Foundations of Math 11 Section 7.3 Quadratic Equations 31 7.3 Quadratic Equations Quadratic Equation Definition of a Quadratic Equation An equation that can be written in the form ax + bx + c = 0 where a, b and c are real numbers with a 0. The solutions of the quadratic equation f (x) = ax + bx + c = 0 are called zeros of the quadratic function f (x) = ax + bx + c. The points where the graph crosses the x-axis are the real solutions, because f (x) is zero at this point. The x-intercepts are called the real roots of the quadratic function. A quadratic function can cross the x-axis either 0, 1, or times. Real Zeros of a Quadratic Function If f (x) is a quadratic function and c is a real zero of f (x), then the following statements are equivalent: 1. x = c is a zero of the function f (x). x = c is a root of the function f (x) 3. x = c is a solution of the equation f (x) = 0. (x c) is a factor of the quadratic equation f (x) 5. (c, 0) is an x-intercept of the graph of f (x) Example 1 f (x) = x x + 3 f (x) = x x +1 f (x) = x x 3 10 8 8 6 6 - -1 0 1 3 - - -1 0 1 3 0 - -1 1 3 - no zeros (no real roots) one zero (double root or two equal real roots) two zeros (two unequal real roots)
3 Chapter 7 Quadratic Functions Foundations of Math 11 We will look at many methods of solving quadratic equations. The first will be with a graphing calculator. Solving Quadratic Equations with Graphing Calculators Example 1 Solve by graphing calculator: x x = 3 Method 1 (by TI-83 Plus calculator) (Find intersection of two graphs.) Graph the equation y = x x and y = 3, and set appropriate window. Y 1 = x x Y = 3 x[ 3, 3] y[ 1, ] (press) nd, CALC, (move cursor down to 5: intersect) ENTER, (move trace cursor close to an intersection, press) ENTER, ENTER, ENTER. Intersect value is displayed at bottom of screen. Repeat for nd intersection point. ( 1, 3) and (1.5, 3) so x = 1 and x = 1.5 Method (by TI-83 Plus calculator) (Find zeros of graph.) First, arrange equation to the form ax + bx + c = 0 x x 3 = 0. Y 1 = x x 3 Graph the equation y = x x 3, and set appropriate window. x[, 3] y[ 5, ] (press) nd, CALC, (move cursor down to : zero) ENTER, (move trace cursor close to left of first zero, press) ENTER, (move cursor to right of first zero) ENTER, ENTER. Repeat for nd intersection point. ( 1, 0) and (1.5, 0) so x = 1 and x = 1.5
Foundations of Math 11 Section 7.3 Quadratic Equations 33 Not all equations are in the form ax + bc + c = 0, but the graphing calculator can still solve them. Example Solve by graphing calculator: x x 5 3 x +1 = 30 x x 5 Graph equation y = x x 5 3 x +1 30 x x 5 and set appropriate window following same steps as Example 1, Method. x = 3 Our next method of solving a quadratic equation is by factoring. We have used this method in past mathematic development. Solving Quadratic Equations by Factoring Any quadratic equation can be written in the standard form of a quadratic ax + bx + c = 0, a 0. If this factors easily, use the zero product theorem to extract the roots. The Zero Product Theorem If a b = 0, then a = 0orb = 0. If the product of two or more numbers is 0, then at least one of the numbers must be 0. e.g., (x )(x + 3) = 0 means x = orx = 3 5x(x + 8)(x 3) = 0 means x = 0orx = 8orx = 3 We will use the zero product rule to solve quadratic equations by factoring. Solving Quadratic Equations by Factoring Step 1: Write the equation in standard form so that the equation equals zero. Step : Factor the quadratic equation, if possible. Step 3: Set each factor which includes a variable equal to zero. Step : Solve the resulting equations. Step 5: Check each solution in the original equation.
3 Chapter 7 Quadratic Functions Foundations of Math 11 Example 1 Solve the equation x x 6 = 0 (x 3)(x + ) = 0 Check x = 3 Check x = x 3 = 0orx + = 0 x x 6 = 0 x x 6 = 0 x = 3or x = 3 3 6 = 0 ( ) ( ) 6 = 0 0 = 0 0 = 0 Example Solve the equation x + 6x = 0 x(x + 3) = 0 Check x = 0 Check x = 3 x = 0orx + 3 = 0 x + 6x = 0 x + 6x = 0 x = 0orx = 3 (0) + 6(0) = 0 ( 3) + 6( 3) = 0 0 = 0 0 = 0 Example 3 Solve the equation 6x 7x 5 = 0 (3x 5)(x +1) = 0 Check x = 5 3 Check x = 1 x = 5 3 or x = 1 6x 7x 5 = 0 6x 7x 5 = 0 6 5 3 7 5 3 5 = 0 1 7 1 5 = 0 0 = 0 0 = 0 Example Solve the equation x(3x +1) = 3x + x = 3x + x = 0 (3x )(x +1) = 0 3x = 0orx +1 = 0 3x = x = 1 x = or x = 1 3 Check x = 3 x(3x +1) = 3 3 3 +1 = ( +1) = 3 = Check x = 1 x(3x +1) = 1( 3( 1) +1)= 1( ) = =
Foundations of Math 11 Section 7.3 Quadratic Equations 35 Example 5 Solve the equation x x 5 3 x +1 = 30 x x 5 x x 5 3 x +1 = 30 x x 5 x (x 5)(x +1) x 5 3 x +1 = 30 (x 5)(x +1) x(x +1) 3(x 5) = 30 x + x 3x +15 = 30 x x 15 = 0 LCD = (x 5)(x +1) Multiply each term by (x 5)(x + 1). Reduce. Simplify. (x 5)(x + 3) = 0 Factor. x 5 = 0orx + 3 = 0 x = 5or x = 3 Put in standard form. Solve. Always check each root in the original rational expression to see if the solution is valid. In the original statement, the equation is undefined for x = 5 or x = 1 (denominator = 0). Hence, reject x = 5 and the solution is x = 3. Check x = 3 x x 5 3 x +1 = 30 x x 5 3 3 5 3 3+1 = 30 ( 3) ( 3) 5 3 8 + 3 = 30 16 3 8 + 1 8 = 15 8 15 8 = 15 8
36 Chapter 7 Quadratic Functions Foundations of Math 11 7.3 Exercise Set 1. Match the zeros of each function on the left with the solutions a, b or c on the right. Function Zeros f (x) = (x 1) a) g (x) = x 1 b) 1 h(x) = x +1 c) 1, 1. Match the roots of each function on the left with the solutions a, b or c on the right. Function Roots f (x) = (x 1) a) No roots g (x) = x 1 b) 1 h(x) = x +1 c) 1, 1
Foundations of Math 11 Section 7.3 Quadratic Equations 37 3. Match the x-intercept of each function on the left with the solutions on the right. Function x-intercept f (x) = x 9 a) No x-intercept g (x) = x + 9 b) 3 h(x) = x 1x + 9 c) 3 i(x) = x +1x + 9 d) 3, 3 j (x) = 9x y e) y 3 k (x) = 9x + y f) y 3 l (x) = 9x 6xy + y g) y 3, y 3 m(x) = 9x + 6xy + y
38 Chapter 7 Quadratic Functions Foundations of Math 11. Solve using the principle of zero product. Check your answers by using a graphing calculator. a) (x + )(x + 5) = 0 b) (x 3)(x ) = 0 c) (x +1)(x 6) = 0 d) x(x 1) = 0 e) x(3 x) = 0 f) x(1 x) = 0 g) 1 x 1 9 + 3x = 0 h) 1 3 x 3 x 1 6 = 0 i) (0.1x + )(0.3x 1) = 0 j) 1 x(x 3)(x + ) = 0 k) 9 x 1 16 10 15 + 3 x = 0 l) 1 3 0.x 0.x + 1 = 0
Foundations of Math 11 Section 7.3 Quadratic Equations 39 5. Solve using a graphing calculator. a) x 6x +1 = 0 b) 3x = x + 3 c) (x )(x +1) = d) (3x 1)(x ) = x + 8 e) 3 x + x 1 = 5 f) + 3 x 135 x = 0
