Answer :
To solve this problem, we need to determine the height of the ball 1.5 seconds after it is thrown. The function that models the height of the ball is given by [tex]\( f(x) = -10x^2 + 20x + 8 \)[/tex], where [tex]\( x \)[/tex] represents the time in seconds after the ball is thrown.
We need to find the value of [tex]\( f(x) \)[/tex] when [tex]\( x = 1.5 \)[/tex].
1. Substitute [tex]\( x = 1.5 \)[/tex] into the function:
[tex]\[ f(1.5) = -10(1.5)^2 + 20(1.5) + 8 \][/tex]
2. Calculate the square of [tex]\( 1.5 \)[/tex]:
[tex]\[ (1.5)^2 = 2.25 \][/tex]
3. Multiply [tex]\( 2.25 \)[/tex] by [tex]\(-10\)[/tex]:
[tex]\[ -10 \times 2.25 = -22.5 \][/tex]
4. Multiply [tex]\( 1.5 \)[/tex] by [tex]\( 20 \)[/tex]:
[tex]\[ 20 \times 1.5 = 30 \][/tex]
5. Add the results together with the constant term [tex]\( 8 \)[/tex]:
[tex]\[ f(1.5) = -22.5 + 30 + 8 \][/tex]
6. Perform the addition and subtraction:
[tex]\[ -22.5 + 30 = 7.5 \][/tex]
[tex]\[ 7.5 + 8 = 15.5 \][/tex]
Thus, the height of the ball 1.5 seconds after it is thrown is [tex]\( 15.5 \)[/tex] meters. Therefore, the correct answer is:
15.5 meters
We need to find the value of [tex]\( f(x) \)[/tex] when [tex]\( x = 1.5 \)[/tex].
1. Substitute [tex]\( x = 1.5 \)[/tex] into the function:
[tex]\[ f(1.5) = -10(1.5)^2 + 20(1.5) + 8 \][/tex]
2. Calculate the square of [tex]\( 1.5 \)[/tex]:
[tex]\[ (1.5)^2 = 2.25 \][/tex]
3. Multiply [tex]\( 2.25 \)[/tex] by [tex]\(-10\)[/tex]:
[tex]\[ -10 \times 2.25 = -22.5 \][/tex]
4. Multiply [tex]\( 1.5 \)[/tex] by [tex]\( 20 \)[/tex]:
[tex]\[ 20 \times 1.5 = 30 \][/tex]
5. Add the results together with the constant term [tex]\( 8 \)[/tex]:
[tex]\[ f(1.5) = -22.5 + 30 + 8 \][/tex]
6. Perform the addition and subtraction:
[tex]\[ -22.5 + 30 = 7.5 \][/tex]
[tex]\[ 7.5 + 8 = 15.5 \][/tex]
Thus, the height of the ball 1.5 seconds after it is thrown is [tex]\( 15.5 \)[/tex] meters. Therefore, the correct answer is:
15.5 meters
