x = ± 6i Complex Numbers A complex number written in standard form is a + bi, where a is the real part and b is the imaginary part. If a is zero, the number is purely imaginary, or just imaginary. If b is zero the number is purely real, or just real. The set of complex numbers is the set of all real numbers and all imaginary numbers. The Set of Complex Numbers Imaginary Irrational Rational Integers Whole Natural

13 The Complex Plane Just like every real number corresponds to a point on a number line, every complex number corresponds to a point in the complex plane.!6 + 9i Imaginary 4! 4i Real Arithmetic of Complex Numbers Adding and subtracting of complex numbers is nothing more than combining like terms. Treat i like any other variable and watch your signs, that s it. ( 9! i) + (!6 + 7i) 3 + 6i ( 3 + 7i)! ( 8! i)!5 + 9i 4i +! 7i! 3i!1! 4 + 3i!8! 3i i i Multiplication Multiplication of complex numbers is basically either distributive property or FOIL type problems. The thing you need to remember and keep in your work is the idea that i = -1. ( + 3i) ( 4! i) 8! 4i + 1i! 6i 8! 4i + 1i! 6 (!1) 8! 4i + 1i i 3( 4 + i) 1 + 6i s ( 4 + 6i) ( 4! 10i) 16! 40i + 4i! 60i 16! 40i + 4i! 60 (!1) 16! 40i + 4i ! 16i 3i ( 4 + i) 1i + 6i 1i + 6!1!6 + 1i Complex Conjugates and Division When we multiply a complex number by its complex conjugate the result is purely real. Complex conjugates are numbers with the same real part, but opposite imaginary parts. Complex division is just multiplying the numerator and denominator by the complex conjugate of the denominator i and 4! 3i!6 + 5i and! 6! 5i + 3i 4 + 5i + 3i 4 + 5i! 4 " 5i 4 " 5i 8! 10i + 1i! 15i 16! 0i + 0i! 5i 8! 10i + 1i! 15!1 16! 0i + 0i! 5!1 8! 10i + 1i ! 0i + 0i i i 41

14 5! 7i! 3i 5! 7i! 3i " ( + 3i) ( + 3i) i! 14i! 1i 4 + 6i! 6i! 9i 4 + 6i! 6i! 9 (!1) i! 14i! 1! i! 14i i! 6i i i 13 Absolute Value The absolute value of a complex number is also called its magnitude or its modulus. The formula is derived from the geometry of plotting complex numbers in a plane and using the distance formula. z = a + bi z = a + b Find the absolute value of each complex number. + 3i = 13!4 + 3i = 5 Section 1.7 Complete the Square 5! 1i = 13 Completing the Square Completing the square is the process of taking something that is not a perfect square trinomial and rewriting it to be one. This can be used for two things: Solving equations (terrible) Rewriting functions in vertex form (fantastic) If its terrible why do we do it? Because the same steps are very useful for other things. x + 0x = 81 ( x + 10) = 81 ( x + 10) = ± 81 x + 10 = ±9 x =!10 ± 9 x =!1 x =!19

15 1. Write the equation with the quadratic and linear term on one side of the equal sign, and constant on the other.. Factor out any leading coefficient other than 1, divide both sides by it. 3. Divide the remaining linear coefficient by. Steps 4. Square the result from step Add the result from step 4 to both sides. 6. Factor perfect square trinomial. 7. Take square root of both sides. 8. Solve. x! 10x + 1 = 0 x! 10x =!1!10 =!5 (!5) = 5 x! 10x + 5 =!1 + 5 ( x! 5) = 4 ( x! 5) = ± 4 x! 5 = ± 4 x! 5 = ± 6 x = 5 ± 6 It s not a coincidence, they should be the same number! 3x! 36x = 0 3x! 36x =!150 =!150 3 x! 1x 3 x! 1x 3 x! 1x =!50!1 =!150 3 =!6 (!6 ) = 36 x! 1x + 36 =! ( x! 6) =!14 ( x! 6) = ±!14 x! 6 = ± 14i x = 6 ± 14i Vertex Form The process we just used to solve equations can be used to rewrite functions from polynomial form to vertex form. The only difference is in step, when we factor out an a greater than 1, you don t divide by it, you just keep it outside the parenthesis, but we will explain that in an example. y = x + 18x y = x + 18x 18 = 9 9 = 81 y + 81 = x + 18x y + 81 = ( x + 9) + 95 y = ( x + 9) + 14 Vertex: (-9, 14) Some people like to add and subtract on the same side instead. + 95! 81 y = x + 18x + 81 y = ( x + 9) + 95! 81 y = x + 4x + 5 y = ( x + 1x) = 6 6 = ! 7 y = x + 1x + 36! 47 y = x + 6 Notice, we didn t subtract 36, we subtracted!36 because we multiplied by the number outside the parenthesis.

