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Quadratic Functions; Parabolas ο Graph of a Quadratic Function The graph of the function π(π) = πππ + ππ + π is a parabola that opens upward (concave up), and the vertex is minimum, if π > π. The parabola opens downward (concave down), and the vertex is a maximum, if π < π. The larger the value of |π|, the more narrow the parabola will be. βπ βπ Its vertex is at the point ( ππ , π ( ππ )). The axis of symmetry of the parabola has equation π = π βπ ππ ππ π(π) = πππ + ππ + π ππππ π > π π(π) = ππ + ππ + π ππππ π < π Vertex: ( βπ Axis of βπ , π ( )) ππ Symmetry π= βπ ππ x x Axis of Symmetry π= βπ Vertex: ( βπ ππ ππ βπ , π ( )) Y Y EX) (a) Determine if the function is quadratic. If it is, (b) determine if the graph is concave up or concave down. (c) Determine if the vertex of the graph is a maximum point or a minimum point. 1. π(π₯) = π₯ 2 + 4π₯ + 4 π = π πππ π = π. Since π > π, The graph is concave up with the minimum vertex. βπ Vertex: ( x-coordinate of the vertex is βπ ππ = βπ π(π) ππ βπ , π ( )) = (βπ, π) ππ = βπ βπ y-coordinate of the vertex is π ( ) = π(βπ) = (βπ)π + π(βπ) + π = π β π + π = π ππ 2. π(π₯) = β2π₯ 2 β 4π₯ + 6 π = βπ πππ π = βπ. Since π < π, The graph is concave down with the maximum vertex. βπ Vertex: ( x-coordinate of the vertex is βπ ππ = βπ β(βπ) π(βπ) = π βπ ππ βπ , π ( )) = (βπ, π) ππ = βπ y-coordinate of the vertex is π ( ) = π(βπ) = βπ(βπ)π β π(βπ) + π = βπ + π + π = π ππ ππ EX) Suppose the monthly revenue from the sale of Carlson 42-inch plasma televisions is given by the function π (π₯) = β0.1π₯ 2 + 600π₯ πππππππ where x is the number of televisions sold. a. Find the vertex and the axis of symmetry of the graph of this function. The function is a quadratic function with π = βπ. π, π = πππ. The x-coordinate of the vertex is βπ ππ = βπππ π(βπ.π) = ππππ, and the axis of symmetry is the line π = ππππ. The y-coordinate of the vertex is πΉ(ππππ) = βπ. π(ππππ)π + πππ(ππππ) = πππ, πππ So the vertex is (3000,900,000). b. Determine if the vertex represents a maximum or minimum point. Vertex: (3000, 900,000) Because π < π, the parabola opens downward (concave down), so the vertex is a maximum point. c. Interpreted the vertex in the context of the application. The x-coordinate of the vertex gives the number of televisions that must be sold to maximize revenue, so selling 3000 sets will result in the maximum revenue. The y-coordinate of the vertex gives the maximum revenue, $900,000. Axis of symmetry ο Vertex Form of a Quadratic Function In general, the graph of the function π = π(π β π)π + π is a parabola with its vertex at the point (π, π). The parabola opens upward if π > π, and the vertex is a minimum. The parabola opens downward if π < π, and the vertex is maximum. The axis of symmetry of the parabola has equation π = π. The a is the same as the leading coefficient in π¦ = ππ₯ 2 + ππ₯ + π, so the larger the value of |π|, the more narrow the parabola will be. EX) (a) Is the vertex a maximum or minimum? (b) Give the coordinates of the vertex 1.π(π₯) = 2(π₯ β 4)2 β 6 π = π πππ π > π, the vertex is a minimum. Vertex: (π, βπ) 2. π(π₯) = β0.5(π₯ + 2)2 + 1 π = βπ. π πππ π < π , the vertex is a maximum. Vertex: (βπ, π) Solving Quadratic Equations ο Zero Product Property For real numbers a and b, the product ππ = π if and only if either π = π or π = π or both a and b are zero EX) Solve the equation 3π₯ 2 + 7π₯ = 6 We firs subtract 6 from both sides of the equation to rewrite the equation with 0 on one side: πππ + ππ β π = π To begin factoring the trinomial πππ + ππ β π, we seek factors of πππ (that is, 3x and x) as the first terms of two binomials and factors of -6 as the last terms of the binomial. The factorization whose inner and outer products combine to 7x is (ππ β π)(π + π), so