Chapter 3 Quadratic Models

Ch 3 Quadratic Equation A quadratic equation involves the square of the variable. It has the form y = ax 2 + bx + c where a, b and c are constants If a = 0 , there is no x-squared term, so the equation is not quadratic

Graph of the quadratic equation y = 2x2 – 5 -3 -2 -1 1 2 3 y 13 -5 To solve the equation 2x2 – 5 = 7 We first solve for x2 to get 2x2 = 12 x2 = 6 x = + = + 2.45 and – 2.45 - 7 5 3 1 -3 -2 - 1 0 1 2 + 3

Extraction of roots (Ex, 1 Pg 145) The formula h= 20 - 16t2 h when t = 0.5 = 20 – 16(0.5) 2 = 20 – 16(0.25) = 20 – 4 = 16ft When h = 0 the equation to obtain 0 = 20 - 16t 2 16t 2 = 20 t 2 = 20/16 = 1.25 t = + = + 1. 118sec - - a (16, 0.5) 20 10 Height b 0.5 1 1.5 t Time

Solving Formulas h r r h Volume of Cylinder V= r2 h V = r2 h Volume of Cone V = 1 r2 h 3 3V= r2 h ( Divide both sides by h ) and find square root r = + 3V - h r r h Volume of Cylinder V= r2 h V = r2 h r = + V - h (Dividing both sides by h )

Compound Interest Formula A = P(1 + r) n Where A = amount, P = Principal, R = rate of Interest, n = No.of years

More Extraction of Roots Equations of the form a( x – p) 2 = q Can also be solved by extraction of roots after isolating the squared expression ( x – p) 2

Pythagorian Formula for Right angled triangle Hypotenuse Height 90 degree C B Base In a right triangle (Hypotenuse) 2 = (Base) 2 +(Height) 2

s represent the length of a side of the square s 2 + s 2 = 16 2 What size of a square can be inscribed in a circle of radius 8 inches ? 8 in 16 inches s 8in s s represent the length of a side of the square s 2 + s 2 = 16 2 2s = 256 s 2 = 128 s = 128 = 11.3 inches

Ch 3.2 Some examples of Quadratic Models Height of a Baseball (Pg 156) H = -16t 2 + 64t + 4 Evaluate the formula to complete the table of values for the height of the baseball t 1 2 3 4 h 52 68 70 60 50 40 30 20 10 Highest point 3) After ½ second base ball height h = -16(1/2) 2 + 64(1/2) + 4 = 32 ft 4) 3.5 second height will be 32 ft 5) When the base ball height is 64 ft the time will be 1.5 sec and 2.5 sec 6) When 20 ft the time is 0.25 and 3.75 sec 7) The ball caught = 4 sec 0 1 2 3 4

Example 4 ( Page 156 ) Press graph Using Graphing Calculator H = - 16 x2 + 64x + 4 Press Y key TblStart = 0 and increment 1 Press 2nd , table Press graph

3.3 Solving Quadratic Equations by Factoring Zero Factor Principle The product of two factors equals zero if and only if one or both of the factors equals zero. In symbols ab = 0 if and only if a = o or b = 0 Example (x – 6) (x + 2) = 0 x – 6 = 0 or x + 2 = 0 x = 6 or x = -2 Check 6, and – 2 are two solutions and satisfy the original equation And x-intercepts of the graph are 6, -2 By calculator, draw the graph

Solving Quadratic Equation by factoring The height h of a baseball t seconds after being hit is given by h = - 16 t 2 + 64t + 4. When will the baseball reach a height of 64 feet ? 64 = - 16 t 2 + 64t + 4 Standard form 16 t 2 – 64t + 60 = 0 4( 4 t 2 – 16t + 15) = 0 Factor 4 from left side 4(2t – 3)(2t – 5) = 0 Factor the quadratic expression and use zero factor principle 2t – 3 = 0 or 2t – 5 = 0 Solve each equation t = 3/2 or t = 5/2 h = - 16 t 2 + 64t + 4 72 64 48 24 0 .5 1 1.5 2 2.5 3 4

Pg 163 Enter window Xmin = -2, Xmax = 8 Ymin = -5 Ymax = 10 Y1 = x2 – 4x + 3 Y2 = 4(x2 – 4x + 3) Enter window Xmin = -2, Xmax = 8 Ymin = -5 Ymax = 10 And enter graph

