3.1-Quadratic Functions & Inequalities

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1 3.1-Quadratic Functions & Inequalities Quadratic Functions: Quadratic functions are polnomial functions of the form also be written in the form f ( ) a( h) k. f ( ) a b c. A quadratic function ma Verte Form of a Quadratic Function: When a quadratic equation is written in the form f ( ) a( h) k the verte is ver eas to determine. In fact, there reall is no work at all because this form of a quadratic function gives us the verte directl. In this form, the verte is: (, ) ( h, k) Eample: Given the equation ( 4) 5 determine the verte. Solution: The verte is ( h, k) (4,5). Eample: Given the equation 5 3 determine the verte Solution: The verte is ( h, k),. 3 4 Rewriting a Quadratic Function in Verte Form: Since the verte form of a quadratic equation is so hand, it would be nice to know how to write an equation in this form. All we need to do is complete the square on the quadratic equation in the form f ( ) a b c to get it in the form f ( ) a( h) k. Eample: Write the equation 4 in verte form and identif the verte. Solution: To complete the square on this equation we will modif the procedure that we learned previousl. This modification will allow us to complete the square while onl working on the right side of b the function. As alwas, find the value of. Then add and subtract this value from the right side of the function. Also, group the terms of the trinomial square in parentheses. f ( ) f ( ) ( 4 4 4) 4 Now write the trinomial square in its factored form and simplif. The result is the equation in verte form. The verte is (-,-). f ( ) ( )

2 Eample: Write the equation 5 6 in verte form and identif the verte. Solution: I will solve without eplanation. 14 3) ( 5 9 9) 6 ( 5 6 The verte is (-3,-14). Eample: Write the equation 6 8 in verte form and identif the verte. Solution: As alwas, the a term must be equal to 1. Since we are working onl on the right side, we must factor the out on the right instead of dividing both sides of the equation b as we did previousl. 3) 4 ( 6 8 Now complete the square inside the parentheses. ) ( 1) ) (( 3) 4 4) 4 (( 3) 4 ( The verte is (-,-). Eample: Write the equation in verte form and identif the verte. Solution: 1) 3( 3 1) ( ) ( The verte is ( ) 1,.

3 Graphing a Quadratic Function: Graphing a quadratic function in verte form is straight forward since Is a transformation of the basic quadratic function f ( ). f ( ) a( h) k Eample: Sketch the graph of the function f ( ) ( ) 7 Solution: The verte is (, 7) Eample: Sketch the graph of the function f ( ) ( 5) 4 Solution: The verte is ( 5,4). Notice that the a value in f ( ) a( h) k is negative indicating a reflection over the -ais. Eample: Sketch the graph of the function f ( ) 4 3 Solution: Rewrite the function in the form f ( ) a( h) k f ( ) f ( ) f ( ) ( f ( ) ( ) ( 1) 1) f ( ) ( 1) f ( ) ( 1) ) Notice that the function is stretched b a factor of.

4 Characteristics of a Parabola: Parabolas have special characteristics that help us to sketch them. Each parabolaa has either a highest or lowest point called a verte. A graph of a quadratic equation also has an ais of smmetr, that is, a vertical line that passes through the verte. A parabola that opens upward has a lowest point with a minimum -value occurring at the verte. A parabola that opens downward has a highest point with a maimum -value occurring at the verte. Note that a parabola etends infinitel far in two directions. A parabola ma open upward or downward. A parabola that opens upward has a leading term thatt is positive. A parabola that opens downward has a leading term that is negative. 3 4 Opens Up 4 1 Opens Down Finding the Verte of a Parabola The verte of a parabola is the turning point of the parabola. There are two methods of finding the verte. The first method is to use the verte form of a quadratic equation. The second method is to use the verte formula. The Verte Formula: The verte is a point on the graph of a parabola a b c and ma be described as an ordered pair. As an ordered pair the verte ma be written as: Eample: Given the parabola b b,, f a a ( ) 8 5 find the verte. Solution: To find the -value of the verte use the verte formula. b a 8 () 8 4 To find the -value of the verte, we substitute the -value back into the quadratic equation and evaluate for. ( ) 8( ) Therefore, the verte is (, 3).

5 Eample: Given the parabola find the verte. Solution: To find the -value of the verte use the verte formula. b a ( 18) (1) 18 9 To find the -value of the verte, we substitute the -value back into the quadratic equation and evaluate for. (9) 18(9) 83 Therefore, the verte is ( 9,) Ais of Smmetr: The ais of smmetr of a parabola is the imaginar line that divides the parabola into two equal halves. Each half of the parabola is a reflection of the other half over the ais of smmetr. The ais of smmetr is a vertical line and will consequentl have an equation of the form a as do all vertical lines. The verte will alwas be on the ais of smmetr. Consequentl, the -value of the verte will also be the ais of smmetr. Eample: Find the verte and ais of smmetr of ( ) 7 Solution: The verte is (, -7). The ais of smmetr is. Eample: Find the verte and ais of smmetr of ( 5) 4 Solution: The verte is (-5, -4). The ais of smmetr is 5.