we have (ππ β π)(π + π) = π Using the zero product property gives ππ β π = π ππ π + π = π Solving these linear equations gives the two solutions to the original equation. π= π ππ π = βπ π ο Quadratic Formula The solutions of the quadratic equation ππ₯ 2 + ππ₯ + π = 0 are given by the formula π= βπ ± βππ β πππ ππ Note that a is the coefficient of π₯ 2 , b is the coefficient of x, and c is the constant term. ο The Discriminant We can also determine the type of solutions a quadratic equation has by looking at the expression ππ β πππ, which is called the discriminant of the quadratic equation ππ₯ 2 + ππ₯ + π = 0. The discriminant is the expression inside in the quadratic formula π₯ = βπ±βπ 2 β4ππ 2π , so it determines if the quantity inside the radical is positive, zero, or negative. Thus, we have ο· ο· ο· ππ β πππ > π, there are two different real solutions. ππ β πππ = π, there are one real solution. ππ β πππ < π, there are no real solution. ο Connections Between Graphs of Quadratic Functions and Solution Methods y y y Graph x a b x x Type of x-Intercepts Graph crosses x-axis twice. Graph touches but does not cross x-axis. Graph does not cross the xaxis. Type of Solutions Equation has two real solutions. Equation has one real solution. Equation has not real solutions. Suggested Solution Methods Use factoring, graphing, or quadratic formula. Use factoring, graphing, or the square root method. Use quadratic formula to verify that solutions are not real. EX) Solve the equations: a. π₯ 2 β 3π₯ + 5 = 0 Using the quadratic formula, with π = π, π = π, πππ π = π, gives π= β(βπ) ± (βπ)π β π(π)(π) π ± ββππ π ± πβππ = = π(π) π π Since ππ β πππ < π, there are no real solution and the solutions are the complex numbers. b. 3π₯ 2 + 4π₯ = 3 Writing 3π₯ 2 + 4π₯ = β3 in the form 3π₯ 2 + 4π₯ β 3 = 0 gives π = π, π = π, πππ π = βπ, so the solutions are π= βπ ± ππ β π(π)(βπ) βπ ± βππ + ππ βπ ± βππ βπ ± πβππ βπ ± βππ = = = = π(π) π π π π Since ππ β πππ > π, there are two real solutions, which is βπ+βππ π and βπββππ π . Piecewise-Defined Functions ο Piecewise-Defined Functions It is possible that a set of data cannot be modeled with a single equation. We can use a piecewise-defined function when there is not a single function that accurately represents the situation. A piecewise-defined function is so named because it is defined with different pieces for different parts of its domain rather than one equation. EX) Excluding fuel adjustment costs and taxes, Georgia Power Company charges its residential power customers for electricity during the months of June through September according to Table. Monthly Kilowatt-hours (kWh) Monthly Charge 0 to 650 More than 650, up to 1000 More than 1000 $7.50 plus $0.04657 per kWh $37.77 plus $0.07738 per kWh above 650 $64.85 plus $0.07976 per kWh above 1000 a. Write the piecewise-defined function C that gives the monthly charge for residential electricity, with input x equal to the monthly number of kilowatt-hours π. ππ + π. ππππππ ππ. ππ + π. πππππ(π β πππ) ππ. ππ + π. πππππ(π β ππππ) πͺ(π) = ππ π β€ π β€ πππ ππ πππ < π β€ ππππ ππ π > ππππ where C(x) is the charge in dollars for x kWh of electricity. b. Find C(950) and explain what it means. To evaluate C(950), we must determine which βpieceβ defines the function when π = πππ. Because 950 is between 650 and 1000, we use the βmiddle pieceβ of the function: πͺ(πππ) = ππ. ππ + π. πππππ(πππ β πππ) = ππ. πππ Companies regularly round charges up to the next cent if any part of a cent is due. This means that if 950 kWh are used in a month, the bill is $60.99. c. Find the charge for using 1560kWh in a month. To find the charge of 1560 kWh, we evaluate C(1560), using the βbottom pieceβ of the function because 1560>1000. Evaluating this gives πͺ(ππππ) = ππ. ππ + π. πππππ(ππππ β ππππ) = πππ. ππππ so the charge for the month is $109.52. Transformations of Graphs and Symmetry ο Graph Transformations ο Vertical Shifts of Graphs If k is a positive real number: The graph of π(π) = π(π) + π can be obtained by shifting the graph of π(π) upward k units. The graph of π(π) = π(π) β π can be obtained by shifting the graph of π(π) downward k units. ο Horizontal Shifts of Graphs If h is a positive real number: The graph of π(π₯) = π(π₯ β β) can be obtained by shifting the graph of π(π) to the right h units. The graph of π(π₯) = π(π₯ + β) can be obtained by shifting the graph of π(π) to the left h units. ο Stretching and Compressing Graphs The graph of π¦ = ππ(π₯) is obtained by vertically stretching the graph of π(π₯) by a factor of |π| if |π| > 1, and vertically compressing the graph of π(π₯) by a factor of |π| if 0 < |π| < 1. ο Reflections of Graphs Across the Coordinate Axes 1. The graph of π¦ = βπ(π₯) can be obtained by reflecting the graph of π¦ = π(π₯) across the x-axis. 2. The graph of π¦ = π(βπ₯) can be obtained by reflecting the graph of π¦ = π(π₯) across the y-axis. 3 EX) Suppose the graph of π¦ = π₯ 2 is shifted to the left 4 units. What is the equation that gives the new graph? 3 π¦ = (π₯ + 4)2 3 EX) Suppose the graph of π¦ = π₯ 2 is shifted down 5 units and to the right 4 units. What is the equation that gives the new graph? 3 π¦ = (π₯ β 4)2 β 5 3 EX) Suppose the graph of π¦ = π₯ 2 is stretched by a factor of 3 and then shifted up 5 units. What is the equation that gives the new graph? 3 π¦ = 3π₯ 2 + 5 3 1 EX) Suppose the graph of π¦ = π₯ 2 is compressed by the factor of and then shifted right 6 unites. What is the equation that gives the new 5 graph? 3 1 π¦ = (π₯ β 6)2 5 ο Symmetry with Respect to the y-axis The graph of π¦ = π(π₯) is symmetric with respect to the y-axis if, for every point (x, y) on the graph, the point (-x, y) is also on the graph. That is, π(βπ) = π(π) for all x in the domain of f. such a function is called an even function. ο Symmetry with Respect to the Origin The graph of π¦ = π(π₯)is symmetric with respect to the origin if, for every point (x, y) on the graph, the point (-x, -y) is also on the graph. That is, π(βπ) = βπ(π) for all x in the domain of f. such a function is called an odd function. ο Symmetry with Respect to the x-axis The graph of an equation is symmetric with respect to the x-axis if, for every point (x, y) on the graph, the point (x, -y) is also on the graph. EX) determine algebraically whether the graph of each equation is symmetric with respect to the x-axis, y-axis, or origin. Confirm your conclusion graphically. a. π¦ = 2π₯ 2 π₯ 2 +1 To test for symmetry with respect to the x-axis, we replace y with βy. Because the result, βπ = πππ ππ + π is not equivalent to the original function, the graph is not symmetric with respect to the x-axis. To test for y-axis symmetry, we replace x with βx: π= π(βπ)π πππ = (βπ)π + π ππ + π which is equivalent to the original equation. Thus, the graph is symmetric to the y-axis. To test for symmetry with respect to the origin, we replace x with βx and y with βy: βπ = π(βπ)π πππ = (βπ)π + π ππ + π which is not equivalent to the original equation. Thus, the graph is not symmetric with respect to the origin. (continuedβ¦) (continuedβ¦) b. π¦ = π₯ 3 β 3π₯ To test for symmetry with respect to the x-axis, we replace y with βy. Because the result βπ = ππ β ππ is not equivalent to the original function, the graph is not symmetric with respect to the x-axis. To test for y-axis symmetry, we replace x with βx: π = (βππ ) β π(βπ) = βππ + ππ which is not equivalent to the original equation. Thus, the graph is not symmetric to the y-axis. To test for symmetry with respect to the origin, we replace x with βx and y with βy: βπ = βππ β π(βπ) = βππ + ππ, π¨π« π = ππ β ππ which is equivalent to the original equation. Thus, the graph is symmetric with respect to the origin. c. π₯ 2 + π¦ 2 = 16 To test for symmetry with respect to the x-axis, we replace y with βy. Because the result ππ + βππ = ππ π¨π« ππ + ππ = ππ This is equivalent to the original equation, so the graph is symmetric with respect to the x-axis. To test for y-axis symmetry, we replace x with βx: (βππ ) + ππ = ππ π¨π« ππ + ππ = ππ This is equivalent to the original equation, so the graph is symmetric to the y-axis. To test for symmetry with respect to the origin, we replace x with βx and y with βy: (βππ ) + (βππ ) = ππ π¨π« ππ + ππ = ππ which is equivalent to the original equation. Thus, the graph is symmetric with respect to the origin. Combining Functions; Composite Functions ο Operations with Functions Operation Formula Example with π(π) = βπ ππ§π π(π) = ππ Sum (π + π)(π₯) = π(π₯) + π(π₯) (π + π)(π₯) = βπ₯ + π₯ 3 Difference (π β π)(π₯) = π(π₯) β π(π₯) (π β π)(π₯) = βπ₯ β π₯ 3 Product (π β π)(π₯) = π(π₯) β π(π₯) (π β π)(π₯) = βπ₯ β π₯ 3 Quotient (π) (π₯) = π(π₯) (π(π₯) β 0) π π(π₯) π (π) (π₯) = βπ₯ π₯3 (π₯ β 0) ο Composite Function The composite function, f of g, is denoted by (π β π)(π) = π(π(π)) The domain of π β π is the subset of the domain of g for which π β π is defined. The composite function π β π is defined by (π β π)(π) = π(π(π)) The domain of π β π is the subset of the domain of f for which π β π is defined. π EX) Find the following: (a) (π + π)(π₯), (b) (π β π)(π₯), (c) (π β π)(π₯), (d) ( ) (π₯), (e) (π β π)(π₯), (f) (π β π)(π₯). π π(π) = ππ β π; π(π) = π β π (a) (π + π)(π) = ππ β π + π β π = ππ β π (b) (π β π)(π) = ππ β π β (π β π) = ππ β π β π + π = ππ β π (c) (π β π)(π) = (ππ β π) β (π β π) = (ππ)(π) + (ππ)(βπ) + (βπ)(π) + (βπ)(βπ) = πππ β πππ β ππ + ππ = βπππ + πππ β ππ π ) (π) π (d) ( = ππβπ πβπ (e) (π β π)(π) = π π(π) = π(π β π) β π = ππ β ππ β π = βππ + π (f) (π β π)(π) = π π(π) = π β (ππ β π) = π β ππ + π = βππ + π Inverse Functions ο Inverse Functions Functions f and g for which π π(π) = π for all x in the domain of g, and π π(π) = π for all x in the domain of f, are called inverse functions. In this case, we denote g by πβπ, read as βf inverse.β The functions f and g are inverse functions if, whenever the pair (a, b) satisfies π¦ = π(π₯), the pair (b, a) satisfies π¦ = π(π₯). Note that when this happens, π π(π) = π and π π(π) = π for all x in the domain of g and f, respectively. EX) The function that can be used to convert a temperature of π₯° Celsius to a Fahrenheit temperature is 9 πΉ(π₯) = π₯ + 32 5 The function that can be used to convert a temperature of π₯° Fahrenheit back to a Celsius temperature is πΆ(π₯) = 5π₯ β 160 9 To see how the two conversion formulas for temperature are related, find πΆ(πΉ(π₯)) and πΉ(πΆ(π₯)) and determine if the functions are inverse functions. We compute πͺ(π(π)) and π(πͺ(π)). Evaluating C at π(π) gives π π ( π + ππ) β πππ ππ + πππ β πππ ππ π πͺ(π(π)) = = = =π π π π and evaluating F at πͺ(π) gives π(πͺ(π)) = π ππ β πππ πππ ππππ β( ) + ππ = β + ππ = π β ππ + ππ = π π π ππ ππ Because πͺ(π(π)) = π and π(πͺ(π)) = π, the two functions F and C are inverse. ο Finding the Inverse of a Function To find the inverse of the function f that is defined by the equation π¦ = π(π₯): 1. Rewrite the equation replacing π(π) with y. 2. Interchange x and y in the equation defining the function. 3. Solve the new equation for y. if this equation cannot be solve uniquely for y, the original function has no inverse function. 