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# examples of quadratic functions word problems

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Question:For the period 1990- 2001, the number of tickets sold (in millions) for Broadway road tours can be modeled by the function y= -10.4x^2 + 132x + 332 where x is the number of years since 1990. In what year were 750 million tickets sold for Broadway road tours?

Answers:y = number of tickets sold (in millions) So, just replace "y" by 750 (millions) : 750 = -10.4x^2 + 132x + 332 -10.4x^2 + 132x + 332 - 750 = 0 -10.4x^2 + 132x - 418 = 0 To find the roots for the quadratic equations : Roots = (-b +/- sqrt(b^2 - 4ac)) / 2a Where : a = -10.4 b = 132 c = -418 Verification of the rationality : sqrt(b^2 - 4ac) sqrt((132)^2 - (4*-10.4*-418)) sqrt(17424 - 17388.8) roots are rationals coz sqrt(b^2-4ac) = 5.933 ---> not negative root 1 = (-b + 5.933) / 2a root 1 = (-132 + 5.933) / (2 *-10.4) root 1 = -126.067 / -20.8 root 1 --------------------> 6.061 root 2 = (-b - 5.933) / 2a root 2 = (-132 - 5.933) / (2 *-10.4) root 2 = -137.933 / -20.8 root 2 --------------------> 6.631 The two roots are differents and give you between 6 and 7. That means you will reach 750 million tickets twice in the year 6 or 1996.

Question:problem: marketing research by a company has shown that the profit, P = in thousands of dollars, is related to the amount spent on advertising, x = in thousands of dollars, by the quadratic function P = f(x) = 230 + 20x - 0.5x^2 table: x: -20 , -10 , 0 , 10 , 20 , 30 , 40 , 50 P=f(x): -310 , -20 , 230 , 380 , 430 , 380 , 230 , -20 questions: A. what are the coordinates of the following points? Which have meaning in this context? Give the meaning or explain why the point has no meaning. 1. y-intercept 2. x-intercept 3. vertex 4. Point with the same y-value as the y-intercept B. when graphed what parts provide a good model of this company's profits. For what values of x does f(x) realistically model the company's profits. Explain. C. What advice do you have for this company as it considers its advertising budget?

Answers:1. y-intercept => let x= 0 f(x) = 230 + 20x - 0.5x^2 f(x) = 230 + 20(0) - 0.5(0)^2 f(x) = 230 y-intercept = 230 2. x-intercept => let y = 0 f(x) = 230 + 20x - 0.5x^2 0 = 230 + 20x - 0.5x^2 use formula to solve 3. vertex f(x) = 230 + 20x - 0.5x^2 f '(x) = 20 - 0.5x Max pt. 20 - 0.5x = 0 => x = 40 => y = 230 + 20(40) - 0.5(40)^2 4. Point with the same y-value as the y-intercept x = 230 + 20x - 0.5x^2 0 = 230 + 19x - 0.5x^2 use formula to solve QED

Question:thank you!! 1.The demand for a product is p=7000-2x dollars, and supply is given by q=.01x^2+2x+1000 dollars, where x is the number of units supplied or demanded. Find the number of units when supply equals demand. 2.The area (in square meters) of a rectangular parking lot is found by equation: y=250x-x^2, where x is equals to the length of the parking lot in meters. If the parking lot must be at least 8000 square meters, find possible values for x. 3.The area (in square meters) of a rectangular parking lot is found by the equation y=100x-x^2, where x is the length of the parking lot in meters. If the parking lot must be smaller than 500 square meters, find possible values for x. 4.An object is thrown upward, and its height in feet is given by s(t)=-16t+80t+3, where t is seconds after being thrown. a)What is the initial height of the object? b)After how many seconds does the object hit the ground? 5) An open box of volume 32 inch^3 is to be made from from a square piece of tin by cutting 2 inch squares, from each corner and turning upsides. Find the dimensions of the piece of tin.

Answers:1.) p = 7000 - 2x q = 0.01x^2 + 2x + 1000 >> Find the number of units when supply equals demand. << if p = q then 7000 - 2x = 0.01x^2 + 2x + 1000 0.01x^2 + 4x - 6000 = 0 <== quadratic 2.) y = 250x - x^2 = area >> If the parking lot must be at least 8000 square meters << 8000 = 250x - x^2 x^2 - 250x + 8000 = 0 <=== quadratic

Question:a. Suppose a parabola has a vertex in Quadrant IV and a < 0 in the equation y=ax^2+bx+c. How many real solutions will the equation ax^2+bx+c=0 have? b. A ball is thrown upward from ground level. Its height h, in feet, above the ground after t seconds is h = 48t - 16t^2. Find the maximum height of the ball. Any help would be appreciated Thank You.

Answers:a. no real solutions If a < 0, the parabola opens down. n this case, the vertex is in Q IV and the parabola opens down, so it will never cross the x-axis (where y = 0). b. By factoring the equation, you can rewrite it as h = -16t(t - 3) The zeros are h = 0 and h = 3 These show the times at which the ball would be on the ground. Because of symmetry, the highest point would occur halfway between these times, that is, 1.5 second. Substitute 1.5 for t and solve to find the height at that point. h = 36 feet