330 Chapter 7 Quadratic Functions Foundations of Math 11 6. Solve each quadratic equation by factoring. Remember to check your solutions. a) x 3x = 0 b) z 3 = 0 c) x x 6 = 0 d) 6x 11x = 3 e) (x 1)(3x + ) = f) 5x = 8x
Foundations of Math 11 Section 7.3 Quadratic Equations 331 6. g) 1y y = 5 h) x +1x = 10 i) 3x 8x = 9 x j) 10x 3x = 1 k) 1x 17x 5 = 0 l) 8a a 1 = 3 m) 3 x 1 + x = 5 n) 1 x x = 8 3
33 Chapter 7 Quadratic Functions Foundations of Math 11 7. Find the x-intercept for the graph of the equation. (The grids are intentionally not included.) a) y = x 3x b) y = x x + 3 c) y = x + x 1 d) y = x + x + 3 e) y = x 7x 3 f) y = x + 7x 3
Foundations of Math 11 Section 7.3 Quadratic Equations 333 8. Use the graph to solve x 3x = 0. 9. Use the graph to solve x + x + 3 = 0 6 y 6 y 0 6 6 x 0 6 6 x 6 6 10. Find the mistake in the working. Then find the correct solution. a) (x )(x + 3) = 6 x = 6orx + 3 = 6 x = 8orx = 3 b) x = 5x x = 5
33 Chapter 7 Quadratic Functions Foundations of Math 11 7. Solving Quadratic Equations The factor method is the easiest method for solving quadratic equations, but cannot be used in all problems because not all expressions factor with integer coefficients. This section will develop methods of solving quadratic equations that cannot factor. Solving Quadratic Equations by the Square Root Method The square root method for solving quadratic equations is used mainly when b = 0 in ax + bx + c = 0. The x is isolated on the left side of the equation, and then the square root is taken. We will first demonstrate this method by first using an example that factors. x 16 = 0 (x )(x + ) = 0 x = 0orx + = 0 x = orx = or x 16 = 0 x = 16 x =± 16 x =± The procedure on the right leads us to the square root property. Solving a Quadratic Equations of the form ax + c =0 Step 1: Isolate the x on left side of the equation and the constant on the right side. Step : Take the square root of both sides, remembering it is ± when taking the square root of the constant. Step 3: Simplify, if possible. Step : Check the solution in the original equation. Thus, the square root property is defined as follows: The Square Root Property The equation x = n has exactly real solutions. x = n and x = n if n > 0 The solutions are often written x = ±n x = 0if n = 0 x = if n < 0 (e.g., x = has no real solution)
Foundations of Math 11 Section 7. Solving Quadratic Equations 335 Solve equations of the form ax + c = 0. Example 1 Solve x 9 = 0 x = 9 x = 9 x =± 9 =±3 Solutions are 3 and 3. Example Solve x + 7 = 0 x = 7 x = 7 x = Solution is the empty set under the real number system. Example 3 Solve x 7 = 0 x = 7 x = 7 x = ± 7 Solutions are 7 and 7. Solve equations of the form a(x + b) = c. Example Solve (x + ) = 10 (x + ) = 10 x + =± 10 x = ± 10 Solutions are 10 and + 10. Example 5 Solve 9(x 1) = 13 (x 1) = 13 9 (x 1) = 1± 13 3 Solutions are 1 13 3 and 1+ 13 3.
