Question

A basketball is thrown upwards. The height f(t), in feet, of the basketball at time t, in seconds, is given by the following function:
f(t) = −16t2 + 94t + 12

Which of the following is a reasonable domain of the graph of the function when the basketball falls from its maximum height to the ground?

2.9375 < t < 6
2 < t < 5
1 < t < 4
0 < t < 3

Answer 1

Answer:

Well I think it is A because domain is the x values.

Step-by-step explanation:

So when you plug this in your calculator (mine is a ti-84 plus ce) you would hit graph. After it graphs it press Zoom, 0 to center it then press 2nd, trace which pulls up parabola menu's. Press 0 and find the left bound, right bound and then press enter which would give you x values of 2.9375 < t< 6

At the same time I don't know if this is right. I never really excelled at parabolas just trying to help.

Answer 2

Check the picture below.

so the domain will be the values that "x" gets, now, the maximum height of the ball is when it reaches the vertex or U-turn up above, well, what is the x-coordinate anyway?

$\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ f(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+94}t\stackrel{\stackrel{c}{\downarrow }}{+12} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left( -\cfrac{94}{2(-16),}\qquad \qquad \right)\implies \left( \cfrac{487}{16},\qquad \qquad \right)\implies (2.9375,\qquad \qquad )$

so then x = 2.9375 at the vertex, now, what is "x" when it hits the ground? recall y = 0 at that instant.

$\bf \stackrel{f(t)}{0}=-16t^2+94t+12\implies 0=-2(8t^2-47t-6) \\\\\\ 0=(8t+1)(t-6)\implies t= \begin{cases} \boxed{6}\\ \begin{matrix} -\frac{1}{8} \\[-0.5em]\cline{1-1}\\[-5pt]\end{matrix} \end{cases}$

so then, the values for "x" or namely the domain from the vertex till the ball hits the ground is 2.9375 < t < 6.