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QUADRATIC FUNCTIONS. Monika V Sikand Light and Life Laboratory Department of Physics and Engineering physics Stevens Institute of Technology Hoboken, New Jersey, 07030. OUTLINE . Graphing Quadratic Functions Solving Quadratic Equations by Factoring
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QUADRATIC FUNCTIONS Monika V Sikand Light and Life Laboratory Department of Physics and Engineering physics Stevens Institute of Technology Hoboken, New Jersey, 07030.
OUTLINE • Graphing Quadratic Functions • Solving Quadratic Equations by Factoring • Solving Quadratic Equations by Finding Square Roots • Complex Numbers • The Quadratic Formula and the Discriminant • Graphing and Solving Quadratic Inequalities • Modeling with Quadratic Functions
A QUADRATIC FUNCTION • A quadratic function has a form y = ax2 + bx + c where a ≠ 0. • The graph of a quadratic function is U-shaped and is called parabola. • The lowest or highest point on the graph of a quadratic function is calledthe vertex. • The graphs of y = x2 and y = -x2 are symmetric about the y-axis, called the axis of symmetry. y = x2 vertex y = -x2 Axis of symmetry
THE GRAPH OF A QUADRATIC FUNCTION • The parabola opens up if a>0 and opens down if a<0. • The parabola is wider than the graph of y = x2 if |a| < 1 and narrower than the graph of y = x2 if |a| > 1. • The x-coordinate of the vertex is -b/2a. • The axis of symmetry is the vertical line x = -b/2a. y = x2 vertex y = -x2 Axis of symmetry
EXAMPLE Graph y = 2x2 -8x +6 Solution: The coefficients for this function are a = 2, b = -8, c = 6. Since a>0, the parabola opens up. The x-coordinate is: x = -b/2a, x = -(-8)/2(2) x = 2 The y-coordinate is: y = 2(2)2-8(2)+6 y = -2 Hence, the vertex is (2,-2).
EXAMPLE(contd.) Axis of symmetry • Draw the vertex (2,-2) on graph. • Draw the axis of symmetry x=-b/2a. • Draw two points on one side of the axis of symmetry such as (1,0) and (0,6). • Use symmetry to plot two more points such as (3,0), (4,6). • Draw parabola through the plotted points. (4,6) (0,6) x (3,0) (1,0) (2,-2) y
VERTEX FORM OF QUADRATIC EQUATION y = a(x - h)2 + k • The vertex is (h,k). • The axis of symmetry is x = h.
GRAPHING A QUADRATIC FUNCTION IN VERTEX FORM Axis of symmetry • Example y = -1/2(x + 3)2 + 4where a = -1/2, h = -3, k = 4. Since a<0 the parabola opens down. • To graph a function, first plot the vertex (h,k) = (-3,4). • Draw the axis of symmetry x = -3 • Plot two points on one side of it, such as (-1,2) and (1,-4). • Use the symmetry to complete the graph. (-3,4) (-1,2) (-5,2) x (-7,-4) (1,-4) y
INTERCEPT FORM OF QUADRATIC EQUATION y = a(x - p)(x - q) • The x intercepts are p and q. • The axis of symmetry is halfway between (p,0) and (q,0).
GRAPHING A QUADRATIC FUNCTION IN INTERCEPT FORM Axis of symmetry • Example y = -(x + 2)(x - 4).where a = -1, p = -2, q = 4. Since a<0 the parabola opens down. • To graph a function, the x - intercepts occur at (-2,0) and (4,0). • Draw the axis of symmetry that lies halfway between these points at x = 1. • So, the x - coordinate of the vertex is x = 1 and the y - coordinate of the vertex is: y = -(1 + 2)(1 - 4)= 9. (1,9) (-2,0) (4,0) x y
WRITING THE QUADRATIC EQUATION IN STANDARD FORM (1). y = -(x + 4)(x - 9) = -(x2 - 9x + 4x - 36) = -(x2 - 5x -36) = -x2 + 5x + 36 (2). y = 3(x -1)2 + 8 = 3(x -1)(x - 1) + 8 = 3(x2 - x - x + 1) + 8 = 3(x2 - 2x + 1) + 8 = 3x2 - 6x + 3 + 8 = 3x2 - 6x + 11
QUADRATIC FUNCTIONS IN REAL LIFE Researchers conducted an experiment to determine temperatures at which people feel comfortable. The percent of test subjects who felt comfortable at temperature x( in degrees Fahrenheit) can be modeled by: y = -3.678x2 + 527.3x - 18.807 What temperature made the greatest percent of test subjects comfortable? At that temperature , what percent felt comfortable?