336 Chapter 7 Quadratic Functions Foundations of Math 11 Example 6 Solve (x + 3) +11 = 0 (x + 3) = 11 (x + 3) = 11 x = Solutions is the empty set under the real number system Solving Quadratic Equations with the Quadratic Formula If we take the standard form of the quadratic equation ax + bx + c = 0 and solve for x, we obtain a formula that can be used to obtain the solution x. This formula will work for any quadratic equation, and is used for the quadratic equations that do not factor or when you are not able to see the factors easily. Steps in Deriving the Quadratic Formula ax + bx + c = 0 ax + bx + = c x + b a x + = c a Isolate the variable on the left side of the equation. Divide by a to make the coefficents of x terms equal to 1. x + b a x + b a = c a + b Divide b a a by and square b a = b a. x + b a = b ac a Write as the square of a binomial on the left side; common denominator on the right side. x + b a =± x = b a ± b ac a Take square root. b ac a Isolate square root; simplify radical. x = b ± b ac a Common denominator. Quadratic Formula The solution of the quadratic equation ax + bc + c = 0, a 0 is given by x = b ± b ac. a
Foundations of Math 11 Section 7. Solving Quadratic Equations 337 Example 1 An example that factors is x + x 6 = 0. Method 1 (by factoring) x + x 6 = 0 Method (by quadratic formula) a =, b = 1, c = 6 (x 3)(x + ) = 0 x 3 = 0orx + = 0 x = b ± b ac a = 1± 1 ()( 6) () x = 3or x = x = 3 = 1± 9 = 1± 7 The solutions are 3,. The solutions are 3,. Example Solve 3x + x = 0 using the quadratic formula. a = 3, b =, c = x = b ± b ac a = ± (3)( ) (3) = = = = = = ± + 8 6 ± 5 6 ± 13 6 ± 13 6 ( 1± 13) 6 1± 13 3 Substitute. Simplify. Collect terms. Simplify radical. Reduce. Factor. Simplify. The solutions are 1 13, 3 1+ 13. 3
338 Chapter 7 Quadratic Functions Foundations of Math 11 Example 3 Solve x 3x =. Rewrite as x 3x + = 0 a =, b = 3, c = x = b ± b ac a = ( 3) ± ( 3) ()() = = 3± 9 3 3± 1 Substitute. Collect terms. Simplify. 1 = empty set under the real number system. Therefore, x = Example Solve bx cx = a. Rewrite as bx cx a = 0 a = b, b = c, c = a x = b ± b ac a Substitute. = ( c) ± ( c) (b)( a) b = c ± c + ab b Collect terms. Simplify. The solutions are c + c + ab b, c c + ab b
Foundations of Math 11 Section 7. Solving Quadratic Equations 339 7. Exercise Set 1. Solve by the square root method. a) x = 16 b) 3y = 39 c) 5z 1 = 0 d) 8x + 3 = 11x e) x + = 1 f) (x ) = 16 g) (x 1) = 1 h) (3x + ) = 18 i) (5x ) 3 = 3 j) (x 1) +18 = 0 k) (x 1) + x + 3 = 1 l) 1 (x 1) 1 = 1 m) (x + 6x + 9) = 5 n) x 3 = 7 o) x + 3 = 5 9 p) x + 1 5 = 0 q) x 1 = x 1 3 r) (x b) = a, a > 0
30 Chapter 7 Quadratic Functions Foundations of Math 11. Solve each of the following equations by using the quadratic formula. Give exact answers, and approximate answers to two decimal places. a) y y 3 = 0 b) x + 5x 3 = 0 c) 3x 6x + = 0 d) 3 y = y e) 5z(z + ) = f) 1x = 3x 8 g) (x )(x 3) = 8 h) x + + 1 x = 1 i) y = 3 y + j) 1 x x = k) 5.13x 7.7x.3 = 0 l) 3x = 8 x 3
Foundations of Math 11 Section 7. Solving Quadratic Equations 31 3. Solve each quadratic equation by the method of your choice. a) 9(3x 5) = 1 b) 5(x +1) 9 = 0 c) x + 5 = 0x d) 5x 5x + 6 = 0 e) (x + 3)(x ) = f) x(x + 3) = 10 g) x(x ) = (9 8x) x h) (x + )(x 1) = 6 i) x 6x = 3 j) x(x + 5) 1 = 0 k) x 3 = 5 x 3 l) 1 x + 1 x + 6 = 1 5