6 Etreme Value of a Parabola: Because the verte is the turning point of the parabola, it also represents the maimum or minimum point of the parabola. This value is also called the etreme value of the parabola. Upward Opening Parabola A parabola opens upward if its leading term is positive. If a parabola opens upward, its etreme value will be a minimum value. The minimum value is the -value of the verte. Eample: Find the etreme value of the parabola ( ) 7 and determine if it opens upward or downward. Solution: Since the leading term is positive the etreme value is a minimum. The minimum value is -7 Downward Opening Parabola A parabola opens downward if its leading term is negative. If a parabola opens downward, its etreme value will be a maimum point. The maimum value is the -value of the verte. Eample: Find the etreme value of the parabola 5( 3) 6 and determine if it opens upward or downward. Solution: Since the leading term is negative the etreme value is a maimum. The maimum value is 6

7 Intercepts of a Quadratic Equation: The intercepts of a quadratic equation are where the graph of the parabola crosses the -ais and the -ais. The procedure for finding the intercepts is the same as for linear equations (or an equations for that matter). To find -intercepts: Let 0 and solve for. This is what we are doing when we solve a quadratic equation.finding the -intercepts. The - intercepts should alwas be epressed as ordered pairs. There ma be as man as two -intercepts. The discriminant helps us determinee how man solutions or -intercepts the equation will have. To find the -intercept: Let 0 and solve for. The -intercept should alwas be epressed as an ordered pair. Theree will onl be one - intercept.

8 Eample: Find the intercepts of the equation Solution: To find the -intercepts, let 0 and solve for ( 3)( 5) 3, 5 To find the -intercept, let 0 and solve for (0) 15 The intercepts are (-3, 0) and (-5, 0) The -intercept is (0, 15) Eample: Find the intercepts of the equation 3. Solution: To find the -intercepts, let 0 and solve for. 0 0 ( 3)( 1) 1, 3 3 To find the -intercept, let 0 and solve for. 3 (0) The intercepts are (-1, 0) and The -intercept is (0, -3) 3,0 Eample: Find the intercepts of the equation. 4 8 Solution: To find the -intercepts, let 0 and solve for ± 7 To find the -intercept, let 0 and solve for (0) 8 8 The intercepts are ( 7,0) and ( 7,0) The -intercept is (0, -8)

9 Quadratic Inequalities A quadratic inequalit in one variable is an inequalit that contains a quadratic epression. The standard form of a quadratic inequalit a b c < 0 looks ver much like the standard form of a quadratic equation where an of the inequalit signs <,, >,, ma be used in place of an equal sign. Eample: The following are all quadratic inequalities. 3 4 < > Solutions of a Quadratic Inequalit: Solving a quadratic inequalit is ver similar to solving a quadratic equation. Remember that the solution to a quadratic equation a b c 0 represents the values where the parabolaa crosses the -ais. The solution to the quadratic inequalit a b c < 0 represents the values where the parabola is below the -ais. The solution to the quadratic inequalit a b c > 0 represents the values where the parabola is above the -ais. Procedure for Solving a Quadratic Inequalit: Step 1: Solve the related quadratic equation using an appropriate method. Step : Use the solutions from Step 1 to divide the number line into either two or three intervals as necessar. Step 3: Select a Test Point in each interval and determine its sign value. Step 4: The solution will consist of the intervals with the appropriate sign value.

10 Eample: Solve the quadratic inequalit 3 10 < 0. Solution: Step 1: Solve the related equation ( 5)( ) 0 5, Step : The values -5 and are called boundar values and the represent where the quadratic is equal to 0. We will use the boundar values to divide the number line into three intervals. These boundar values will be the endpoints of the intervals. The three intervals are (, 5),( 5, ),(, ). Step 3: Select a test point in each of these intervals and evaluate this point in the original quadratic Interval Test Point Evaluate in Quadratic Sign Value (, 5) -6 ( 6) 3( 6) >0 or positive ( 5,) 0 (0) (, ) 3 (3) (0) <0 or negative 3 (3) >0 or positive Step 4: Since we are interested in knowing where the quadratic is <0 the solution will be the interval where the sign value is negative or <0. Consequentl the solution to the quadratic inequalit is ( 5,).

