Quadratic Functions and Transformations Quadratic Function (or Parabola) Standard Form: 𝑓 𝑥 =𝑎 𝑥 2 + 𝑏𝑥+𝑐 ; 𝑎≠0 Vertex Form: 𝑓 𝑥 =𝑎 (𝑥−ℎ) 2 +𝑘 ; 𝑎≠0 The vertex is at (h,k) and the axis of symmetry is at x=h. The parent function is f(x) = x2 and we can make any other parabola by applying the transformations we have studied. For a>0, y is a minimum at k, for a<0, y is a maximum at k.

Interpreting the Vertex Form For the following parabolas, find: Vertex and Axis of Symmetry Minimum or Maximum value Domain and Range y = 3(x-4)2 – 2 y = -2(x+1)2 + 4

Using Vertex Form For the following parabolas: f(x) = -2(x-1)2 + 3 Graph by hand: Identify a, h, k Plot the vertex Plot two symmetric points Sketch by hand Describe a series of transformations that result in the parabola, starting from y=x2. f(x) = -2(x-1)2 + 3 f(x) = -2(x+1)2 + 3 f(x) = 2(x+2)2 - 5

Using Vertex Form (cont’d) Write the equation of the following parabolas in vertex form: Hint: Identify (h,k), substitute in the given point and solve for “a” in the vertex form of the equation: y = a(x-h)2+k Vertex (1,2) through point (2, -5) Vertex (0,5) through point (1, -2) Do Problem 39 page 199, using a graphing calculator.

Hwk 16 Hwk 16: Pages 199-201: 28, 30, 32, 36, 38, 40, 42

Standard Form of a Quadratic Function Standard Form: f(x) = ax2+bx+c ; a ≠0 Properties of the parabola: If a>0, opens upward; if a<0, opens downward. Axis of symmetry: x = - (b/2a) Vertex: (h, k) is: h = -(b/2a) and k = f(h) y-intercept is at (0,c) Exercise: Verify the above by converting: f(x) = a(x - h)2 + k (vertex form) into standard form. (In other words: find a, b, and c in terms of a, h, k.)

Using Standard Form Find the axis of symmetry, vertex, and graph: f(x) = x2 + 2x + 3 f(x) = -2x2 + 2x -5 Confirm with a graphing calculator Convert the following parabolas to vertex form: f(x) = 2x2 + 10x +7 f(x) = -x2 + 4x -5

Modeling the New River Gorge Bridge An arch bridge spans a river using an arch that can be modeled as y = -.000498x2+.847x where y = height above some level on both river banks from which the bridge starts to rise, and x is the horizontal distance from the left bank from where the bridge starts to rise. (Both x and y are in feet.) Find the maximum height of the bridge. Find the horizontal extent of the bridge above the arch.

Modeling with Quadratic Functions How many examples of quadratic functions have you seen in Physics? Give one from kinematics. Give one from the work-energy unit.

Writing an Equation of a Parabola Standard form of a parabola: y = ax2 + bx + c How many points on a parabola would we need in order to be able to write its equation? Write the equation of the following parabolas : (0,0), (-1,-2), (1,6) (0,0), (1,-2), (-1,-4)

Quadratic Regression Quadratic Regression finds the best fit parabola for a set of data points. Here our temperatures at various times during a day: Time (24 hour clock) (x) (8:00) 8 (10:00) 10 (12:00) 12 (2:00P) 14 (4:00P) 16 (6:00P) 18 Temp (◦F) (y) 52 64 72 78 81 76

Using the TI-nspire for Quadratic Regression Start a new Document:Prob3Pg211 Choose Lists and Spreadsheets (4) Enter data Menu: Statistics(4)-> Stat Calculations(1) ->Quadratic Regression(6)->Put in x,y, tab to OK. Now Let’s plot the points and the best fit parabola: Add a Page: ctrl + Menu: Add data&Statistics(5) Menu: Plot Properties(2) Add x, Plot Properties(2) Add y Superimpose best fit: Menu: Analyze(4)->Plot Function(4)-> Enter function given by best fit.

