advertisement

Quadratics Definition or Depiction • Linear Function: – f(x) = mx+b – Makes a line – Constant change in y for each change in x • Quadratic Function: – Standard Form: f(x) = x2+bx+c or – Vertex Form: f(x) = a(x-h)2+k – Makes a parabola Parent Function • f(x) = x2 – Centered at (0, 0) – Symmetric – (graph) • Can move (Transform) – Left or Right – Up or Down – Stretching or Compressing Transformations (Alg 2.1) • In Vertex Form • F(x) = a(x – h)2 + k – a is vertical stretch/compression (skinnier/fatter) • Reflection over x & Direction of opening (- opens down (max), + opens up(min)) – h is translation left/right (x coordinate) (it is –h) – k is translation up/down (y coordinate) – Horizontal stretch/compression is with the x • Vertex is (h, k) • Domain & Range Impacts Example: Translating Quadratic Functions Use the graph of f(x) = x2 as a guide, describe the transformations and then graph each function. g(x) = (x – 2)2 + 4 Identify h and k. g(x) = (x – 2)2 + 4 h k Because h = 2, the graph is translated 2 units right. Because k = 4, the graph is translated 4 units up. Therefore, g is f translated 2 units right and 4 units up. Example: Translating Quadratic Functions Use the graph of f(x) = x2 as a guide, describe the transformations and then graph each function. g(x) = (x + 2)2 – 3 Identify h and k. g(x) = (x – (–2))2 + (–3) h k Because h = –2, the graph is translated 2 units left. Because k = –3, the graph is translated 3 units down. Therefore, g is f translated 2 units left and 4 units down. Domain & Range • Domain is possible x values – In quadratic functions – domain is typically All Real Numbers (ARN) – Written as [x: (-∞, ∞)] or ARN • Range is possible y values – Range will be limited by min/max in vertex – Range will be [y: (k, ∞)] or [y: (-∞, k)] Domain & Range Caution! The minimum (or maximum) value is the y-value at the vertex. It is not the ordered pair that represents the vertex. Transformations Reflections and Stretch/Compression • Across the x or y axis • a or b value stretches or compresses the graph Example: Reflecting, Stretching, and Compressing Quadratic Functions Using the graph of f(x) = x2 as a guide, describe the transformations and then graph each function. 1 2 ( ) g x =- x 4 Because a is negative, g is a reflection of f across the x-axis. Because |a| = , g is a vertical compression of f by a factor of . Example: Reflecting, Stretching, and Compressing Quadratic Functions Using the graph of f(x) = x2 as a guide, describe the transformations and then graph each function. g(x) =(3x)2 Because b = , g is a horizontal compression of f by a factor of . Application Example! Example 5 The minimum braking distance d in feet for a vehicle on dry concrete is approximated by the function (v) = 0.045v2, where v is the vehicle’s speed in miles per hour. The minimum braking distance dn in feet for a vehicle with new tires at optimal inflation is dn(v) = 0.039v2, where v is the vehicle’s speed in miles per hour. What kind of transformation describes this change from d(v) = 0.045v2, and what does this transformation mean? Example 5 Continued Examine both functions in vertex form. d(v)= 0.045(t – 0)2 + 0 dn(t)= 0.039(t – 0)2 + 0 The value of a has decreased from 0.045 to 0.039. The decrease indicates a vertical compression. Find the compression factor by comparing the new a-value to the old a-value. a from dn(t) a from d(v) = 0.039 0.045 = 13 15 Example 5 Continued The function dn represents a vertical compression of d by a factor of . The braking distance will be less with optimally inflated new tires than with tires having more wear. Check Graph both functions on a graphing calculator. The graph of dn appears to be vertically compressed compared with the graph of d. 15 0 15 0 Transformations – All Together • Example: f(x) = -4(x + 3)2 – 2 – Transformations: • Graph: Vertex Form from a Graph/Point • Vertex Form Equation: f(x) = a (x – h)2 + k • From Graph or given another point: – Determine vertex • Plug x coordinate for h and y coordinate for k – Identify