Question

select the correct answer.
a ball is thrown up from a height of 400 feet, and the height (in feet) of the ball at any time t, in seconds, is given by the function shown below.
$h(t) = -16t^2 + 400$
find the equation for the axis of symmetry of the given function.
a. $t = 6$
b. $t = 5$
c. $t = 0$
d. $t = 4$

Explanation:

Step1: Recall the axis of symmetry formula for a quadratic function

A quadratic function is in the form \( y = ax^2+bx + c \), and the axis of symmetry is given by the formula \( t=-\frac{b}{2a} \) (here we use \( t \) as the variable instead of \( x \) since the function is in terms of \( t \)).

Step2: Identify \( a \) and \( b \) from the given function

The given function is \( h(t)=- 16t^{2}+400 \). Comparing with \( h(t)=at^{2}+bt + c \), we have \( a=-16 \), \( b = 0 \) (since there is no term with \( t \) to the first power), and \( c = 400 \).

Step3: Substitute \( a \) and \( b \) into the axis of symmetry formula

Using the formula \( t=-\frac{b}{2a} \), substitute \( a=-16 \) and \( b = 0 \):
\[
t=-\frac{0}{2\times(-16)}
\]
Simplify the right - hand side:
\[
t = 0
\]

Answer:

C. \( t = 0 \)