Domain And Range: Object Tossed From Building
Let's break down how to find the domain and range of the function h(t) = -4.92t^2 + 17.69t + 575, which models the height of an object tossed from a tall building. This is a classic problem involving quadratic functions, and understanding the context is key. Guys, we're dealing with a real-world scenario here, so we need to think about what the numbers actually mean. The domain represents the possible values for t (time in seconds), and the range represents the possible values for h(t) (height in meters).
First, in determining the domain, we need to consider the practical limitations of our scenario. Time t cannot be negative since we're starting at the moment the object is tossed. So, t must be greater than or equal to zero. But where does it end? The object will eventually hit the ground, so there’s a maximum time. To find this maximum time, we need to determine when the height h(t) becomes zero. This involves solving the quadratic equation -4.92t^2 + 17.69t + 575 = 0. This might look a bit intimidating, but we have tools for this, like the quadratic formula. Remember, the quadratic formula is t = [-b ± sqrt(b^2 - 4ac)] / (2a). In our case, a = -4.92, b = 17.69, and c = 575. Plugging these values in, we get t = [-17.69 ± sqrt(17.69^2 - 4(-4.92)(575))] / (2 * -4.92). Calculating the discriminant (the part under the square root), we have 17.69^2 - 4(-4.92)(575) = 312.9361 + 11304 = 11616.9361. The square root of this is approximately 107.78. So, t = (-17.69 ± 107.78) / -9.84. This gives us two possible values for t: t ≈ -9.15 and t ≈ 12.75. Since time cannot be negative, we discard the negative value. Therefore, the object hits the ground at approximately t = 12.75 seconds. This means our domain is approximately 0 ≤ t ≤ 12.75. So, the domain is all the possible times from when the object is thrown until it hits the ground.
Range Calculation
Next, let's focus on the range. The range represents the possible heights the object reaches. Since the coefficient of the t^2 term is negative (-4.92), the parabola opens downward, meaning the function has a maximum value. This maximum value represents the highest point the object reaches. To find the maximum height, we need to find the vertex of the parabola. The t-coordinate of the vertex is given by t = -b / (2a). Using our values, t = -17.69 / (2 * -4.92) ≈ 1.80 seconds. This is the time at which the object reaches its maximum height. To find the maximum height itself, we plug this value of t back into the function: h(1.80) = -4.92(1.80)^2 + 17.69(1.80) + 575. Calculating this, we get h(1.80) ≈ -15.94 + 31.84 + 575 ≈ 590.90 meters. So, the maximum height the object reaches is approximately 590.90 meters. Now, what about the minimum height? The object starts at a height of 575 meters (when t = 0) and eventually hits the ground at height 0 meters. Therefore, the range of the function is approximately 0 ≤ h(t) ≤ 590.90. In simpler terms, the range is the set of all possible heights the object can be at, from the ground up to its highest point.
Putting It All Together
So, to recap, the domain represents the time interval during which the object is in the air, from when it's thrown to when it hits the ground. We calculated this to be approximately 0 ≤ t ≤ 12.75 seconds. The range represents the possible heights the object reaches, from the ground up to its maximum height. We found this to be approximately 0 ≤ h(t) ≤ 590.90 meters. Remember, guys, the domain and range are crucial for understanding the behavior of a function, especially in real-world applications like this one. They tell us the boundaries within which our model is valid and meaningful. When dealing with word problems, always consider the context to help determine these boundaries. Thinking about what the numbers represent makes the math much more intuitive and less like just memorizing formulas! Understanding domain and range is like understanding the playing field – you need to know the boundaries to play the game effectively.
Visualizing the Parabola
Let's visualize this parabola to solidify our understanding. Imagine a graph with time (t) on the horizontal axis and height (h(t)) on the vertical axis. Our parabola opens downwards, meaning it has a peak (the vertex). The vertex, as we calculated, is approximately at the point (1.80, 590.90). The parabola starts at a height of 575 meters when t = 0, and it curves upwards to the vertex before curving back down to the ground at t ≈ 12.75 seconds. The domain is the interval along the horizontal axis that the parabola covers, from t = 0 to t ≈ 12.75. The range is the interval along the vertical axis that the parabola covers, from h(t) = 0 to h(t) ≈ 590.90. Seeing the graph helps to connect the algebraic calculations with the physical reality of the object's trajectory. It also highlights the significance of the vertex as the point of maximum height.
Importance of Rounding
Finally, let’s quickly touch on the importance of rounding. The question specifically asks us to round to the nearest hundredth. This is crucial because it reflects the level of precision that is appropriate for our model. In real-world applications, we rarely need infinite precision. Rounding to the nearest hundredth (two decimal places) gives us a reasonable level of accuracy without cluttering our results with unnecessary digits. It's a good practice to always pay attention to the instructions regarding rounding, as it demonstrates attention to detail and a practical understanding of numerical approximations. Rounding, in essence, helps us communicate our results in a clear and meaningful way. It's about striking a balance between accuracy and readability. So, always double-check your rounding instructions! Guys, remember, math is not just about getting the right answer; it's about understanding the process and communicating your results effectively. That includes knowing when and how to round appropriately. So, keep practicing, keep visualizing, and keep thinking about the real-world implications of your mathematical models!