another point on graph – Plug x and y coordinates into new vertex form • Solve for a • Plug a, h, & k into Vertex Form Equation Example • Vertex is (4, 4) – Write f(x) = a(x-4)2+4 • Additional point – (8, 2) • Insert – 2 = a(8-4)2 + 4 • Solve for a • Rewrite Vertex Equation: Standard Form (Alg 2.2) • F(x) = ax2 + bx + c – a determines which way parabola opens • a is the same in vertex and standard form • + opens up (happy) - opens down (sad) – Axis of Symmetry x = - b/(2a) – Y coordinate of vertex is when x=-b/(2a) – Vertex is (-b/(2a), f(-b/(2a))) Developing the standard form of a quadratic Another useful form of writing quadratic functions is the standard form. The standard form of a quadratic function is f(x)= ax2 + bx + c, where a ≠ 0. The coefficients a, b, and c can show properties of the graph of the function. You can determine these properties by expanding the vertex form. f(x)= a(x – h)2 + k f(x)= a(x2 – 2xh +h2) + k Multiply to expand (x – h)2. f(x)= a(x2) – a(2hx) + a(h2) + k Distribute a. f(x)= ax2 + (–2ah)x + (ah2 + k) Simplify and group terms. a= b= c= Graphing Standard Form No Calculator • • • • • • • Determine whether parabola opens up/down Find AOS (x value of vertex) Find y at the AOS (y value of vertex) Plot Vertex Plot y axis intercept = the value of c Plot a point across AOS from y intercept Fill in the curve Example: Graphing Quadratic Functions in Standard Form Consider the function f(x) = 2x2 – 4x + 5. a. Determine whether the graph opens upward or downward. Because a is positive, the parabola opens upward. b. Find the axis of symmetry. The axis of symmetry is given by Substitute –4 for b and 2 for a. The axis of symmetry is the line x = 1. . Example: Graphing Quadratic Functions in Standard Form Consider the function f(x) = 2x2 – 4x + 5. c. Find the vertex. The vertex lies on the axis of symmetry, so the x-coordinate is 1. The y-coordinate is the value of the function at this x-value, or f(1). f(1) = 2(1)2 – 4(1) + 5 = 3 The vertex is (1, 3). d. Find the y-intercept. Because c = 5, the intercept is 5. Example 2A: Graphing Quadratic Functions in Standard Form Consider the function f(x) = 2x2 – 4x + 5. e. Graph the function. Graph by sketching the axis of symmetry and then plotting the vertex and the intercept point (0, 5). Use the axis of symmetry to find another point on the parabola. Notice that (0, 5) is 1 unit left of the axis of symmetry. The point on the parabola symmetrical to (0, 5) is 1 unit to the right of the axis at (2, 5). Summary of Standard Form Properties of Quadratic Example • Given: f(x) = -3x2 + 6x – 4 • Find: – Opens Up or Down – Maximum or minimum – Axis of Symmetry – Vertex • Coordinate and Minimum or Maximum – Y intercept Real life Application: Example The highway mileage m in miles per gallon for a compact car is approximately by m(s) = –0.025s2 + 2.45s – 30, where s is the speed in miles per hour. What is the maximum mileage for this compact car to the nearest tenth of a mile per gallon? What speed results in this mileage? Example Continued The maximum value will be at the vertex (s, m(s)). Step 1 Find the s-value of the vertex using a = –0.025 and b = 2.45. 2.45) ( b s === 49 2a 2 (-0.025) Example Continued Step 2 Substitute this s-value into m to find the corresponding maximum, m(s). m(s) = –0.025s2 + 2.45s – 30 Substitute 49 for r. m(49) = –0.025(49)2 + 2.45(49) – 30 m(49) ≈ 30 Use a calculator. The maximum mileage is 30 mi/gal at 49 mi/h. Roots (Solutions or Zeros) • Roots are what make the equation = zero – f(x) = 0 or y=0 • One method to determine – Plug equation into calculator ( Y= ……..) – Look at Table (2nd Graph) for when y equals 0 – Take the x value – Works when roots are whole numbers • Example: Find roots of f(x) = x2 + 5x - 6 Factoring (Alg 2.3) • Set equation equal to 0 (f(x) = 0) • Factor - Separate Sheet – Common Factors – Product-Sum – Perfect Square – Difference of Squares • Example: f(x) = x2 - 4x - 12 Factoring (Alg 2.3) • • • • Lead coefficient not 1 (a not 1) Product – Sum – Group – Reduce(combine) Box