Preston Meadow's Height Enhancement Blog

Help with word problems using the functions of the second degree?

If you could help me with any of these problems would be greatly appreciated. Thank you. Now we are using the standard form of parabola: y = a (xh) ^ 2 + k An object is fired upwards from the top of a tower 200 meters to a speed of 80 feet per second. The height of the object (t) seconds after firing is given by h (t) =- 16t ^ 2 80 t 200. Find the maximum height reached by the object and the time it reaches the height. problem 2: A rectangle is inscribed in an isosceles triangle. The base of the triangle is 8 inches and height is 10 inches. Find the dimensions of the rectangle with the inscription of the MGA.

For Problem 1 There are 2 ways to do this 1 is to find the equation derived from putting it to zero and it will be the minimum or maximum function. Since you are obviously not in the calculation of another way to solve is completing the square and put it in standard form. in order to do everything that this division by 16 to get: h (t) / 16 = t ^ 2 + 5t -6.25 + 200/16 You can simplify + 6.25 to be h (t) =- 16 (T-2.5) ^ 2 + 200 + 100 So you have h (t) = 300 – x, where x = 16 (T-2.5) ^ 2 in order to maximize h (t) minimize x Because X can not be negative it to 0 and becomes 2.5 tons, with a maximum height of 300 The second problem is a bit difficult to conceptualize. To obtain the equation of the surface of the rectangle you need to relate the length of the rectangle to the width of the rectangle, triangle. The duration will vary L. The width will be (10-10/8L) multiply both of them to obtain the area of the rectangle. 10L – 10 / 8 L ^ 2 = To complete the square to put it in standard form -10 / 8 (L ^ 2 – 8L 16 -16) = A -10 / 8 (L ^ 2 – 8L 16) + 20 A -10 / 8 (L -4) ^ 2 = 20 A Max when L = 4 and A-20 = L = 4 Therefore the width = 10 to 10 / 8 L = 10-10/8 * 4 = 10-5 = 5 A = 20 = W * L for W = 20 / L = 5

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