Math 1201 Unit 4 Analysis

Math Assignment Unit 3

Timothy Weber Mathematical Department, University of the People MATH 1201: College Algebra Abhitosh Kedia May 2, 2024 Math Assignment Unit 3 Task 1 (i): Mathematical Understanding of Quadratic Concepts To understand the concepts applied in this analysis, refer to section 2.1, pages 151-172 from the textbook by Stitz and Zeager (2013). Task 1 (a) Domain and Range of h(t). The equation to evaluate is h(t)=-0.5t2+210. The domain of h(t) in a real-world context like this is the set of all possible values of t during which the jump occurs. Physically, this means from the time the jumper leaps off the bridge (t = 0 seconds) until he touches the river or ends the jump. Since t represents time, it must be non-negative. Solving …show more content…
This situation likely represents a point close to the minimum height reached by the jumper before rebounding, given the small distance above the water. Task 1 (f): Time When the Bungee Jumper Touches the River. The Bungee jumper touches the river when h(t)=0. Setting the height function to zero: -0.5t2+210=0 0.5t2=210 t2=210/0.5=420 t=420 Calculating 420: t20.49 seconds The bungee jumper touches the river approximately 20.49 seconds after jumping. These calculations provide insights into the dynamics of the bungee jump, illustrating how algebraic concepts can be applied to understand real-world scenarios, as elaborated in the reading material by Stitz and Zeager (2013). Task 1 (ii)(a): Graph of the Height Function. The graph is a downward-opening parabola, starting at h=210 meters when t=0 (the jump's starting point). As time progresses, the height decreases, reaching its lowest point near the river surface, and then potentially rising slightly again if the bungee cord is elastic and causes the jumper to rebound. Task 1(ii)(b): Height Increasing or Decreasing The height increases from the start if we consider the scenario from the perspective of the jumper being …show more content…
Practically, in the model given, where v_0=0, the height only decreases from the beginning. The height decreases immediately as t increases from 0, reflecting the effect of gravity pulling the jumper downwards. Task 1(ii)(c): Axis of Symmetry The axis of symmetry in a parabolic function ax2+bx+c is given by x=-b/2a. For the height function: t=0/(2x-0.5)=0. The axis of symmetry is at t=0, which is also the line through the vertex of the parabola. This axis divides the parabola into two symmetrical halves. In this scenario, since the initial velocity v_0=0, the symmetry axis at t=0 shows that the maximum height occurs at the beginning of the jump. Task 1(ii)(d): t and h Intercepts To find where the function intersects the t-axis, set h(t)=0: -0.5t2+210=0. Solving this as previously done, 20.49 seconds. This represents the time at which the jumper hits the river (height equals zero). This is where the function intersects the h-axis, which is when t=0: h(0)=210 This point represents the jumper's initial height at the jump's start. The t-intercept represents the time it takes for the jumper to reach the river, and the h-intercept represents the jumper's starting height. These intercepts provide critical timings and initial conditions for

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