3 Chapter 7 Quadratic Functions Foundations of Math 11 7.5 Application of Quadratic Functions & Equations The concept of maximum and minimum values of a quadratic function is used to solve many types of problems. Solving for the vertex, and knowing if the vertex is a maximum or minimum value, is the key to solving quadratic formula problems. The next few examples will demonstrate this concept. Example 1 A rectangular pen is to be built along the side of a barn to house chickens. Find the maximum area that can be enclosed with 60 m of fencing if the barn is one side of the enclosure. Let w = width of pen Then 60 w = length of pen w Pen w 60 w A = w(60 w) = 60w w Vertex = b a, c b a = 60 60,0 ( ) ( ) = (15, 50) or w = 60 a = 60 ( ) = 15, A = 60(15) (15 ) = 50 Therefore, area is a maximum at the vertex (15, 50) where w = 15 m and A = 50 m, or l = 60 (15) = 30 m A = l w = 15 30 = 50 m Example Mary stands on the top of a building and fires a gun upwards. The bullet travels according to the equation h = 16t + 38t + 50, where h is the height of the bullet off the ground in metres at t seconds after it was fired. a) How far is Mary above the ground when she shoots the gun? b) What is the bullet s maximum height above the ground? c) How long does it take for the bullet to reach its greatest height? a) When t = 0, h = 50 m. b) Vertex b a, c b a = 38 38,50 ( 16) ( 16) = (1, 35) or t = b a = 38 ( 16) = 1, h = 16(1) + 38(1) + 50 = 35 The bullet s maximum height is 35 m. c) t = 1, bullet reaches greatest height in 1 seconds.
Foundations of Math 11 Section 7.5 Application of Quadratic Functions & Equations 33 Example 3 Bob s Rent-a-Wreck rents 300 cars at $0 per day. For each $1 increase in cost of renting, five fewer cars are rented. For what rate should the cars be rented to produce the maximum income, and what is that income? Let R(x) = income from renting cars. If there is no change in rates, then R(x) = 0 300 = $1 000 in income. Let x = increase in rate (in $) The new cost of renting a car is (0 + x). The number of cars rented is (300 5x). R(x) = (0 + x)(300 5x) = 1 000 00x + 300x 5x = 5x +100x +1 000 Vertex = b a, c b a = 100 100,1 000 ( 5) ( 5) = (10,1 500) or x-coordinate of vertex is b a = 100 ( 5) = 10, then R(x) = 5x +100x +1 000 R(10) = 5(10 ) +100(10) +1 000 = 1 500 Maximum at (10, 1 500) so maximum income $1 500 occurs when x = 10, so a car should be rented for 0 + 10 = $50.00 per day to produce the maximum income of $1 500. Example Find two numbers whose difference is 100 and the sum of whose squares is minimum. Let x = larger number and y = smaller number. Then x y = 100 x = y +100 Sum = s = x + y = ( y +100) + y = y + 00y +10 000 + y = y + 00y +10 000 y = b a = 00 () = 50 x y = 100 x = 50 Thus, x = 50 and y = 50. The two numbers are 50 and 50.
3 Chapter 7 Quadratic Functions Foundations of Math 11 Applications of Quadratic Equations Now that we can solve any type of quadratic equation, there are many word problems we can solve. When solving the equation, first try to factor the equation; if this is not possible, use the square root property or, as a last resort, the quadratic equation. Strategy for Solving Word Problems 1. Read the problem carefully, many times. Know what you need to solve for and what you are given.. Let x represent the unknown variable, and respresent everything else in terms of x. 3. If possible, make a diagram to illustrate your problem.. Write an equation relating your unknown quantities to what you are given. 5. Solve the equation. 6. Check your solutions in terms of the original problem to make sure your answer makes sense. Example 5 The sum of a number and twice its reciprocal is 9. Find the number. Let x = the number 1 = reciprocal of number x = twice the reciprocal x x + x = 9 x x+ x = 9 x + = 9x x 9x + = 0 (x )(x 1) = 0 x = 0orx 1 = 0 x = orx = 1 Note: x 0. Clear fraction. Write in standard form. Solve by factoring. Check + 1 = 9 also 1 + 1 = 9 Therefore, the number can be 1 or.
