Question 1030179
The diver's height, {{{s}}} (in feet), above the water can be modeled by
{{{s(t)=-16t^2+12t+10}}} for {{{0<=t<T}}} ,
where {{{t}}} is the time since the diver jumped off the diving board, and
{{{T}}} is the total time (in seconds) that the diver is in the air.
Negative times make no sense, because the model equation/function cannot predict how high above the water the diver was at various times before jumping off the board.
Times after the diver hits the water make no sense either. The model will not work for those times.
 
(a) The diver hits the water at {{{t=T}}}, when the diver's height, {{{s}}} (in feet), above the water is
{{{s(T)=0}}} , so
{{{-16T^2+12T+10=0}}}
To solve that quadratic equation we could
use the quadratic formula, or
factor the quadratic polynomial, or
"complete the square".
FACTORING:
We notice that {{{-16*10=-160}}} can be factored many ways,
but one of them, {{{20)(-8)=-160}}} , has factors that add to the middle coefficient:
{{{20-8=12}}} .
So we re-write the polynomial (and the equation) as
{{{-16T^2-8T+20T+10=0}}} , and factor by parts:
{{{-16T^2-8T+20T+10=0}}}-->{{{-8T(2T+1)+20(2T+1)=0}}}-->{{{(-8T+10)(2T+1)=0}}} .
The solutions to that equation,
coming from {{{system(2T+1=0,"or",-8T+10=0)}}}
are {{{T=-1/2}}} and {{{T=10/8=1.25}}} .
Since the function was defined for a domain of {{{0<=t<=T}}} ,
the only solution that makes sense is {{{highlight(T=1.25)}}} .
The diver is in the air for {{{highlight(1.25)}}} seconds.
COMPLETING THE SQUARE:
{{{-16T^2+12T+10=0}}}-->{{{16T^2-12T=10}}}-->{{{(4T)^2-12T=10}}}-->{{{(4T)^2-2(4T)(3/2)+(3/2)^2=10+(3/2)^2}}}-->{{{(4T-3/2)^2=10+9/4}}}-->{{{(4T-3/2)^2=49/4}}} .
So, either {{{4T-3/2=-sqrt(49/4)=-7/2)}}} or {{{4T-3/2=sqrt(49/4)=7/2)}}} .
{{{4T-3/2=7/2)}}}-->{{{4T=7/2+3/2}}}-->{{{4T=10/2}}}-->{{{T=10/8=highlight(1.25)}}} .
The other choice, {{{4T-3/2=-7/2)}}} , yields a negative value for {{{T}}} ,
which is not in the domain where the model function should be defined,
and it would not make sense to say that the diver is in the air for a negative number of seconds after jumping off the board.
 
(b) {{{s(t)=-16t^2+12t+10}}} is a quadratic function with the negative leading coefficient {{{-16<0}}} .
APPLYING FORMULAS:
A formula that your teacher may have wanted to make you memorize says that a quadratic function
{{{f(x)=ax^2+bx+c}}} with a negative leading coefficient, {{{a<0}}} ,
has a maximum for {{{x=(-b)/"2 a"}}} .
In this case, we have {{{t}}} instead of {{{x}}} , and {{{s(t)}}} instead of {{{f(x)}}} .
Then, we have {{{a=-16<0}}} , and {{{b=12}}} .
So, the maximum height happens at {{{t=(-12)/(2*(-16))=12/32=3/8=highlight(0.375)}}} .
At that time, the height is
{{{s(3/8)=-16(3/8)^2+12(3/8)+10=-16*9/64+12*3/8+10=-9/4+9/2+10=(-9+18+40)4=49/4=highlight(12.25)}}} .
JUST USING YOUR AWESOME KNOWLEDGE OF BASIC ALGEBRA, BECAUSE YOU'D RATHER REASON THAN MEMORIZE AND APPLY FORMULAS:
{{{s(t)=-16t^2+12t+10}}}
{{{s(t)=-16(t^2-(3/4)t)+10}}}
{{{s(t)=-16(t^2-(3/4)t+(3/8)^2-(3/8)^2)+10}}} 
{{{s(t)=-16(t^2-(3/4)t+(3/8)^2-9/64)+10}}}
{{{s(t)=-16((t-3/8)^2-9/64)+10}}}
{{{s(t)=-16(t-3/8)^2-(9/64)(-16)+10}}}
{{{s(t)=-16(t-3/8)^2+9/4+10}}}
{{{s(t)=-16(t-3/8)^2+49/4}}}
Since {{{(t-3/8)^2>=0}}} , {{{s(t)=-16(t-3/8)^2+49/4<=49/4=highlight(12.25)}}} ,
and {{{s(t)=12.25}}} only when {{{t-3/8=0}}}<-->{{{t=3/8=highlight(0.375)}}} .
 
Either way you can get to the answer,
the maximum height is {{{highlight(12.25)}}} feet, and it occurs after {{{highlight(0.375)}}} seconds.