4. Replace y with πβπ (π). ο Graphs of Inverse Functions The graphs of a function and its inverse are symmetric with respect to the line π = π. EX) (a) Find the inverse function of π(π₯) = 2π₯β1 3 Using the steps for finding the inverse of a function, we have π= ππβπ π= ππβπ 1. Replace π(π) with y. π 2. Interchange x and y. π ππ = ππ β π 3. Solve for y. ππ + π = ππ ππ+π π =π πβπ (π) = (b) Graph π(π₯) = 2π₯β1 3 ππ+π π and its inverse function on the same axes. 4. Replace y with πβπ (π) πβπ (π) = ππ+π π π=π π(π) = ππβπ π Exponential Functions ο Exponential Function If b is a positive real number, π β 1, then the function π(π) = ππ is an exponential function. The constant b is called the base if the function and the variable x is the exponent. ο Exponential Growth Function Equation: π(π₯) π = π πππ , π > π, π > π, π > π *x-intercept: none *y-intercept: (0, a) *Domain: all real numbers *Range: all real numbers y > 0 *Horizontal asymptote: x-axis (the line y = 0) *Shape: increasing on domain and concave up x ο Exponential Decay Function Equation: π = π πβππ , π > π, π > π, π > π π(π₯) or π = π(ππ ), π < π < π, π > π *x-intercept: none *y-intercept: (0, a) *Domain: all real numbers x *Range: y>0 *Horizontal asymptote: x-axis (the line y = 0) *Shape: decreasing on domain and concave up EX) determine if the following function is a growth exponential or a decay exponential. π¦ = 3βπ₯ π π We can write the function π = πβπ in the form π = ( ) because π πβπ = πβπ π π π =( ) π π In general, a function of the form π(π) = π(πβπ ) with b > 1 can also be written in the form π(π) = π(ππ ) 0 < c < 1 and π = . π Functions with equations of this form can be used to model exponential decay. ο Growth and Decay For initial amount π > 0, the equation π = ππππ for π > 1 defines a growth function if π > 0 and a decay function if π < 0. ο Transformations of Graphs of Exponential Functions Graphs of exponential function, like other types of functions, can be shifted, reflected, or stretched. Transformations, which were introduced in Chapter 4.1, can be applied to graphs of exponential functions. Ex) a. Explain how the graph of π¦ = 3 π₯β4 compares to the graph of π¦ = 3 π₯ The graph of π = ππβπ has the same shape as π = ππ , but it is shifted 4 units to the right. b. Explain how the graph of π¦ = 3 π₯ β 4 compares to the graph of π¦ = 3 π₯ The graph of π = ππ β π has the same shape as π = ππ , but it is shifted 4 units down. c. Explain how the graph of π¦ = 2 + 3 π₯β4 compares to the graph of π¦ = 3 π₯ The graph of π = π + ππβπ has the same shape as π = ππ , but it is shifted 4 unis to the right and 2 units up. d. Explain how the graph of π¦ = 5(3π₯ ) compares to the graph of π¦ = 3 π₯ As we saw in Chapter 4.1, multiplication of functions by a constant greater than 1 stretches the graph of the function by a factor equal to that constant. Each of the y-values of π = π(ππ ) is 5 times the corresponding y-value of π = ππ . Logarithmic Functions; Properties of Logarithms ο Logarithmic Function For x > 0, b > 0, and π β 1, the logarithmic function to the base b is π = ππππ π which is defined by π = ππ That is, ππππ π is the exponent to which we must raise the base b to get the number x. This function is the inverse function of the exponential function π = ππ . EX) 1. Write the logarithmic equations in exponential form; 2π¦ = πππ5 π₯ To rewrite ππ = ππππ π in exponential form, note that the base of the logarithm, 5, becomes the base of the exponential expression, and 2y equals the exponent in the equivalent exponential form. That is, the exponential form is πππ = π 2. Write the exponential equations in logarithmic form; 92π₯ = π¦ To rewrite πππ = π in logarithmic form, note that the base of the logarithm will be 9, the same as the base of the exponential expression. Also not that 2x is the exponent in πππ = π, so y is the value of the logarithm. That is, the exponential form of πππ = π is ππ = ππππ π. ο Natural Logarithms The logarithmic function with base e is π = ππππ π, defined by π = ππ for all positive numbers x and denoted by πππ = ππππ π ο Basic Properties of Logarithms For π > 0, π β 1, 1. ππππ π = π 2. ππππ π = π 3. ππππ ππ = π 4. ππππππ = π 5. For positive real number M and N, if M = N, then ππππ π΄ = ππππ π΅ ο Additional Logarithmic Properties For π > 0, π β 1, k a real number, and M and N positive real numbers, Product Property 6. ππππ (π΄π΅) = ππππ π΄ + ππππ π΅ Quotient Property 7. ππππ ( π΅ ) = ππππ π΄ β ππππ π΅ Power Property 8.