Question

A ball is thrown into the air with an upward velocity of 32 ft/s. Its height h in feet after t seconds is given by the function h = -16t^2 + 32t + 6. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. What is the ball’s maximum height?

A) 1 s; 22 ft
B) 2 s; 22 ft
C) 2 s; 6 ft
D) b1 s; 54 ft

Answer 1

find the vertex of the parabola

Answer 2

oh wait your function isn't a parabola

Answer 3

and it should be
interesting

Answer 4

\[\text{ maybe you meant } h=-16t^2+32t+6 ?\]

Answer 5

wait let me recheck

Answer 6

yeah sorry forgot the ^2

Answer 7

\[h=at^2+bt+c \\ h=a(t^2+\frac{b}{a}t)+c \\ h=a(t^2+\frac{b}{a}t+(\frac{b}{2a})^2)+c-a(\frac{b}{2a})^2 \\ h=a(t+\frac{b}{2a})^2+c-a (\frac{b}{2a})^2 \text{ is \in vertex form } \\ \text{ so the vertex is } (\frac{-b}{2a},c-a (\frac{b}{2a})^2)\]

Answer 8

so if a<0 then you have a max
and if a>0 then you have a min

we have the first case so
\[c-a(\frac{b}{2a})^2 \text{ the max } \\ \text{ and } \\ \frac{-b}{2a} \text{ will be where the max occurs }\]

Answer 9

I'm kinda confused.. I don't really get it

Answer 10

do you need an example of writing in vertex form?

Answer 11

there's too much letters, I really got confused

Answer 12

\[h=2t^2+5t+5\]
we are going to follow the same steps above:
\[h=(2t^2+5t)+5 \\ h=2(t^2+\frac{5}{2}t)+5 \\ h=2(t^2+\frac{5}{2}t+(\frac{5}{2 \cdot 2})^2)+5 -2 \cdot (\frac{5}{2 \cdot 2})^2 \\ h=2(t+\frac{5}{2 \cdot 2 })^2+5-2 (\frac{5 }{ 2 \cdot 2})^2 \\ \text{ so the vertex here is } (\frac{-5}{2 \cdot 2 },5-2 (\frac{5}{2 \cdot 2})^2) \\ \text{ you can simplify } (\frac{-5}{4},5-2 (\frac{25}{16})) \\ \text{ still simplifying } (\frac{-5}{4},5-\frac{25}{8}) \\ \text{ sill simplifying } (\frac{-5}{4}, \frac{40-25}{8}) \\ \text{ and finally } (\frac{-5}{4},\frac{15}{8})\]

Answer 13

i used completing the square to do this if it wasn't obvious
\[x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2\]

Answer 14

\[h=-16t^2+32t+6 \\ h=(-16t^2+32t)+6\]
first step look at the terms in ( )
and factor out what is in front of your first term inside

Answer 15

notice that is -16

Answer 16

\[h=-16(t^2-2t)+6 \\ h=-16(t^2-2t+?)+6+16(?) \text{ this step notice I added in a zero } \\ \text{ this zero it to keep the equation equivalent and to complete the square } \\ \text{ the zero I added in was of the form } -16(?)+16(?) \]

Answer 17

compare
\[t^2-2t+? \text{ to the left hand side of } \\ t^2+kt+(\frac{k}{2})^2=(t+\frac{k}{2})^2\]
what do we need to make k
and therefore the ? mark is ?

Answer 18

oops I should have ended that question with punctuation because that might be confusing :p

Answer 19

you know since I'm already using a ? for a constant