INTERNET MAT 117. Solution for the Review Problems. (1) Let us consider the circle with equation. x 2 + 2x + y 2 + 3y = 3 4. (x + 1) 2 + (y + 3 2

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1 INTERNET MAT 117 Solution for the Review Problems (1) Let us consider the circle with equation x 2 + y 2 + 2x + 3y = 0. (a) Find the standard form of the equation of the circle given above. (i) Group the x and y terms together and take the constant term to the other side. x 2 + 2x + y 2 + 3y = 3 4 (ii) Complete the square by adding the square of half of the coefficients of x and y to both sides of the equation. Add 1 and to both sides of the equation = 9 4 x 2 + 2x y 2 + 3y = (iii) Complete the square, combine the constants. See the standard form below: (x + 1) 2 + (y ) = 10 4 = 5 2 (b) Find the center and the radius of the circle given above. Center: ( 1, 3 2 ). Radius: 5/2 = (2) (a) Find the slope intercept form of the line passing through ( 2, 5) and parallel to the line 5x 3y = 8. (i) Find the slope intercept form of the line 5x 3y = 8 by expressing y in terms of x. 3y = 5x + 8 y = 5 3 x 8 3 Thus, the slope of the line 5x 3y = 8 is 5. Hence, the slope 3 of a parallel line to the line 5x 3y = 8 is also 5 3.

2 2 (ii) Now we have to find the equation of the line passing through the point ( 2, 5) that has the slope 5. The slope intercept 3 form of any line with slope 5 3 can be written in the form of y = 5 3 x + b. The line passing through the point ( 2, 5). Substitute x = 2 and y = 5 and solve the equation for b. 5 = b. b = 25 3 So the equation of the line passing through the point ( 2, 5) and has the slope 5 3 is y = 5 3 x (b) Find the slope intercept form of the line passing through ( 2, 5) and perpendicular to the line 5x 3y = 8. (i) Find the slope intercept form of the line 5x 3y = 8 by expressing y in terms of x 3y = 5x + 8 y = 5 3 x 8 3 Thus, the slope of the line 5x 3y = 8 is 5. Hence, the slope 3 of a perpendicular line to the line 5x 3y = 8 is 3 5. (ii) Now we have to find the equation of the line passing through the point ( 2, 5) that has the slope 3. The slope intercept 5 form of any line with slope 3 5 can be written in the form of y = 3 5 x + b. The line passing through the point ( 2, 5). Substitute x = 2 and y = 5 and solve the equation for b. 5 = b. b = 19 5

3 So, the equation of the line passing through the point ( 2, 5) and that the slope 3 5 is y = 3 5 x (3) Find the domain of the following functions. Give your answer in interval notation. x 1 (a) g(x) = 5 x (i) The denominator cannot be zero, so x 5. (ii) The expression under the radical cannot be negative, so 1 x. Hence 1 x and x 5. With interval notation, [1, 5) (5, ). (b) f(x) = x x 2 (i) The denominator cannot be zero, so x that is x 2 and x 2. With interval notation (, 2) ( 2, 2) (2, ). (4) A small business buys a computer for $ 4,000. After 4 years the value of the computer is expected to be $ 200. For accounting purposes, the business uses linear depreciation to assess the value of the computer at a given time. This means the if V is the value of the computer at the time t, then a linear equation is used to relate V and t. (a) Find a linear equation that relates V and t. Slope of the line passing through the points (0, 4000) and (4,200) is 950. The equation of the line passing through the give points with slope 950 is V (t) = 950t (b) Find the depreciated values of the computer 3 years from the date of purchase. Substitute t = 3 years in to the equation obtained in part a. Then V = So, it is $1,150. (5) Let f(x) = x 2 1 and g(x) = x + 2 (a) Find ( f g )(x) and give the domain of f g. ( f g )(x) = x2 1 x + 2.

4 4 (i) The denominator can not be zero and the expression can not be negative under the radical, so x > 2. (ii) The domain of ( f )(x) is ( 2, ) since x > 2. g (b) Find (f g)(x) and (g f)(x) (f g)(x) = x = x + 1 (g f)(x) = x = x (6) Let f(x) = 2x + 4. (a) Find f 1 (x). y = 2x + 4 (i) Solve the equation for x. (Express x in terms of y) y 4 = 2x y 4 2 = x or 4 y = x 2 (ii) Interchange the variables x and y. 4 x = y 2 f 1 (x) = 4 x 2 (b) Graph f and f 1 on the same coordinate axis. 6 4 y = (4 x)/2 y = 2x

5 (7) Use the graph of f(x) below to sketch the graph of g(x) it if is known that g(x) = f(x 1) Draw g(x) on the same set of axes as the given graph f(x 1) f(x) f(x 1)