11 Eample: Solve the quadratic inequalit 6 7. Solution: Step 1: Rewrite inequalit with a zero on the right side and then solve the related equation. 6 7 ( 5)( 1) 0 5, 1 Step : The values 5 and 1 are the boundar values and the represent where the quadratic is equal to 0. We will use the boundar values to divide the number line into three intervals. These boundar values will be the endpoints of the intervals. The three intervals are (,1],[1,5],[5, ). Note that because of the sign the boundar values ma be part of the solution and we use brackets in the interval notation. Step 3: Select a test point in each of these intervals and evaluate this point in the original quadratic 6 5. Interval Test Point Evaluate in Quadratic Sign Value (,1] 0 (0) 6 (0) 5 5 5>0 or positive [ 1,5] () [ 5, ) 6 (6) () 5 3-3<0 or negative 6 (6) 5 5 5>0 or positive Step 4: Since we are interested in knowing where the quadratic is 0 the solution will be the interval where the sign value is positive or >0. Consequentl the solution to the quadratic inequalit is (,1],[5, ).

12 Eample: Solve the quadratic inequalit 5 0. Solution: Step 1: Solve the related equation ± 5 ± 5i Step : because the boundar values are imaginar, the parabola does not cross the -ais. Consequentl, the solution is either (, ) or no solution. Because the leading term is positive the parabolaa is an upturning parabola and must therefore be above the -ais. Since the inequalit is looking for points where the quadratic is below the -ais, there is no solution. Eample: Solve the quadratic inequalit 36 < 0. Solution: Step 1: Solve the related equation ± 36 ± 6i Step : because the boundar values are imaginar, the parabola does not cross the -ais. Consequentl, the solution is either (, ) or no solution. Because the leading term is negative the parabola is a down turning parabola and must therefore be below the -ais. Since the inequalit is looking for points where the quadratic is below the -ais, the solution is (, ).

13 3.1-Applications Eample: Mike wants to enclose a rectangular area for his rabbits alongside his large barn using 30 feet of fencing. What dimensions will maimize the area fenced in if the barn is used for one side of the rectangle? What is the maimum area? Solution: This problem involves two geometr formulas; PLW and ALW. Begin with the perimeter formula b letting the length of the rectangular area. Then the width ma be determined as follows: P L w 30 w w 30 In the above formula onl one width is required since the barn occupies one side. Now substitute the value of length and width into the area formula. A LW A (30 ) A 30 To maimize this we need to find the verte: b b,, f a a ( ) Use a -, and b Evaluate the function at 7.5 A( ) 30 A(7.5) 30(7.5) (7.5) A(7.5) 11.5 The maimum area is 11.5 sq ft and will have a length of 7.5 ft and a width of 15.

14 Eample: If an archer shoots an arrow straight upward with an initial velocit of 160 ft/sec from a height of 8 ft., then its height above the ground in feet at time t in seconds is given b the function h ( t) 16t 160t 8. What is the maimum height reached b the arrow? How long does it take to reach the maimum height? How long does it take for the arrow to reach the ground? Solution: The maimum height will be the verte of the parabola. Since the leading term is negative, the parabola will be down turning and so this makes sense. The verte is a point on the graph of a parabola and ma therefore be described as an ordered pair. As an ordered pair the verte ma be written as: b b,, f a a ( ) Use the values a -16, and b 160 and substitute into the formula to find the -value of the verte ( 16) Now find the function value at 5. h( t) 16t h(5) 16(5) h(5) h(5) t 8 160(5) 8 The maimum height on the graph is the -value of the verte and is therefore 408 ft. The time to reach the maimum height is the -value of the verte. It takes 5 seconds to reach the maimum height. To determine when the arrow hits the ground we need to determine the time when the height is zero. Therefore, let h (t) 0 and solve h( t) 16t 160t 8 for t. h( t) 16t 0 16t 0 t 160t 8 0t 1 160t 8 Solving this equations gives the solutions: 51 t 5 ± t 5 ± 5.05 t It will take seconds for the arrow to hit the ground. The other solution would be which of course makes mathematical sense but not in the contet of the problem.

15 Eample: Fling too fast or too slow wastes fuel. For the Spirit of St. Louis miles per pound of fuel M was a function of airspeed A in miles per hour, modeled b the formula M A 0.17 A At what airspeed does the Spirit of St. Louis obtain its best fuel burn (greatest number of miles per pound of fuel)? What is the greatest fuel burn? Solution: Notice that the function is a quadratic that epresses fuel burn M as a function of airspeed A. Consequentl, we need to maimize the fuel burn b finding the verte of the parabola. Since rewriting this function in verte form would be an unpleasant affair, I will use the verte formula. b b,, f a a ( ) Use a , b ( ) Evaluate the function at M (97.4) (97.4) M (97.4) (97.4) The maimum fuel burn of 1.17 miles per pound of fuel is obtained at 97.4 miles per hour.

