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Graph the relationship \( h=10 t-t^{2} \)

Ask by Martinez Tucker. in the United States
Jan 21,2025

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Answer

The graph of \( h = 10t - t^{2} \) is a downward-opening parabola with its vertex at \( (5, 25) \). It crosses the t-axis at \( t = 0 \) and \( t = 10 \).

Solution

Certainly! Let's analyze and graph the quadratic relationship \( h = 10t - t^{2} \). ### 1. Understanding the Equation The equation \( h = 10t - t^{2} \) is a quadratic equation in the form: \[ h = -t^{2} + 10t \] This represents a parabola that opens **downward** because the coefficient of \( t^{2} \) is negative. ### 2. Key Features of the Graph #### a. **Vertex** The vertex of a parabola \( h = at^{2} + bt + c \) occurs at: \[ t = -\frac{b}{2a} \] For \( h = -t^{2} + 10t \): - \( a = -1 \) - \( b = 10 \) So, \[ t = -\frac{10}{2(-1)} = 5 \] To find the corresponding \( h \)-value: \[ h = 10(5) - (5)^2 = 50 - 25 = 25 \] **Vertex:** \( (5, 25) \) #### b. **Axis of Symmetry** The axis of symmetry is the vertical line that passes through the vertex: \[ t = 5 \] #### c. **Intercepts** - **T-intercepts (where \( h = 0 \)):** Set \( h = 0 \): \[ 0 = 10t - t^{2} \] \[ t^{2} - 10t = 0 \] \[ t(t - 10) = 0 \] So, \( t = 0 \) and \( t = 10 \). **T-intercepts:** \( (0, 0) \) and \( (10, 0) \) - **H-intercept (where \( t = 0 \)):** When \( t = 0 \), \( h = 0 \). **H-intercept:** \( (0, 0) \) (same as one of the t-intercepts) #### d. **Direction of Opening** Since the coefficient of \( t^{2} \) is negative (\( a = -1 \)), the parabola opens **downward**. ### 3. Table of Values | \( t \) | \( h = 10t - t^{2} \) | |---------|-----------------------| | 0 | 0 | | 1 | 10(1) - 1 = 9 | | 2 | 20 - 4 = 16 | | 3 | 30 - 9 = 21 | | 4 | 40 - 16 = 24 | | 5 | 50 - 25 = 25 | | 6 | 60 - 36 = 24 | | 7 | 70 - 49 = 21 | | 8 | 80 - 64 = 16 | | 9 | 90 - 81 = 9 | | 10 | 100 - 100 = 0 | ### 4. Sketching the Graph Using the key features and the table of values, here's a step-by-step guide to sketch the graph: 1. **Plot the Vertex:** \( (5, 25) \). 2. **Plot the T-intercepts:** \( (0, 0) \) and \( (10, 0) \). 3. **Draw the Axis of Symmetry:** A vertical dashed line through \( t = 5 \). 4. **Plot Additional Points:** Use the table above to plot points like \( (2, 16) \), \( (3, 21) \), etc. 5. **Connect the Points:** Draw a smooth, downward-opening parabola passing through all the points. ### 5. Graph Representation While I can't provide a visual graph directly, here's a rough sketch using text: ``` h | 25| * 20| * * 15| * * 10| * * 5 | * * 0 |*---------------------*----- t 0 2 4 5 6 8 10 ``` - The highest point is at \( (5, 25) \). - The parabola crosses the \( t \)-axis at \( t = 0 \) and \( t = 10 \). ### 6. Interpretation This quadratic relationship could represent, for example, the height \( h \) of an object in meters at time \( t \) seconds if it's thrown upwards with an initial velocity, considering gravity's effect. ### 7. Summary of Key Points - **Vertex:** \( (5, 25) \) - **Axis of Symmetry:** \( t = 5 \) - **T-intercepts:** \( t = 0 \) and \( t = 10 \) - **Opening Direction:** Downward This comprehensive analysis should help you visualize and understand the graph of the equation \( h = 10t - t^{2} \).

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The Deep Dive

To graph the relationship \( h = 10t - t^2 \), start by identifying the shape and key features of the parabola. This equation is a quadratic in \( t \) and opens downwards because the coefficient of \( t^2 \) is negative. The vertex, which is the highest point, can be found using the formula \( t = -\frac{b}{2a} \), where \( a = -1 \) and \( b = 10 \). Plugging in the values gives \( t = 5 \); substituting back into the equation gives \( h = 25 \). Plotting the vertex and a few points around it will yield a nice curve, revealing how height varies with time! Another fun way to explore this equation is to consider how it might represent a projectile's motion, like throwing a ball. As time increases, the height climbs to a maximum of 25 units at 5 seconds, after which gravity causes the height to decrease. If you set \( h = 0 \), you can find when the ball hits the ground by solving \( 10t - t^2 = 0 \), leading to \( t(10 - t) = 0 \). Therefore, the ball returns to the ground at \( t = 0 \) and \( t = 10 \). This real-life analogy of projectile motion is a great way to connect math with the physics of everyday phenomena!

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