6 6 (8) Consider the graph of the function f(x) below. Answer the following questions: (a) Use interval notation to state the domain of f(x). [ 2, 4] (b) Use interval notation to state the range of f(x). [ 1, 4] (c) State the interval where the function f(x) increases. [0, 1) (d) State the interval where the function f(x) decreases. [ 2, 0] [1, 4] (e) State the values of f(1) and f(4). f(1) = 2 f(4) = 1

7 (9) Let P(x) = x 3 3x 2 9x 5 (a) Determine all the possible rational zeros for P(x). By the Rational Zeros Theorem if p is a zero of P(x) then p divides q 5 and q divides 1, so p in the form of q factor of 5 factor of 1. The factors of 5 are 1,-1, 5, -5 and the factors of 1 are 1 and -1. Thus, the possible rational zeros p are 1, -1, 5, -5. q (b) Factor the polynomial P(x) completely using the procedure of long division or synthetic division. 5 is a zero, that is P(5)=0. Using synthetic division ( or long division) we obtain x 3 3x 2 9x 5 = (x 5)(x 2 + 2x + 1) = (x 5)(x + 1) 2. (c) Find all the zeros of P(x) and state the multiplicity for each zero. Give the exact answer. No decimals. x = 5 multiplicity 1. x = 1 multiplicity 2. (d) Find where P(x) approaches when x and when x. P(x) as x. P(x) as x. (e) Sketch the graph of P(x) indicating the zeros x = 1 x =

8 8 (10) Consider the polynomial P(x) = x(x + 2) 2 (x 3) 3. (a) Find all zeros of P(x) and state their multiplicities. x = 2 multiplicity 2. x = 0 multiplicity 1. x = 3 multiplicity 3. (b) Where P(x) approaches when x and when x. P(x) as x. P(x) as x. (c) Sketch the graph of the function P(x) indicating the x-intercepts (the zeros) x= 2 x=0 x= (11) Find the quotient (Q(x)) and the remainder (R(x)) using the long division.. 3x 4 5x 3 + 4x + 3 x 2 + x + 3

9 9 +3x 2 8x 1 x 2 +x +3 3x 4 5x 3 +0x 2 +4x +3 3x 4 +3x 3 +9x 2 8x 3 9x 2 +4x +3 8x 3 8x 2 24x x 2 +28x +3 x 2 x 3 29x +6 Q(x) = 3x 2 8x 1. R(x) = 29x + 6. (12) An object is projected upward from the top of a building. The height of the object in meters is described by the function h(t) = 4.9t t + 10, where t is in seconds and corresponds to the moment the object is projected. (a) Determine the height of the building. In the t = 0 moment the object is in the top of the building. So, when t = 0 then h = 10 meters. Thus, the height of the building is 10 meters. (b) Algebraically determine for what value of t the object reaches the maximum height and determine this maximum height.(hint use the vertex formula.) Check your answer with your calculator. Note that the graph of the height describes by the moving object is an open downward parabola. So, the maximum height is at the vertex of the parabola. Using the vertex formula with b = 40 and a = 4.9 the object reaches the maximum height when t = b = 2a seconds. Substituting t = seconds into h(t) = 4.9t t we obtain that the maximum height is meters. (c) Algebraically determine when the object reaches the ground. (Hint use the quadratic formula). Check your answer with your calculator. The object reaches the ground when h = 0. So, we have to substitute h = 0 and solve the equation for t using the quadratic formula with c = 10, b = 40 and a = 4.9. Since the variable t describes the time we just consider the positive solution. The solution for the equation 0 = 4.9t t + 10 is t = seconds. (13) The function p(x) = x 2 +46x 360 models the daily profit in hundreds of dollars for a company that manufactures x computers daily. (You may show your work algebraically or graphically, which includes a sketch of the graph.)

10 10 (a) How many computers should be manufactured each day to maximize profit? Note that the graph of the function which describes the daily profit is an open downward parabola. So, the maximum profit is at the vertex of the parabola. Using the vertex formula with b = 46 and a = 1 the manufacturer reaches the maximum profit when x = b = 46. So, x = 23 computers should be manufactured each day 2a to maximize profit. (b) What is the maximum daily profit? Substituting x = 46 into p(x) = x x 360. we obtain that the maximum profit is $ 16,900, since the profit function describes the profit in hundreds of dollars. (14) (a) Consider the rational function R(x) = x2 x 6 2x 2 2. (i) Find the vertical asymptotes of R(x) if there is any. Factor the numerator and the denominator if you can. x 2 x 6 2x 2 2 = (x 3)(x + 2) 2(x 1)(x + 1) If there is no common factor the vertical asymptotes of a rational function are there where the denominator is 0. So, the vertical asymptotes are x = 1 and x = 1. (ii) Find the horizontal asymptote of R(x) if there is any. If the degree of the numerator=the degree of the denominator then the horizontal asymptote is the ratio of the leading coefficients. So, the horizontal asymptote is y = 1 2. (iii) Find the x-intercepts of R(x) if there is any. The x intercepts of a rational function are there where the numerator is 0. So, the x intercepts are x = 3 and x = 2. (iv) Find the y-intercept of R(x) if there is any. Substitute x = 0 (if you can) and solve it for y. So, the y-intercept is y = 6 2 = 3. (b) Consider the rational function R(x) = x + 3 x 2 + x 2. (i) Find the vertical asymptotes of R(x) if there is any. Factor the numerator and the denominator if you can. x + 3 x 2 + x 2 = x + 3 (x 1)(x + 2)