ππππ π΄π = π ππππ π΄ π΄ EX) Rewrite the following expressions as a single logarithm. a. πππ3 π₯ + 4 πππ3 π¦ = ππππ π + ππππ ππ Use Logarithmic Property 8. = ππππ πππ Use Logarithmic Property 6. 1 b. log π + 3 log π 2 = π₯π¨π ππ/π β π₯π¨π ππ = π₯π¨π ( ππ/π ππ Use Logarithmic Property 8. ) Use Logarithmic Property 7. c. ln(5π₯) β 3 ln(π§) = π₯π§(ππ) β ππ(ππ ) Use Logarithmic Property 8. ππ = π₯π§( π ) Use Logarithmic Property 7. π EX) a. Use logarithmic properties to estimate ln(5π) if ln 5 β 1.61. π₯π§(ππ) = π₯π§ π + π₯π§ π β π. ππ + π = π. ππ 15 b. Find ππππ ( ) if ππππ 15 = 2.71 and ππππ 8 = 2.079 8 ππ ππππ ( ) = ππππ ππ β ππππ π = π. ππ β π. πππ = π. πππ π 152 c. Find ππππ ( ππππ ( β8 πππ βπ ) if ππππ 15 = 2.71 and ππππ 8 = 2.079 ) = πππππ ( πππ βπ ) π = π ππππ πππ β ππππ ππ = π πππππ ππ β π ππππ π π π = π π(π. ππ) β (π. πππ) π = ππ. ππππ Exponential and Logarithmic Equations ο Solving Exponential Equations Using Logarithmic Forms To solve an exponential equation using logarithmic form: 1. Rewrite the equation with the term containing the exponent by itself on one side. 2. Divide both sides by the coefficient of the term containing the exponent. 3. Change the new equation to logarithmic form. 4. Solve for the variable. EX) Solve the equation 3000 = 150(1014π‘ ) for t by converting it to logarithmic form. 1. The term containing the exponent is by itself on one side of the equation. 2. We divide both sides of the equation by 150. ππππ = πππ(ππππ ) ππ = ππππ 3. We rewrite this equation in logarithmic form. ππ = πππππ ππ or ππ = π₯π¨π ππ 4. Solving for t gives the solution: π= π₯π¨π ππ β π. πππππ π ο Change of Base Formula If π > 0, π β 1, π > 0, π β 1, and π₯ > 0, then ππππ π = ππππ π ππππ π In particular, for base 10 and base e, ππππ π = π₯π¨π π π₯π¨π π and ππππ π = π₯π§ π π₯π§ π ο Solving Exponential Equations Using Logarithmic Properties To solve an exponential equation using logarithmic properties: 1. Rewrite the equation with a base raised to a power on one side 2. Take the logarithm, base e or 10, of both sides of the equation. 3. Use a logarithmic property to remove the variable from the exponent. 4. Solve for the variable. EX) Solve the following exponential equations: a. 4096 = 82π₯ 1. This equation has the base 8 raised to a variable power on one side 2. Taking the logarithm, base 10, of both sides of the equation ππππ = πππ gives π₯π¨π ππππ = π₯π¨π πππ 3. Using the Power Property of Logarithms removes the variable x from the exponent: π₯π¨π ππππ = ππ± π₯π¨π π 4. Solving for x gives the solution. π₯π¨π ππππ =π π π₯π¨π π π=π b. 6(43π₯β2 ) = 120 1. We first isolate the exponential expression on one side of the equation by dividing both sides by 6. π πππβπ πππ = π π πππβπ = ππ 2. Taking the natural logarithm of both sides leads to the solution. π₯π§ πππβπ = π₯π§ ππ (ππ± β π) π₯π§ π = π₯π§ ππ ππ β π = ππ = π₯π§ ππ π₯π§ π π₯π§ ππ +π π₯π§ π π₯π§ ππ +π π = π₯π§ π β π. πππ π Exponential Functions and Investing ο Future Value of an Investment with Annual Compounding If $P is invested at an interest rate r per year, compounded annually, the future value S at the end of t years is πΊ = π·(π + π)π ο Future Value of an Investment with Periodic Compounding If $P is invested for t years at the annual interest rate r, where the interest is compounded k times per π years, then the interest rate per period is π‘ , the number of compounding periods if kt, and the future value