16 Eample: If a football is kicked straight up with an initial velocit of 18 ft/sec from a height of 5 ft, then its height above the earth is a function of time given b h ( t) 16t 18t 5. At what time does the ball reach this height? What is the maimum height reached b the ball? At what time does the ball hit the ground? Solution: Find the verte of the parabola in the form (t, h (t)) using the given function. The time that the ball reaches its maimum height is given b the t-value of the verte and the maimum height of the ball is given b the h(t) value of the verte. b t a Use this value to find h(t). h( t) 16t h(4) 16(4) 18t 5 18(4) 5 h(4) h(4) 61 The ball reaches a maimum height of 61 ft at 4 seconds after being kicked. To find the time that the ball hits the ground we must solve the equation 0 16t 18t 5 Using the quadratic equation we obtain 18 ± 19.4 t 3 t t 8.03 Therefore, the ball hit the ground approimatel 8.03 seconds after being kicked.

17 Eample: An air charter compan s weekl revenue in dollars is given b R( ) where is the number of hours flown during a month. 1. What is the maimum revenue?. For what number of hours is the revenue maimized? 3. On what interval is R() increasing? Decreasing? 4. Over what interval is R ( ) > 0? Solution: Find the verte of the parabola in the form (, R()) using the given function. b a Use this value to find R(). R( ) R(500) 5(500) 5000(500) R(500) 1,50,000,500,000 R(500) 1,50,000 The maimum revenue is $1,50,000 which is earned at 500 hours flight time per month. Because the parabola is down turning, the function is increasing over the interval ( 0,500) and decreasing over the interval ( 500, ). Consequentl, it is to the compan s advantage to fl as close to 500 hours per month as possible. To find where R( ) > 0 we must first find the -intercepts b solving the equation ( 1000) Because this is a down turning parabola, R( ) > 0 over the interval (0, 1,000).

18 Eample: Eagle View tour compan gives a jeep tour of the south rim of the Grand Canon to one person for $78.00 To increase their business, the compan has offered Westwind Air Tours a special rate that would lower the price b $ per person for each additional person, up to 30 people. 1. Write the price per person p as a function of the number of people.. Write the revenue as a function of the number of people on the tour. 3. What is the maimum revenue for the tour? Solution: The price per person is given b the function. To find the constant, add the discount () to the normal rate (78). Then, subtract the discount times the number of people (). P( ) 80 The revenue function is the price per person P() times the number of people. R( ) P( ) R( ) (80 ) R( ) 80 The maimum revenue is found b finding the verte of the parabola of the revenue function. b a Use this value to find R(). R( ) 80 R(0) 80(0) (0) R(0) R(0) 800 The maimum revenue is $ which is earned when 0 people are on the tour.

19 Eample: A soldier in basic training fires a rocket propelled grenade straight up from ground level with an initial velocit of 56 ft/sec. Find the maimum height reached b the RPG. How long does it take the RPG to reach its maimum height? How long does it take to reach the ground? Solution: The maimum height will be the verte of the parabola. Since the leading term is negative, the parabola will be down turning and so this makes sense. The function that models this situation is h( t) 16t 56t. I will use the verte form of the equation to determine the verte. h( t) 16t h( t) 16( t h( t) 16( t 56t 16t) h( t) 16( t 8) 16t 64 64) 104 The verte of the parabola is (8, 104). Therefore, the RPG reaches its maimum height of 104 ft. in 8 sec. To find the time it takes to reach the ground we need to solve the equation. h( t) 16t 0 16t 16t( t 16) 0 t 0, t 16 56t 56t The RPG reaches the ground in 16 seconds. Notice that it was also on the ground at zero seconds as would be epected.

20 Eample: A compan s weekl revenue in dollars is given b R( ) 000, where is the number of items produced during a week. A) For what values is R() > 0? B) For what number of items is revenue maimized? What is the maimum revenue? C) On what interval is R() increasing? Decreasing? Solution: To find where R()>0 we need to solve the inequalit 000 > 0. First solve the corresponding equation to find the -intercepts. 000 ( 1000 ) 0 0, 1,000 Because this is a down-turning parabola, the graph will be above the -ais between these values, R()>0 over the interval (0, 1,000). A) To find the number of items for which revenue is maimized we need to find the verte. R( ) 000 R( ) ( R( ) ( R( ) ( 1000) R( ) ( 500) ,000 50,000) ,000) 500, ,000 The number of items for which revenue is maimized is 500 and the revenue on that number will be $500,000. B) Since the verte is (500, 500,000) and the parabola is down turning, the function will be increasing from (,500) and decreasing from ( 500, )

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