16 Section 1.8 Use the Quadratic Formula and the Discriminant The Quadratic Formula ax + bx + c = 0 ax + bx =!c! a x + b " # a x $ % & = 'c x + b a x =!c a! b $ " # a% & = b 4a x + b a x + b 4a =!c a + b 4a x + b a x + b 4a =!4ac + b 4a 4a x + b a x + b 4a = b! 4ac 4a! x + b $ " # a% & = b ' 4ac 4a The Quadratic Formula x + b a x + b 4a =!c a + b 4a x + b a x + b 4a =!4ac + b 4a 4a x + b a x + b 4a = b! 4ac 4a! x + b $ " # a% & = b ' 4ac 4a x + b a = ± x + b a = ± x =!b a ± x =!b ± b! 4ac 4a b! 4ac a b! 4ac a b! 4ac a The Quadratic Formula x =!b ± b! 4ac a The quadratic formula works for all quadratic equations, no restrictions, no limitations. Can t factor, use this. Can t solve with square roots, this works. Complex numbers, this will do it just fine. EVERYTHING WORKS! The Discriminant The radicand of the formula is called the discriminant. It tells us how many solutions we have, not what they are, just how many we have. b! 4ac < 0 b! 4ac = 0 b! 4ac > 0 complex solutions 1 real solution real solutions Remember, the solutions, zeros, x intercepts are all the same thing, so knowing how many solutions you have, is also knowing how many x intercepts there are. The Discriminant continued b! 4ac = 0 b! 4ac > 0 b! 4ac < 0

17 x! 5x = 7 x! 5x! 7 = 0 a = 1 b =!5 c =!7 x =!b ± b! 4ac a x =!!5 x = ±!5 ( 1) 5 ± x = 5 ± 53! 4 1 (!7) 16x! 3x = 17x! 5 16x! 40x + 5 = 0 a = 16 b =!40 c = 5 x =!b ± b! 4ac a ±!40 ( 16) x =!!40 x =! ± 1600! x = 40 3 x = 5 4 ( 5) x + 6x + 10 = 0 a = 1 b = 6 c = 10 x =!b ± x =! 6 b! 4ac a ± 6 ( 1)! 4 1 ( 10) x = x = x =!6 ± 36! 40!6 ±!4!6 ± i x =!3 ± i Basic Physics Formula English units Si Units Dropped Object Launched Object h =!16t + h 0 h =!4.9t + h 0 h =!16t + v 0 t + h 0 h =!4.9t + v 0 t + h 0 h0 is initial height v0 is initial vertical velocity The forces at Matulewicz Marshmallow are trying to attack Prusik Chocolate Chip Cookies. In defense of your cookie factory, you launch giant balls of cookie dough from the roof. You launch a ball upward with an initial velocity of 55 feet per second off a roof that is 30 feet high. How long before it hits the ground? continued h =!16t + v 0 t + h 0 v 0 = 55 h 0 = 30 h =!16t + 55t + 30 a =!16 b = 55 c = 30 t =!b ± b! 4ac a t =! 55 t = t = ± ( 55)! 4 (!16)( 30) (!16)!55 ± !3!55 ± 4945!3 t! "0.479 t! About seconds

18 Section 1.9 Graph and Solve Quadratic Inequalities Graphing Quadratic Inequalities Remember the whole < below and > above rule from graphing linear inequalities, well it 1. Graph the parabola using the appropriate line type. Solid for! and " and dashed use it. for < or >. works here too, so. Test a point inside the parabola, (0, 0) is the best point if it is available. 3. If you get a true statement, shad that region of the graph. If you get a false statement, shade the other region of the graph. y <!x! 4x Graph y <!x! 4x Vertex: (!, 4) x y x y Systems Just like in systems of linear inequalities, you care about the region of the graph where the shadings overlap. That could be inside one and outside the other, outside both, inside both, its just a matter of finding it. Graph # y > x! 3 $ %& y "!x + 4x +

19 Inequalities A quadratic inequality in one variable can be written in one of 4 formats: ax + bx + c > 0 ax + bx + c! 0 ax + bx + c < 0 ax + bx + c " 0 The first two produce OR inequalities, the second two produce AND inequalities when a is positive. Solve using a table x + x! 8 x + x! 8 " 0 x x + x! 8 x + x! 8 = 0 ( x + 4) x! = x! "4 OR x # Solve by graphing x! 6x < 0 y = x! 6x 0 < x < 6 Solve algebraically 3x! 9x! 1 < 0 = 0 3 x! 3x! 4 ( x + 1) = 0 3 x! 4 x = 4 x =!1!1 < x < 4 A truck 11 feet tall and 8 feet wide is traveling under an arch modeled by the function y =!0.065x + 1.5x , where x and y are measured in feet. Will the truck fit under the arch? = = 10 =!0.065( 10) + 1.5( 10) = 1 ( 10,1) x =!b a =! 1.5!0.065 y 10 y 10 A truck 11 feet tall and 8 feet wide is traveling under an arch modeled by the function y =!0.065x + 1.5x , where x and y are measured in feet. Will the truck fit under the arch? ( 10,1)!0.065x + 1.5x < 11 x! 0x + 84 = 0!0.065x + 1.5x! 5.5 < 0 ( x! 6) x! 14 < 0!0.065 x! 0x + 84 x! 0x + 84 > 0 = 0 x = 6 x = 14 x < 6 OR x > 14

20 A truck 11 feet tall and 8 feet wide is traveling under an arch modeled by the function y =!0.065x + 1.5x , where x and y are measured in feet. Will the truck fit under the arch? 10, = 14 10! 4 = 6 x < 6 OR x > 14 Yes, the truck will barely fit, but it will fit.