Foundations of Math 11 Section 7.5 Application of Quadratic Functions & Equations 35 Example 6 Ray and Ann ride a bicycle a distance of km each morning. They both finish at the same time but Ann starts 1 minute before Ray, and Ray travels 1 km/h faster than Ann. At what speed are they travelling? Ann s time minus Ray s time = 1 minute 1 = 60 hour. Remember, distance = speed time so Speed (km/hr) Distance (km) Ann x Time (hrs) x t = d s with d = km. Ann s time minus Ray s time = 1 60 hour. Ray x + 1 x +1 x x +1 = 1 60 60x(x +1) x x +1 = 1 60 Multiply by LCD. 0(x +1) 0x = x(x +1) Simplify. 0x + 0 0x = x + x x + x 0 = 0 (x +16)(x 15) = 0 Factor. x = 16 or 15, reject x = 16 km hr Therefore, Ann travels 15 km/h, and Ray travels 16 km/h. Example 7 Kate canoes 1 mile upstream and back in a total time of 1 hour. The current in the river is miles per hour. Find the speed of Kate s canoe in still water. distance = speed time Let x = speed of canoe in still water t u + t d = 1 d v + d v = 1; speed upstream = x, speed downsteam = x + 1 (x )(x + ) x + 1 x + = 1 x x = 0 x = ( ) ± ( ) (1)( ) (1) = ± 0 = ± 5 = 1± 5 x = 1+ 5 3., reject x = 1 5 (negative speed) Therefore, Kate canoes at 3. mph in still water.
36 Chapter 7 Quadratic Functions Foundations of Math 11 Example 8 The cold water tap can fill a tub one minute faster than the hot water tap. The two taps together can fill the tub in minutes. How long does it take each tap to fill the tub on its own? Let x = number of minutes it takes for cold water tap to fill the tub alone. x + 1 = number of minutes it takes for hot water tap to fill the tub alone. Cold Hot Total Time (min.) x x + 1 Rate (per min) 1 x 1 x +1 1 Portion of tub filled in one minute by cold water 1 x 1 x(x +1) x + 1 x +1 = 1 + + Portion of tub filled in one minute by hot water 1 x +1 = = Multiply by LCD. Portion of tub filled by both taps in one minute 1 (x +1) + x = x(x +1) Simplify. x + + x = x + x x 7x = 0 x = 7 ± ( 7) (1)( ) (1) Quadratic equation. Quadratic formula. = 7 ± 65 Cold water tap takes 7.53 minutes. Hot water tap takes 8.53 minutes. = 7.53, 0.53 Reject negative time.
Foundations of Math 11 Section 7.5 Application of Quadratic Functions & Equations 37 7.5 Exercise Set Maximum and minimum problems 1. The Acme Automobile Company has found that the revenue from sales of cars is a function of the unit price p that it charges. If the revenue R is R = 1 p + 000 p, what unit price p should be charged to maximize revenue? What is the maximum revenue?. A cattle ranch with 6000 metres of fencing wants to enclose a rectangular feedlot that borders on a river. If the cattle will not go in the river, what is the largest area that can be enclosed? 3. A cannon is fired from a cliff 100 m above the water. The height h of the cannon above the water if given by h = 0.005x + x +100, where x is the horizontal distance of the cannon from the base of the cliff. How far from the base of the cliff is the height of the cannon shell a maximum? Find the maximum height of the cannon shell.. The sum of two integers is 10 and the sum of their squares is a minimum. Find the integers.
38 Chapter 7 Quadratic Functions Foundations of Math 11 5. A rancher has 100 m of fencing to enclose two adjacent rectangular corrals. What dimensions will produce a maximum enclosed area if the common sides are of equal length? 6. Semiahmoo Fish and Game Club charges its members $00 per year. For each membership increase over 60 members, the membership cost is decreased by $.00. What number of members leads to the maximum revenue for the club? Application of Quadratic Equations 7. The length of a rectangle is m more than the width. The area is 30 m. Find the length and width. 8. The hypotenuse of a right triangle is 16 cm long. One leg is cm longer than the other. Find the length of the legs.