11 If there is no common factor the vertical asymptotes of a rational function are there where the denominator is 0. So, the vertical asymptotes are x = 1 and x = 2. (ii) Find the horizontal asymptote of R(x) if there is any. If the degree of the numerator<the degree of the denominator then the horizontal asymptote is always 0. So, the horizontal asymptote is y = 0. (iii) Find the x-intercepts of R(x) if there is any. The x intercepts of a rational function are there where the numerator is 0. So, the x intercept is x = 3. (iv) Find the y-intercept of R(x) if there is any. Substitute x = 0 (if you can) and solve it for y. So, the y-intercept is y = 3 2. (c) Consider the rational function R(x) = x3 x x (i) Find the vertical asymptotes of R(x) if there is any. Factor the numerator and the denominator. x 3 x x(x 1)(x + 1) = x x If there is no common factor the vertical asymptotes of a rational function are there where the denominator is 0. Since there is no real solution for x = 0, there is no vertical asymptote. (ii) Find the horizontal asymptote of R(x) if there is any. If the degree of the numerator>the degree of the denominator then there is no horizontal asymptote. (iii) Find the x-intercepts of R(x) if there is any. The x intercepts of a rational function are there where the numerator is 0. So, the x intercepts are x = 0, x = 1 and x = 1. (iv) Find the y-intercept of R(x) if there is any. Substitute x = 0 (if you can) and solve it for y. So, the y-intercept is y = (15) The number of deer in a state forest can be modeled using the the rational function N(t) = 4500( 6t t ) where t is the time in years after the herd was first introduced.

12 12 (a) Approximately how many deer are there 5 years after they were introduced into the forest? Substitute t = 5 into the formula N(t) = t t Approximately 8940 deer. (b) In the long run, how many deer will there be in this state forest? Explain the reasoning behind your answer. Also explain what possible real-world factors might limit the number of deer in the forest. You have to find the horizontal asymptote of the rational function N(t) = t The rational function approaches to the horizontal asymptote as t increases. As t N(t) = t deer. The answer is 9000 deer.

13 13 (16) Solve for x. (a) ln x + ln(x 15) = ln 34. ln(x(x 15)) = ln 34 x(x 15) = 34 x 2 15x 34 = 0 (x + 2)(x 17) = 0 Possible solutions are x = 2 and x = 17. Checking the possible answers, we find that the only solution is x = 17. (b) log x + log(x + 3) = 1. log(x(x + 3)) = 1 The equivalent exponential form is (x(x + 3)) = 10 x 2 + 3x 10 = 0 (x + 5)(x 2) = 0 So, the solution is x = 2. (c) e 2x 3e x 10 = 0. Substitute y = e x. Then the equation becomes y 2 3y 10 = 0 (y 5)(y + 2) = 0 So, y = e x = 5 or y = e x = 2. The solution for y = e x = 5 is x = ln 5. There is no solution of y = e x = 2. So, the solution is x = ln 5.

14 14 (17) (a) Use logarithms to find the solution correct to 3 decimal places. 2 2x 3 = 3 2x+1 Take the natural logarithm (ln) of both sides of the equation. ln 2 2x 3 = ln 3 2x+1 Using the 3rd law of logarithm we obtain (2x 3) ln2 = (2x + 1) ln 3 ln 2 = and ln 3 = with 3 decimal places accuracy. So, (2x 3)0.693 = (2x + 1)1.099 After distribution we have 1.386x = 2.198x Now solve the equation for x = 0.812x x = (b) Use the Laws of Logarithms to rewrite the following expression in a form with no logarithm of a product, quotient or power. log (x 4 y 3 y log (x 4 3 z2) = log (x4y3/2 z ) = log (x4 )+log (y 3/2 ) log z = 4 log x+3/2 log y log z So, the answer is 4 logx + 3/2 log y log z. z 2) (c) Rewrite the following expression as a single logarithm. 2 log (x + 1) 3 log (y 5) + log z 2 log (x + 1) 3 log (y 5)+log z = log (x + 1) 2 log (y 5) 3 +log z = log[ (x + 1)2 z (y 5) 3 ] The answer is log[ (x + 1)2 z (y 5) 3 ].