that results is given by π ππ πΊ = π· (π + ) dollars π EX) a. Write the equation that gives the future value of $1000 invested for t years at 8% compounded annually. Substituting π· = ππππ and π = π. ππ in πΊ = π·(π + π)π gives πΊ = ππππ(π + π. ππ)π or πΊ = ππππ(π. ππ)π b. Write the equation that gives the future value of $1000 invested for t years at 8% compounded daily. π ππ Substituting π· = ππππ and π = π. ππ, and π = πππ in πΊ = π· (π + ) π πΊ = ππππ (π + gives π. ππ ππππ ) πππ ο Future Value of an Investment with Continuous Compounding If $P is invested for t years at an annual interest rate r compounded continuously, then the future value S is given by πΊ = π·πππ dollars EX) a. What is the future value of $2650 invested for 8 years at 12% compounded continuously? The future value of this investment is πΊ = ππππππ.ππ(π) = ππππ. ππ dollars b. How much interest will be earned on this investment? The interest earned on this investment is the future value minus the original investment: $ππππ β $ππππ = $ππππ Annuities; Loan Repayment ο Future Value of an Ordinary Annuity If R dollars are contributed at the end of each period for n periods into an annuity that pays interest at rate i at the end of the period, the future value of the annuity is πΊ = πΉ[ (π + π)π β π ] π EX) Find the 5-year future value of an ordinary annuity with a contribution of $500 per quarter into an account that pays 8% per year compounded quarterly. π.ππ The payments and interest compounding occur quarterly, so the interest rate per period is = π. ππ and the number of π compounding periods is π(π) = ππ. Substituting the information into the formula for the future value of an annuity gives πΊ = πππ [ (π + π. ππ)ππ β π ] = ππ, πππ. ππ π. ππ Thus, the future value of this investment is $12,148.68 ο Present Value of an Ordinary Annuity If a payment of $R is to be made at the end of each period for n periods from an account that earns interest at a rate of i per period, then the account is an ordinary annuity, and the present value is π¨ = πΉ[ π β (π + π)βπ ] π EX) A couple who wants to purchase a home has $30,000 for a down payment and wants to make monthly payments of $2200. If the interest rate for a 25-year mortgage is 6% per year on the unpaid balance, what is the price of a house they can buy? The amount of money that they can pay for the house is the sum of the down payment and the present value of the payments that π% π.ππ they can afford to pay. The present value of the ππ(ππ) = πππ monthly payments of $2200 with interest at = = π. πππ ππ Per month is π¨ = ππππ [ π β (π + π. πππ)βπππ ] = πππ, πππ. ππ π. πππ They can buy a house costing $341,455 + $30,0000 = $371,455.10 ππ ο Amortization Formula If a debt of $A, with interest at a rate of i per period, is amortized by n equal periodic payments made at the end of each period, then the size of each payment is πΉ=π¨ π π β (π + π)βπ EX) A couple that wants to purchase a home with a price of $230,000 has $50,000 for a down payment. If they can get a 25-year mortgage at 9% per year on the unpaid balance, a. What will be their equal monthly payment? The amount of money that they must borrow is $230,000 - $50,000 = $180,000. The number of monthly payments is 12(25) = 300, and the interest rate is payment is πΉ = πππ, πππ π% ππ = π.ππ ππ = π. ππππ per month. The monthly π. ππππ β ππππ. πππ π β (π + π. ππππ)βπππ so the required payment would be $1510.55 b. What is the total amount they will pay before they own the house outright? The amount that they must pay before owning the house is the down payment plus the total of the 300 payments: $50,000 + 300($1510.55) = $503,168 c. How much interest will they pay? The total interest is the total amount paid minus the price for the house. The interest is $503,168 - $230,000 = $273,168