Understanding Quadratic Functions through Real-Life Example of a Bridge
Explore the practical application of quadratic functions in analyzing the shape of a bridge arch. Learn how to find the equation of a parabolic arch from given points, determine the lengths of vertical supports, and sketch the graph. This detailed explanation provides insight into quadratic models in real-world scenarios.
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18 September 2024 Quadratic models LO: Find the equation of a quadratic functions in real life situations. www.mathssupport.org
Quadratic functions are everywhere www.mathssupport.org
Quadratic functions are everywhere www.mathssupport.org
Quadratic functions are everywhere www.mathssupport.org
Quadratic functions are everywhere www.mathssupport.org
Quadratic functions are everywhere www.mathssupport.org
Consider the following bridge. AB is the longest vertical support of the bridge its length is 40m a parabolic arch, which contains The vertical supports are 10m apart. (0, 40). A (100, 0). B (-100, 0). If axes are drawn on the diagram of the bridge with x-axis the road and y-axis on AB, find the equation of the parabolic arch. Hence, determine the lengths of the other vertical supports. www.mathssupport.org
Finding the equation of the function from its graph Using the information provided in the problem, we sketch the graph. y Since the x-intercepts are given, start with the equation in factorised form f(x) = a(x p)(x q) (0, 40) -100 100 x f(x) = a(x 100)(x + 100) We know that y = 40 when x = 0 Substitute these values into the equation to find a 40 = a (0 + 100)(0 100) 40 = -10 000a a = f(x) = 0.004(x 100)(x + 100) f(x) = 0.004x2 + 40 x-intercepts (p, 0) = (-100, 0) (q, 0) = (100, 0) -0.004 y-intercept (0, 40) www.mathssupport.org
Finding the equation of the function from its graph The equation of the parabolic arch is f(x) = 0.004x2 + 40 y Since the vertical supports are 10 m apart, we are going to calculate f(x) For x = 10 or -10 f(x) = 0.004(10)2 + 40 For x = 30 or -30 f(x) = 0.004(30)2 + 40 (0, 40) -100 100 x = 39.6 For x = 20 or -20 f(x) = 0.004(20)2 + 40 = 38.4 = 36.4 For x = 40 or -40 f(x) = 0.004(40)2 + 40 For x = 60 or -60 f(x) = 0.004(60)2 + 40 For x = 80 or -80 f(x) = 0.004(80)2 + 40 For x = 50 or -50 f(x) = 0.004(50)2 + 40 For x = 70 or -70 f(x) = 0.004(70)2 + 40 For x = 90 or -90 f(x) = 0.004(90)2 + 40 = 30 = 33.6 = 25.6 = 20.4 = 14.4 = 7.6 www.mathssupport.org
A vegetable gardener has 40 metres of fencing to enclose a rectangular garden plot where one side is an existing brick wall. If the width is x metres, garden in terms of x find the length and the area of the x If the gardener has 40 metres of fencing the three sides add up to 40 40 = x + x + y y 40 2x 40 2x = y Calculate the area of the rectangle A = x(40 - 2x) x A = 2x2 + 40x www.mathssupport.org
A vegetable gardener has 40 metres of fencing to enclose a rectangular garden plot where one side is an existing brick wall. Find the width of the garden with an area of 192 m2 A = 2x2 + 40x x 192 = 2x2+ 40x 2x2 40x + 192 = 0 y 40 2x x= 12 x= 8 If the width is 8 metres the length is 24 x If the width is 12 metres the length is 16 www.mathssupport.org
A vegetable gardener has 40 metres of fencing to enclose a rectangular garden plot where one side is an existing brick wall. Find the maximum area the garden can have The easiest way to find the maximum area is to graph the function y= 2x2 + 40x where y is the area and x is the width Using your GDC you can calculate the vertex x y 40 2x (10, 200) Is the highest point on the graph, and tells you the maximum area occurs when the width of the garden is 10 metres, the area is 200 m2 x A = 2x2+ 40x www.mathssupport.org
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