Stopping Distance The following data relates speed to stopping distance: Let’s use the TI-nspire to construct a Quadratic Regression model for this data. Can we estimate the coefficient of kinetic friction (µk ) from this data? speed (mph) 10 20 30 40 50 stopping distance (ft) 15.1 39.9 75.2 120.5 175.9

Identifying Linear and Quadratic Data Linear Data: “first differences” in the y-values are the same as we increment x by the same amount. Here the first differences are 3. Quadratic Data: “second differences” (differences in the first differences” in the y-values are the same as we increment x by the same amount. Here, the second differences are 6. x -2 -1 1 2 y=3x+1 -5 4 7 x -2 -1 1 2 y=3x2+1 13 4

Which of the following are sets of data are Quadratic? x 1 2 3 4 5 6 7 y 12 22 36 54 76 102 x -2 -1 1 2 3 4 y -8 8 27 64

Factoring Quadratic Expressions We can factor trinomial quadratic expressions into the product of two binomial expressions. When a = ±1: We need to find the numbers whose product is c and sum is b in x2+bx+c. x2+ 9x +20, x2+14x-72, -x2+13x+12 Special case 1: “Perfect square trinomial” A quadratic expression that can be factored into a form: f(x) = x2+2kx+k2 = (x+k)2 , f(x) = x2-2kx+ k2 = (x-k)2 (Note: b = 2k, c = k2 = 𝑏 2 2 Special case2: “Difference of two squares” f(x) = x2-k2 = (x-k) (x+k) = x2-kx +xk – k2 = x2-k2

Completing the Square This technique allows us to: rewrite any quadratic expression as a perfect square plus a constant term. solve any quadratic equation by using one formula Ex1: Rewrite f(x) = 3x2 -12x as a perfect square plus a constant term. Factor out 3: f(x) = 3(x2-4x ) Add a term to the part in parentheses to turn it into a perfect square and then subtract the same amount to keep f(x) unchanged. We need to add: c = 𝑏 2 2 or −4 2 2 : f(x) = 3(x2-4x +4 -4 ) = 3( x2-4x +4 -4 )= 3( (x-2)2 -4) = 3(x-2)2 -12

Completing the Square Ex 2: f(x) = 3x2 -12x =0 two ways 1) By factoring: f(x) = 3x( x-4) = 0 So x =0, +4 are roots. 2) By completing the square: From last slide we saw: f(x) = 3(x-2)2 -12 =0 Divide by 3: (x-2)2 - 4=0 Then: (x-2)2 = 4 Then x-2 = ±2 Which gives two linear equations: x -2 = 2 giving x=4 x-2 = -2 giving x=0 But completing the square will work when you can’t factor.

The Quadratic Formula By completing the square, we can come up with a formula that will solve any quadratic equation of the form 𝑎 𝑥 2 + 𝑏𝑥+𝑐=0

The Quadratic Formula By completing the square, we can come up with a formula that will solve any quadratic equation of the form 𝑎 𝑥 2 + 𝑏𝑥+𝑐=0

The Quadratic Formula The result is 𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 Solve the following equations using the quadratic formula. x2+6x+9=0 2x2 –x = 4

The Discriminant We call the value: 𝑏 2 −4𝑎𝑐 the discriminant of a quadratic equation. Since it appears within a radical and appears after the ± sign, its value tells us the following: If 𝑏 2 −4𝑎𝑐 > 0, we have two real solutions. If 𝑏 2 −4𝑎𝑐 = 0, we have one real solution. If 𝑏 2 −4𝑎𝑐 < 0, we have no real solutions. We will soon define a number: 𝑖= −1 which will expand our work into “complex” numbers and we’ll see that when 𝑏 2 −4𝑎𝑐 < 0, we have complex solutions.

The Different Numbers Natural or Counting Numbers: 1, 2, 3, …. Whole Numbers: 0, 1, 2, 3, …. Integers: …., -3, -2, -1, 0, 1, 2, 3, …. Rational Numbers: numbers that can be expressed as ratios of two integers (fractional form); includes decimal numbers that have an infinitely repeating pattern. Irrational Numbers: decimals that can’t be expressed in fractional form. (e.g.: π, 2 ) Real Numbers: Includes rational and irrational numbers.