Method Example: f(x) = 6x2 - 13x + 6 Roots from Factors • • • • Find the factors (from previous) Set each factor = 0 Solve for x Example: f(x) = x2 + 3x – 18 • Example: (2x-3)(x+5) Roots From Square Roots • Find roots by taking square root of both sides – If there is no linear term (no bx) – Using square roots to find solutions • Example: x2 – 25 = 0 • Example: 3x2 – 4 = 68 • Example: 25 = (x + 3)2 – If in form of an equation set f(x) = 0 • Example: f(x) = 2x2 - 32 Solutions by Square Roots • If written in vertex form (f(x) = a(x-h)2 + k) • Move k over • Divide by a • Take square root of both sides – Example: f(x) = 2(x-2)2 - 8 • Can work the same technique by setting up vertex form – Done by working to vertex form & setting up perfect square Complex Numbers • The square root of -1 is an imaginary number – No number multiplied by itself can be negative • Negative * negative = positive • Positive * positive = positive – i represents square root of -1 • To take square root of a negative number – Factor into -1 * positive number – Factor positive portion with greatest square (reduce) – Solve the positive portion and replace -1 with i Complex Numbers (cont) • 2 Parts of number (5 + 2i) – Real part (constant without the i : 5) – Imaginary part (part with the i : 2i) • Square root of negative can lead to both parts – Example: x2 = -25 – Example: 2x2 + 144 = 0 – Example: (x – 5)2 = -16 Complex Numbers (cont) • Complex Conjugates – – – – Write in form a + bi Leave the sign with the constant Flip the sign with the imaginary (i) part Allows multiplication (FOIL) to achieve quadratic • Examples: 9 – i i + 6 -4i Setting up a Perfect Square Complete the Square (Alg 2.4) • Add term to make a perfect trinomial – x2 + bx + _____ – Need to fill in blank (c) with (b/2)2 – Factors to (x + (b/2))2 – Example: x2 – 2x + ____ – Example: x2 + 5x + _____ Setting up the Square (cont.) • If there is already a value for c – Move c to opposite side of equation – Add (b/2)2 to both sides of the equation – Example: x2 + 4x – 6 = 0 – Example: y = 2x2 + 6x - 8 Using CTS to Solve Quadratics Check It Out! Example Solve the equation by completing the square. 3x2 – 24x = 27 x2 – 8x = 9 x2 –8x + =9+ Divide both sides by 3. Set up to complete the square. Add to both sides. Simplify. Check It Out! Example Continued Solve the equation by completing the square. Factor. Take the square root of both sides. Simplify. x – 4 =–5 or x – 4 = 5 x =–1 or x = 9 Solve for x. Using CTS to set up Vertex Form • If equation is in f(x) = x2 + bx + c – Parenthesize variables: f(x) = (x2+bx ) + c – Add (b/2)2 inside parentheses (x2 + bx + (b/2)2) – Subtract (b/2)2 from outside () + c – (b/2)2) – Rewrite parentheses into square (x+(b/2))2 – Combine constants outside • Example: (a=1) f(x) = x2 + 10x -13 Using CTS - Vertex Form: a not 1 • If equation is in f(x) = ax2 + bx + c – Parenthesize variables: f(x) = (ax2+bx ) + c – Factor out the a: f(x) = a(x2+(b/a)x ) + c – Add (b/2a)2 inside: a(x2 + (b/a)x + (b/2a)2) – Subtract a(b/2a)2 from outside () + c – a(b/2a)2) – Rewrite parentheses into square (x+(b/2a))2 – Combine constants outside • Example: f(x) = 4x2 – 8x + 5 Quadratic Formula • Keep track of numbers – always works Example Find the zeros of f(x)= x2 – 8x + 10 using the Quadratic Formula. x2 – 8x + 10 = 0 Set f(x) = 0. Write the Quadratic Formula. Substitute 1 for a, –8 for b, and 10 for c. Simplify. Write in simplest form. Example Continued Check Solve by completing the square. x2 – 8x + 10 = 0 x2 – 8x = –10 x2 – 8x + 16 = –10 + 16 (x + 4)2 = 6 Quadratic Formula • The Discriminant: – Value under the square root (b2 – 4ac) • Tells how many real roots the equation has – Discriminant is > 0 : there are 2 real roots – Discriminant = 0 : there is one real root (double) – Discriminant is < 0 : 2 COMPLEX roots (No Real) The graph shows related functions. Notice that the number of real solutions for the equation can be changed by changing the value of the constant c. Quadratic Formula – Finding Roots • When the Discriminant is > 