Foundations of Math 11 Section 7.5 Application of Quadratic Functions & Equations 39 9. Find two consecutive odd whole numbers such that the sum of their squares is 130. 10. The length and width of a rectangular sheet of plywood is ft by 8 ft. How much must be added equally to the length and width to double the area? 11. Two planes travel at right angles to each other after leaving an airport at the same time; 1 hour later, they are 390 km apart. If one plane travels 10 km/h faster than the other, what is the speed of the slower plane? 1. A boat takes 1 hour longer to go 36 km up a river than to go down the river. If the boat travels 15 km/h in still water, what is the speed of the current?
350 Chapter 7 Quadratic Functions Foundations of Math 11 13. John takes hours to weed the garden alone, and Mike requires 6 hours to do the same job. How long does it take these brothers to do the job together? 1. From each corner of a square piece of cardboard, remove a square of sides cm. Turn up the edges to form an open box. If the box is to hold 00 cm 3, what are the dimensions of the original piece of cardboard? 15. Paul drives 300 km. If he increases his speed by 10 km/h, it takes 1 hour less time. Find Paul s original speed. 16. A carpenter and her helper can do a job in hours. Working alone, the carpenter can do the job in 6 hours. How long does it take the helper to do the job alone?
Foundations of Math 11 Section 7.5 Application of Quadratic Functions & Equations 351 17. A gardener surrounds a m 8m flower bed with a border of mulch of uniform width. If there is enough mulch to cover 8 m, how wide is the border? 18. A canoer who paddles 5 km/h in still water takes 1 hour longer to paddle 1 km upstream than he does to make the return trip downstream. Find the speed of the current.
35 Chapter 7 Quadratic Functions Foundations of Math 11 7.6 Chapter Review Section 7.1 1. Graph the quadratic function. Plot at least points other than vertex. a) y = 1 (x 1) 1 x y Vertex: Max/min: Axis of symmetry: Domain: Range: y 6 6 0 6 6 x b) y = (x ) 1 x y Vertex: Max/min: Axis of symmetry: Domain: Range: y 6 6 0 6 6 x
Foundations of Math 11 Section 7.6 Chapter Review 353. Determine an equation for the parabola. a) y b) 6 6 y 0 6 6 x 0 6 6 x 6 6 Section 7. 3. Graph the following quadratic functions. Plot at least points other than the vertex. a) y = x x b) y = 1 x + x + 3
35 Chapter 7 Quadratic Functions Foundations of Math 11. Determine a quadratic function with the given information. a) Vertex ( 1, ) and point ( 3, ) b) Vertex (, ) and goes through the origin Section 7.3 5. Solve by factoring. a) 1 x(x 0.1)(0.3x + 6) = 0 b) 1 x 3 1 3 x + = 0 c) 16x x + 9 = 0 d) (x )(x + 3) = 6 e) 6x x 0 = 0 f) (x 3)(x + 6) = (x 9)(x + ) g) x x + = x x +1 h) x x +1 x x 1 = 3
Foundations of Math 11 Section 7.6 Chapter Review 355 Section 7. 6. Solve by square root property. a) (3x + ) = b) 3(x 3) + 6 = 0 c) (x 1) + = 0 d) (x 1) = 0 e) 3 (x + 3) = 6 f) (5x + 3) = 8
356 Chapter 7 Quadratic Functions Foundations of Math 11 7. Solve by the quadratic formula. a) x x = 3 b) x(7x +1) = c) x 3 = 9 x d) 3x = x Section 7.5 8. A ball is thrown upward from ground level. Its height in metres after t seconds is given by h(t) = 16t + 0t. Find the maximum height of the ball and the number of seconds the ball is in the air. 9. Find two numbers whose sum is 9 and whose product is as large as possible.
Foundations of Math 11 Section 7.6 Chapter Review 357 10. Kate travelled 10 km at a certain speed, then drove 100 km at a speed 10 km/h slower than her beginning speed. If the total time of the trip was hr, what was her beginning speed? 11. The Fraser River flows at a rate of 5 mph. A fisheries boat travels 60 miles upriver and returns in a total time of 8 hours. What is the speed of the boat in still water?
358 Chapter 7 Quadratic Functions Foundations of Math 11