Complex Numbers Invented by Gerolamo Cardano (1500’s) and given practical usage by Leonhard Euler (1700’s). Initially invented to solve polynomial equations that otherwise couldn’t be solved. (ex: 𝑥 2 + 1 = 0). We define: 𝑖 2 = -1 or 𝑖= −1 Then z = a + bi is a complex number where: a = “the real part” and b = “the imaginary part” If a=0 , and b≠0 then z is a pure imaginary number. Complex numbers can be represented graphically with a plotted on the horizontal (real) axis and b plotted on the vertical (imaginary) axis. The absolute value of a complex number is: 𝑧 = 𝑎 2 + 𝑏 2

Graphically Represent and find the Absolute Value:

Algebra with Complex Numbers Addition/Subtraction: Add or Subtract real and imaginary components separately. (What type of addition/subtraction is this similar to in Physics?) (4-3i) + (-4+3i) = (2+4i) - (6+2i) = Multiplication: Multiply the same way as you would binomials, but: when you get a term involving i2 it becomes real and negated since: 𝑖 2 = -1. (3 + 4i) (4 - 2i) = (2 + 3i) (2 - 3i) = (a + bi) (a - bi) =

Algebra with Complex Numbers (cont’d) Note that the product: (2 + 3i) (2 - 3i) or, in general: (a + bi)(a - bi) gave us a real result. We call such a pair of numbers complex conjugates. Division: Multiply the numerator and the denominator by the complex conjugate of the denominator. The new denominator is real and you can finish the problem with distribution of the denominator. 9+12𝑖 3𝑖 = 5−2𝑖 3+4𝑖 = 8−7𝑖 8+7𝑖 =

Quadratic Equations and Complex Solutions Every quadratic equation has two complex solutions which are sometimes real. Look at the quadratic formula: 𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 When does the solution have an imaginary part? (Hint: look at the discriminant.) When the solution has an imaginary part, how do the two solutions relate to each other? So if the solutions are not real, they occur in complex conjugate pairs.

Some Exercises Solve using factoring: 5 𝑥 2 +20 = 0 Solve using the quadratic formula: 2 𝑥 2 −3𝑥+5=0

Powers of 𝒊 Notice: 𝑖 0 =1 𝑖 1 =𝑖 𝑖 2 =−1 𝑖 3 =−𝑖 𝑖 4 =1 𝑖 5 =𝑖 etc. SO: When you have a power: 𝑖 𝑛 you can divide n by 4 and get the answer from the remainder.

Hwk 20 Pages 253-255: 22, 24, 28, 30, 40, 45 (just do a,b,c,d for point A), 50, 56, 60, 66, 68.

Solving Quadratic Inequalities Factoring and reasoning applied to possible signs of the factors: 2 𝑥 2 −14𝑥 <0 2 𝑥 2 −14𝑥>0 Use the quadratic formula to find roots, then look at the boundaries on either side of the roots. 𝑥 2 −6𝑥+5<0 Use a graphing calculator to plot the quadratic function and identify the zeros and regions of desired inequality.

Solving Quadratic Systems Linear-Quadratic System of Equations Quadratic System of Equations Solve by: Graphing and identifying points of intersection Algebra using substitution. Solve: 𝑦=− 𝑥 2 +5𝑥+6 and 𝑦=𝑥+6 𝑦=− 𝑥 2 −𝑥+6 and 𝑦=𝑥+3 𝑦=− 𝑥 2 −𝑥+12 and 𝑦= 𝑥 2 +7𝑥+12 𝑦= 𝑥 2 −4𝑥+5 and 𝑦= −𝑥 2 +5

Quadratic System of Inequalities Graph and identify regions of overlap: Solve: 𝑦<− 𝑥 2 −9𝑥−2 and 𝑦> 𝑥 2 −2

Hwk 21 Pages 262-264: 18, 29, 30, 47, 61