SOLUTION Since a = -3.678 is negative, the graph of the quadratic function open down and the function has a maximum value. The maximum value occurs at: The corresponding value of y is: Hence, The temperature that made the greatest percent of test subjects comfortable was about 72. At that temperature about 92% of the subjects felt comfortable.
REAL LIFE EXAMPLE 6(page 252) The Golden Gate Bridge in San Francisco has two towers that rise 500 feet above the road and are connected by suspension cables as shown. Each cable forms a parabola with equation where x and y are measured in feet. What is the distance d between the towers? What is the height l above the road of a cable at its lowest point?
GOLDEN GATE BRIDGE y 500ft l x d 200ft Not down to scale
SOLUTION Hence the vertex of the parabola is (2100,8)
SOLUTION(contd.) The vertex of the parabola is (2100,8), so the cable’s lowest point is 2100 feet from the left tower shown above. Since the heights of the two tower’s are the same, the symmetry of the parabola implies that the vertex is also 2100 feet from the right tower. Therefore the towers are d = 2(2100) = 4200 feet apart. The height l above the road of a cable at its lowest point is the y-coordinate of the vertex. Since the vertex is (2100,8), this height is l= 8 feet.
FACTORING QUADRATIC EXPRESSION The expression x2 + bx + c is a trinomial because it has three terms. We can use factoring to write it as product of two terms or binomials such as x2 + bx + c = (x+m)(x+n) = x2 + (m+n)x + mn Example: x2 + 8x + 15 = (x + 3)(x + 5)
EXAMPLE Problem: Factor x2 - 12x - 28 Solution: x2 - 12x - 28 = (x+m)(x+n) where mn = -28 and m+n = -12 The table shows that m = 2 and n = -14. So, x2 - 12x - 28 = (x + 2)(x - 14)
SPECIAL FACTORING PATTERNS Difference of two squares: a2 - b2 = (a+b)(a-b) Example: x2 - 9 = (x+3)(x-3) Perfect square Trinomial: a2 + 2ab + b2 = (a+b)2 Example: x2 + 12x +36 = (x+6)2 3. Perfect square Trinomial: a2 - 2ab + b2 = (a-b)2 Example: x2 -8x +16 = (x-4)2
FACTORING MONOMIALS FIRST Monomial is an expression that has only one term. Factor the quadratic expression: 5x2 - 20 = 5(x2 -4) = 5(x+2)(x-2) 6p2 + 15p + 9 = 3(2p2 + 5p + 3) = 3(2p + 3)(p + 1)
SOLVING QUADRATIC EQUATIONS Solve: Solve: The solution is 5. The solutions are -6 and 3.
REAL LIFE EXAMPLE You have made a rectangular stained glass window that is 2 feet by 4 feet. You have 7 square feet of clear glass to create a border of uniform width around the window. What should the width of the border be? x x x x 4+2x 4 x x x x 2 2+2x
SOLUTION Let the width of the border be = x Area of the border = 7 Area of the border and window = (2+2x)(4+2x) Area of the window = 24 = 8 Area of border = Area of border & window - Area of window 7 =(2+2x)(4+2x) - 8 0 = 4x2 + 12x -7 0 = (2+7x)(2x-1) 2+7x = 0 or 2x-1 = 0 x = -3.5 o x = 0.5 Rejecting negative value, -3.5. Hence the border’s width is 0.5ft
ZERO PRODUCT PROPERTY Let A and B be real numbers or algebraic expressions. If AB=0, then either A = 0 or B = 0 Solve (a)x2 + 3x -18 = 0 (x + 6)(x - 3) = 0. Hence either x + 6 = 0 or x - 3 = 0. The solutions arex = -6 or x = 3
ZERO PRODUCT PROPERTY Let A and B be real numbers or algebraic expressions. If AB=0, then either A = 0 or B = 0 Solve (a)2t2 - 17t + 45 = 3t - 5 2t2 - 20t + 50 = 0 t2 - 10t + 25 = 0 (t - 5)2 = 0 t - 5 = 0 t = 5 Hence the solutions is 5.
