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Topic: U2L5 Quadratic Word Problems

Topic: U2L5 Quadratic Word Problems. EQ: Can I solve and interpret the solutions of a quadratic function in the context of a problem?. By giving you the equation. There are basically two types of quadratic word problems:. Those that ask you to find the vertex.

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Topic: U2L5 Quadratic Word Problems

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  1. Topic: U2L5 Quadratic Word Problems EQ: Can I solve and interpret the solutions of a quadratic function in the context of a problem?

  2. By giving you the equation There are basically two types of quadratic word problems: Those that ask you to find the vertex Those that ask you to find the roots

  3. By giving you the information to find the equation There are basically two types of quadratic word problems: Those that ask you to find the vertex Those that ask you to find the roots

  4. We have seen that the vertex of a quadratic in general form is given by The y-value of the vertex is either the maximum value the function can have or it’s the minimum value the function can have. y-value’s a min when a >0 y-value’s a max when a<0

  5. Example 1 A small business’ profits over the last year have been related to the price of the only product. The relationship is R(p) = -0.4p2 +64p-2400, where R is the revenue measured in thousands of dollars and p is the price of the product measured in dollars. a. What price would maximize the revenue? The word maximize screams “FIND THE VERTEX!!” Thus, a price of $80 would maximize the revenue.

  6. Example 1 A small business’ profits over the last year have been related to the price of the only product. The relationship is R(p) = -0.4p2 +64p-2400, where R is the revenue measured in thousands of dollars and p is the price of the product measured in dollars. b. What is the maximum revenue possible? The answer to this question is the y-value of the vertex. The maximum revenue is $160 000.

  7. Example 1 A small business’ profits over the last year have been related to the price of the only product. The relationship is R(p) = -0.4p2 +64p-2400, where R is the revenue measured in thousands of dollars and p is the price of the product measured in dollars. c. How much money would they lose if they gave the product away? This question is talking about a price of 0 or p = 0 The business would lose $2 400 000.

  8. Example 2Ex A lifeguard has 75m of rope to section off the supervised area of the beach. What is the largest rectangular swimming area possible? 2w + L = 75 A = wL L= 75-2w A=w(75-2w) A = 75w-2w2 A = 75(18.75)-2(18.75)2 A =1406.25-703.125 A= 703.125

  9. We know that we can find the roots of a quadratic function by setting one side equal to zero and Factoring (sometimes) Using the quadratic root formula This ALWAYS works for a quadratic in general form and is easy to do.

  10. Example 1 A duck dives under water and its path is described by the quadratic function y = 2x2 -4x, where y represents the position of the duck in meters and x represents the time in seconds. a. How long was the duck underwater? The duck is no longer underwater when the depth is 0. We can plug in y= 0 and solve for x. The duck was underwater for 4 seconds So x = 0 or 4

  11. Example 1 A duck dives under water and its path is described by the quadratic function y = 2x2 -4x, where y represents the position of the duck in metres from the water and x represents the time in seconds. b. When was the duck at a depth of 5m? We can plug in y= -5 and solve for x. We cannot solve this because there’s a negative number under the square root. We conclude that the duck is never 5m below the water.

  12. Example 1 A duck dives under water and its path is described by the quadratic function y = 2x2 -4x, where y represents the position of the duck in metres from the water and x represents the time in seconds. We conclude that the duck is never 5m below the water. b. When was the duck at a depth of 5m? We can check this by finding the minimum value of y.

  13. Example 1 A duck dives under water and its path is described by the quadratic function y = 2x2 -4x, where y represents the position of the duck in metres and x represents the time in seconds. c. How long was the duck at least 0.5m below the water’s surface? We can plug in y= -0.5 and solve for x. This will give us the times when the duck is at 0.5 m below. The duck was 0.5m below at t = 0.14s and at t = 1.87s Therefore it was below 0.5m for 1.73s

  14. Roots Word Problems Example 2:no equation Two numbers have a difference of 18. The sum of their squares is 194. What are the numbers? Let’s define our variables: S = one of the numbers G= the other number S – G = 18 The question indicates two equations relating these two variables. S2+ G2 = 194 Again, we have a substitution situation. Solve the simpler equation for a variable and plug it in to the other equation. S=18+G (18+G)2 + G2 = 194

  15. Roots Word Problems Example 3: no equation Two numbers have a difference of 18. The sum of their squares is 194. What are the numbers? (18+G)2 + G2 = 194 We need to solve this equation for G. Then, use the quadratic formula. Let’s FOIL and make one side equal to 0. 324+36G+ G2 + G2 = 194 2G2 +36G+324-194 =0 2G2 +36G+130=0 S = 18 + (-5) = 13 or S = 18+(-13)=5

  16. Roots Word Problems Example 3: no equation Two numbers have a difference of 18. The sum of their squares is 194. What are the numbers? (18+G)2 + G2 = 194 The numbers are either -5 and 13 or -13 and 5.

  17. Roots Word Problems: Try one A rectangle is 8 feet long and 6 feet wide. If the same number of feet increases each dimension, the area of the new rectangle formed is 32 square feet more than the area of the original rectangle. How many feet increased each dimension? 6 So 80 = (6+x)(8+x) The new area is 32+6∙8 =32+48 =80 80=48+14x+x2 0=x2 +14x-32 8 The dimensions of the new rectangle are x =2 or x = -16 6+x and 8+x

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