We are given the height of the ball as a function of time [tex]\( f(x) = -16(x^2 - 5x - 6) \)[/tex]. We need to determine when the height of the ball is 0, which means we must solve for [tex]\( x \)[/tex] in the equation [tex]\( f(x) = 0 \)[/tex].
The equation representing the height of the ball is:
[tex]\[ f(x) = -16(x^2 - 5x - 6) \][/tex]
First, we set the equation equal to zero to find the x-intercepts (the times when the ball touches the ground):
[tex]\[ -16(x^2 - 5x - 6) = 0 \][/tex]
To solve for [tex]\( x \)[/tex], we remove the factor of -16 by dividing both sides of the equation by -16:
[tex]\[ x^2 - 5x - 6 = 0 \][/tex]
Next, we solve the quadratic equation [tex]\( x^2 - 5x - 6 = 0 \)[/tex] using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex], we have:
[tex]\[ a = 1, \quad b = -5, \quad c = -6 \][/tex]
Plugging these values into the quadratic formula gives:
[tex]\[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-6)}}{2(1)} \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{25 + 24}}{2} \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{49}}{2} \][/tex]
[tex]\[ x = \frac{5 \pm 7}{2} \][/tex]
This gives us two possible solutions:
[tex]\[ x = \frac{5 + 7}{2} = \frac{12}{2} = 6 \][/tex]
[tex]\[ x = \frac{5 - 7}{2} = \frac{-2}{2} = -1 \][/tex]
Since [tex]\( x \)[/tex] represents the number of seconds since the ball was thrown, we discard the negative value [tex]\( x = -1 \)[/tex], as it doesn't make sense in the context of time. Thus, the only feasible solution is:
[tex]\[ x = 6 \][/tex]
Therefore, the ball takes [tex]\( 6 \)[/tex] seconds to reach the ground.
The correct answer is:
C. 6