FINDING ZEROES OF QUADRATIC FUNCTIONS Find zeros of y = x2 -x -6 Solution: y = x2 - 3x + 2x - (32) y = x(x-3) + 2(x-3) y = (x + 2)(x - 3) Hence the zeros of the function are -2 and 3.
SQUARE ROOT A number r is a square root of a number s if r2 = s. A positive number s has square roots denoted by Example : Since 32 = 9 and (-3)2 = 9. The square roots of 9 are:
PROPERTIES OF SQUARE ROOTS (a>0, b>0) PRODUCT PROPERTY: QUOTIENT PROPERTY:
Solving a quadratic equation Solve: Solve: Hence the solutions are: Hence the solutions are: and and
REAL LIFE EXAMPLE A stunt man working on the set of a movie is to fall out of a window 100 feet above the ground. For the stunt man’s safety, an air cushion 26 feet wide by 30 feet long by 9 feet high is positioned on the ground below the window. For how many seconds will the stunt man fall before he reaches the cushion? A movie camera operating at a speed of 24 frames per second records the stunt man’s fall. How many frames of film show the stunt man falling?
SOLUTION The stunt man’s initial height is ho = 100 feet, so his height as a function of time can be modeled by function h = -16t2 + 100. Since the height of the cushion is 9 feet above the ground, the the time taken by the stunt man to reach the cushion is: h = -16t2 + 100 9 = -16t2 + 100 -91 = -16t2 91/16 = t2 or t ≈ 2.4. Thus, it takes about 2.4 seconds for the stunt man to reach the cushion. The number of frames of film that show the stunt man falling is given by the product (2.4sec)(24 frames/sec), or about 57 frames.
COMPLEX NUMBER • A complex number written in standard form is a number a+bi where a and b are real numbers. • The number a is the real part of the complex number and number bi is the imaginary part. • If b≠0, then a+bi is an imaginary number. If a=0 and b≠0, then a+bi is a pure imaginary number. • A complex plane has a horizontal axis called the real axis and a vertical axis called the imaginary axis.
THE SQUARE ROOT OF A NEGATIVE NUMBER 1. If r is a positive real number, then where i is the imaginary unit defined as 2. By property (1), it follows that
SOLVING A QUADRATIC EQUATION Solve: Hence, the solutions are and
PLOTTING COMPLEX NUMBERS y Plot 2-3i in the complex plane. To plot 2-3i , start at the origin, move 2 units to the right and then move 3 units down. 0 x 2-3i
ADDING AND SUBTRACTING COMPLEX NUMBERS Difference of complex numbers: (a+bi) - (c+di) = (a-c) + i(b-d) Example: (7-5i) - (1-5i) = (7-1) + i(-5+5) = 6 + 0i = 6 Sum of complex numbers: (a+bi) + (c+di) = (a+c) + i(b+d) Example: (4-i) + (3+2i) = (4+3) + i(-1+2) = 7 + i
MULTIPLYING THE COMPLEX NUMBERS Write the expression as a complex number in standard form. 5i(-2+i) = -10i + 5i2 = -10i + 5(-1) = -5-10i (7-4i)(-1+2i) = -7 + 14i + 4i - 8i2 = -7 + 18i - 8(-1) = 1 + 18i
DIVIDING COMPLEX NUMBERS Write the quotient in standard form. Solution: Multiply the numerator and denominator by the complex conjugate of the denominator.
ABSOLUTE VALUES OF COMPLEX NUMBER Find the absolute value of each complex number.
THE QUADRATIC FORMULA Let a, b, and c be real numbers such that a≠0. The solutions of the quadratic equation ax2 + bx +c = 0 are:
SOLVING QUADRATIC EQUATION WITH TWO REAL SOLUTIONS The solutions are:
SOLVING QUADRATIC EQUATION WITH ONE REAL SOLUTIONS Hence, the solution is 3.
SOLVING QUADRATIC EQUATION WITH TWO IMAGINARY SOLUTIONS The solutions are: 1+i and 1-i