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Answer :
Let's carefully evaluate the options to determine which function best represents the height of the T-shirt as a function of time. Each function represents a quadratic equation in the form of a parabola:
1. Option 1: [tex]\( f(t) = -16(t-1)^2 + 24 \)[/tex]
- This function is [tex]\( f(t) = a(t - h)^2 + k \)[/tex], where:
- [tex]\( a = -16 \)[/tex] (gravity's effect in feet per second squared)
- [tex]\( h = 1 \)[/tex] (horizontal shift to the right by 1 unit)
- [tex]\( k = 24 \)[/tex] (vertical shift upward by 24 units)
- The parabola opens downwards because [tex]\( a \)[/tex] is negative. The vertex, which gives the maximum height, is at [tex]\( t = 1 \)[/tex] second and the height at this point is 24 feet.
2. Option 2: [tex]\( f(t) = -16(t+1)^2 + 24 \)[/tex]
- Here, the vertex is shifted to the left by 1 unit.
- The maximum height of 24 feet occurs at [tex]\( t = -1 \)[/tex] second, which doesn't make sense in a real-world scenario since time can't be negative.
3. Option 3: [tex]\( f(t) = -16(t-1)^2 - 24 \)[/tex]
- This function shifts the vertex to the right by 1 unit and downward by 24 units.
- The maximum height would be below ground level at -24 feet, which is not appropriate for this context either.
4. Option 4: [tex]\( f(t) = -16(t+1)^2 - 24 \)[/tex]
- This shifts the vertex left by 1 unit and downward by 24 units.
- As with Option 2, it gives a height below ground at the vertex time of [tex]\( t = -1 \)[/tex] second.
By examining the context of the real-world scenario where a T-shirt is being thrown, Option 1 makes the most sense. It describes a parabola where the maximum height reached is 24 feet at 1 second, a plausible scenario for throwing an object.
Therefore, the function that best represents the situation is:
[tex]\( f(t) = -16(t-1)^2 + 24 \)[/tex].
1. Option 1: [tex]\( f(t) = -16(t-1)^2 + 24 \)[/tex]
- This function is [tex]\( f(t) = a(t - h)^2 + k \)[/tex], where:
- [tex]\( a = -16 \)[/tex] (gravity's effect in feet per second squared)
- [tex]\( h = 1 \)[/tex] (horizontal shift to the right by 1 unit)
- [tex]\( k = 24 \)[/tex] (vertical shift upward by 24 units)
- The parabola opens downwards because [tex]\( a \)[/tex] is negative. The vertex, which gives the maximum height, is at [tex]\( t = 1 \)[/tex] second and the height at this point is 24 feet.
2. Option 2: [tex]\( f(t) = -16(t+1)^2 + 24 \)[/tex]
- Here, the vertex is shifted to the left by 1 unit.
- The maximum height of 24 feet occurs at [tex]\( t = -1 \)[/tex] second, which doesn't make sense in a real-world scenario since time can't be negative.
3. Option 3: [tex]\( f(t) = -16(t-1)^2 - 24 \)[/tex]
- This function shifts the vertex to the right by 1 unit and downward by 24 units.
- The maximum height would be below ground level at -24 feet, which is not appropriate for this context either.
4. Option 4: [tex]\( f(t) = -16(t+1)^2 - 24 \)[/tex]
- This shifts the vertex left by 1 unit and downward by 24 units.
- As with Option 2, it gives a height below ground at the vertex time of [tex]\( t = -1 \)[/tex] second.
By examining the context of the real-world scenario where a T-shirt is being thrown, Option 1 makes the most sense. It describes a parabola where the maximum height reached is 24 feet at 1 second, a plausible scenario for throwing an object.
Therefore, the function that best represents the situation is:
[tex]\( f(t) = -16(t-1)^2 + 24 \)[/tex].
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