Quadratic Equations whose solutions are given Example 31, Page 166 Solutions are – 3 and ½, The equation should be in standard form with integer coefficients [ x – (-3)] (x – ½) = 0 (x + 3)(x – ½) = 0 x2 – ½ x + 3x – 3/2 = 0 x2 +5x – 3= 0 2 2 2(x2 + 5x –3) = 2(0) 2 2 2 x2 + 5x – 3 = 0

3.3 ,No 48, Page 169 I = kCx – k x2 = 0.0002 (6000)x – 0.0002x2 = 1.2x – 0.0002x2 x – intercept is 6000 i.e neither decrease nor increase Larger Increase x 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 5500 6000 6500 7000 I 550 1350 1600 1750 1800 -650 -1400 Population 2000 will increase by 1600 1800 1750 1600 1350 1000 500 Population 7000 will decrease by 1400 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 5500 6000 6500 7000

3.4 Graphing Parabolas Special cases The graph of a quadratic equation is called a parabola Vertex y-intercept x-intercept x-intercept x-intercept y-intercept x-intercept Axis of symmetry Axis of symmetry

Using Graphing Calculator (Page 171) Enter Y Y = x2 Y = 3 x2 Y = 0.1 x2 Graph Enter equation Enter Graph

a) The x-coordinate of the vertex is xv = -b/2a= -(-16.2)/2(-1.8) Example 3, Pg 177, Finding the vertex of the graph of y = -1.8x2– 16.2x Find the x-intercepts of the graph a) The x-coordinate of the vertex is xv = -b/2a= -(-16.2)/2(-1.8) To find the y-coordinate of the vertex, evaluate y at x = -4.5 yv = -1.8(-4.5)2 – 16.2(-4.5) = 36.45 The vertex is (-4.5, 36.45) b) To find the x-intercepts of the graph, set y = 0 and solve - 1.8 x2 – 16.2x = 0 Factor -x(1.8x + 16.2) = 0 Set each factor equal to zero - x = 0 1.8x + 16.2 = 0 Solve the equation x = 0 x = -9 The x-intercepts of the graph are (0,0) and (-9,0) 36 24 12 - 10 - 5 0 2

3.6 Quadratic Formula The solutions of the equation ax 2 + bx + c = 0 with a = 0 are given by - b + b2 – 4ac Complex Numbers i2 = - 1 or i = For a > 0, = = i 2a Discriminant D = b2 - 4ac If D > 0, the equation has two unequal real solutions If D = 0, the equation has one real solution of multiplicity two If D < 0, the equation has two complex (conjugate) solutions

3.6, Page 195, No. 15 Let w represent the width of a pen and l the length of the enclosure in feet Then the amount of chain link fence is given by 4w + 2l = 100 b) 4w +2l = 100 2l = 100 – 4w l = 50 – 2w c) The area enclosed is A = wl = w(50 – 2w) = 50w –2w2 The area is 250 feet, so 50w – 2w2 = 250 0 = w2– 25w + 125 Thus a = 1, b = -25 and c = 125 W = -(-25) + (- 25) 2 - 4(1)(125) - 2(1) The solutions are 18.09, 6.91 feet d) l =50 – 2(18.09) = 13.82 feet and l = 50 – 2(6.91) = 36.18 feet The length of each pen is one third the length of the whole enclosure, so dimensions of each pen are 18.09 feet by 4.61 feet or 6.91 feet by 12.06 feet

Ex 16, Pg 195 r = ½ x The area of the half circle = 1 r2 2 = 1/2 (1/2 x )2 = 1/8 x2 The area of the rectangle = x2 – 2x = 1 x2 + x2 - 2x 8 h = x - 2 Total area = 120 square feet 120 = = 1 x2 + x2 - 2x 8 8(120) = 8 ( 1 x2 + x2 - 2x ) 0 = x2 + 8 x2 - 16x – 960 , 0 = ( + 8) x2 - 16x – 960 0 = 11.142 x2 – 16x - 960 use quadratic formula , x = 10.03 ft , h = 10.03 – 2 = 8.03ft The overall height of the window is h + r = h + ½ x = 8.03 + ½ (10.03) = 13.05 ft x