0 – Example: x2 + 5x – 6 • When the Discriminant = 0 – Example: x2 – 12x + 36 • When the Discriminant is < 0 – Example: x2 + 4x + 8 Complex Number Operations • Adding Complex Numbers – Add the real numbers – Add the imaginaries – Keep them separate • Expansion of i – i2 = -1 – i3 = -i – i4 = 1 – And so on …………. Examples Complex Number Operations • Multiplying Complex Numbers – FOIL the terms – remember i2 = -1 – Combine like terms (reals, i’s) • Dividing Complex Numbers – Multiply top & bottom by complex conjugate (from Intro to Complex Numbers) – Eliminates complex on bottom – Results in multiplying rules on top – Simplify Writing Equations from Roots • • • • • • Write roots as solutions Set roots = 0 Combine factors & set = 0 Multiply binomials (FOIL) Check via calculator Example: Roots 2 & -4 • Example: Roots -3/2, 3 Quadratic Inequalities • Change to Equality • Make = 0 – Add/Subtract term to make standard equation • Factor (if able) • Find the roots from the factors (or CTS, Quad Form) – These set the boundaries of the inequality – If Greater Than or Less Than • Open Circles – If Greater Than/Equal To or Less Than/Equal To • Closed Circles Quadratic Inequalities • Plot on Number Line – Open vs Closed Circle • Take value above, between & below – Check for True or False – And is between – written low root<x<high root – Or is outside – written x<low root or x>high root Quadratic Inequalities • Using a calculator – Graphing • Plot Y1 = Quadratic Equation • Plot Y2 = Value • Look at graph where equation meets criteria – Table • Plot Y1 = Quadratic Equation • Look at where table meets criteria Real World Example A business offers educational tours to Patagonia, a region of South America that includes parts of Chile and Argentina . The profit P for x number of persons is P(x) = –25x2 + 1250x – 5000. The trip will be rescheduled if the profit is less $7500. How many people must have signed up if the trip is rescheduled? Setup 1 Understand the Problem The answer will be the number of people signed up for the trip if the profit is less than $7500. List the important information: • The profit will be less than $7500. • The function for the profit is P(x) = –25x2 + 1250x – 5000. Setup to Solve 2 Make a Plan Write an inequality showing profit less than $7500. Then solve the inequality by using algebra. Apply Appropriate MATH Concept 3 Solve Write the inequality. –25x2 + 1250x – 5000 < 7500 Find the critical values by solving the related equation. –25x2 + 1250x – 5000 = 7500 –25x2 + 1250x – 12,500 = 0 –25(x2 – 50x + 500) = 0 Write as an equation. Write in standard form. Factor out –25 to simplify. Do the Math 3 Solve Use the Quadratic Formula. Simplify. x ≈ 13.82 or x ≈ 36.18 Continue the Math 3 Solve Test an x-value in each of the three regions formed by the critical x-values. Critical values 5 10 15 20 25 30 Test points 35 Finish the Math 3 Solve –25(13)2 + 1250(13) – 5000 < 7500 7025 < 7500 –25(30)2 + 1250(30) – 5000 < 7500 10,000 < 7500 –25(37)2 + 1250(37) – 5000 < 7500 7025 < 7500 Try x = 13. Try x = 30. x Try x = 37. Write the solution as an inequality. The solution is approximately x > 36.18 or x < 13.82. Because you cannot have a fraction of a person, round each critical value to the appropriate whole number. Answer the Question 4 State the Answer The trip will be rescheduled if the number of people signed up is fewer than 14 people or more than 36 people. Focus & Directix • Focus is a point inside the parabola • Directrix is a straight line outside the parabola • The parabola is formed by points equidistant from both • Visual Example: online at – http://www.mathwarehouse.com/quadratic/para bola/focus-and-directrix-of-parabola.php Focus & Directrix • Focus is on Axis of Symmetry – h or x coordinate of the vertex • Vertex is ½ way between focus & directrix • The vertical stretch defines the distance from the Vertex to the Focus – If focus above vertex: positive p (+p) – If focus below vertex: negative p (-p) – Distance (p) = 1/(4a) – Flatter parabola = farther focus/directrix