15 (18) Find the time required for an investment of 2500 dollars to grow to 9000 dollars at an interest rate of 6.5 percent per year, compounded quarterly. 15 A(t) = P(1 + r/n) nt P=$ 2500 A(t)=$ 9000 r = n = 4 Find t (number of years) 9000 = 2500( /4) 4t dividing both sides of the equation by = ( /4) 4t 3.6 = ( ) 4t taking the natural logarithm (ln) of both sides of the equation ln 3.6 = ln(( ) 4t ) using the 3rd Law of the Logarithm t = ln 3.6 = (4t) ln( ) ln 3.6 = years 4 ln( ) (19) You are in a group of city planners that is trying to determine whether or not to expand your water supply facilities. To aid in your decision, you will use the information from the last two census figures for the city, showing a population of 80,000 at the start of 1990 and a population of 88,300 at the start of Knowing that you can currently supply enough water for 95,000 people use the exponential model Q(t) = Q 0 e kt to determine during what year you will no longer have enough water to meet the needs of your city. When t = 0 then Q 0 = 80, 000 When t = 10 then Q(10) = 88, , 330 = 80, 000e k = e k10 ln ln = k10 = = k

16 16 The exponential model is Q(t) = 80, 000 e t Find t when Q(t) = 95, , 000 = 80, 000 e t = e t ln = t ln = = t So, =2007, when the city no longer has enough water. (20) Find the half-life of a radioactive substance if 200 grams of the substance decays to 180 grams in 2 year. Q(t) = Q 0 e kt Q 0 = 200, when t = 2 Q(2) = 180. Find k. 180 = 200 e k2 0.9 = e k2 ln0.9 = k2 ln = = k Thus, the equation describing the procedure is Q(t) = 200 e t To find the half-life, substitute Q(t) = 100 and calculate t (the time during the radio active material loses half of its original mass). 100 = 200 e t 1/2 = e t ln 1/2 = t ln 1/2 = = t is the half-life

17 (21) (a) Find the exponential function f(x) = a x whose graph goes through the point. (2, 10.3). Substitute x = 2 and y = 10.3 and solve the equation for a = a 2 Note, the the equation above is not exponential. It is just a simple quadratic equation. Taking the square root of both side of the equation we get a = (10.3) = So, the solution is y = x. (b) For the logarithmic function log 3 (x 4) = y find the domain and the x-intercept. To find the domain of log 3 (x 4) = y, note that we can take the logarithm only positive numbers. So, x 4 > 0 that is x > 4. So the domain is (4, ). To find the x intercept substitute y = 0 and solve the equation log 3 (x 4) = 0. The equivalent exponential form is x 4 = 3 0 = 1, that is x=5. So, the x intercept is x = 5. (22) Find all the solutions of the following system of equations. (a) 8x 3y = 3 5x 2y = 1 We use the elimination method. 8x 3y = 3 multiply by -2 5x 2y = 1 multiply by 3 Then, we get 16x + 6y = 6 15x 6y = 3 Add the equations above. Then you eliminate the variable y. x = 3 x = 3 17

18 18 (b) Substitute x = 3 either of the original equations and solve it for y. We obtain y = 7. So, the solution is x = 3 and y = 7. x 2y = 2 y 2 x 2 = 2x + 4 We use the substitution method. Let express x from the first equation. x = 2 + 2y Substitute into the second equation y 2 (2 + 2y) 2 = 2(2 + 2y) + 4 Combining and rearranging the terms we obtain Simplifying by 3 which is 3y y + 12 = 0 y 2 + 4y + 4 = 0 (c) (y + 2) 2 = 0 So, y = 2. From the first equation x = 2 + 2y, thus x = 2. The solution is x = 2 and y = 2. x + 4y = 8 3x + 12y = 2 We use the elimination method. x + 4y = 8 multiply by -3 Then, we get 3x + 12y = 2 3x 12y = 24 3x + 12y = 2 Add the equations above. We obtain, 0 = 22

19 (d) which is impossible. It means there is no solution of this system of equations. Geometrically it means that the two lines represented by the equations are parallel, so they never intersect. 3x 2 + 4y = 17 2x 2 + 5y = 2 We use elimination method. 3x 2 + 4y = 17 multiply by We obtain, 2x 2 + 5y = 2 multiply by 3 6x 2 8y = 34 6x y = 6 Add the equations above. Then we have, 7y = 28 y = 4 The substitute y = 4 into the first equation. We have x 2 = 11 So x = 11 and x = 11. The solutions are x = 11 and y = 4 or x = 11 and y = 4