Question 756464
The equation must be
{{{S(t)=-16t^2+64t+170}}}
A quadratic equation of the form {{{y=ax^2+bx+c}}} with {{{a<>0}}} is the equation of a parabola in standard form. It has a minimum at the vertex if {{{a>0}}}, and it has a maximum at the vertex if {{{a>0}}}.
The equation can be written in vertex form as {{{y=a(x-h)^2+k}}} based on the coordinates of the vertex: {{{x=h}}} and {{{y=k}}}
 
{{{S(t)=-16t^2+64t+170}}} is the equation of a parabola, and the function {{{S(t)}}} has a maximum at the vertex of that parabola.
You may remember formulas to find the coordinates of that vertex.
Otherwise, you can always transform the equation into the vertex form.
 
REMEMBERING FORMULAS:
The axis of symmetry of {{{y=ax^2+bx+c}}} is the line {{{x=-b/2a}}} and contains the vertex. In other words, thw x-coordinate of the vertex is {{{h=-b/2a}}}.
The axis of symmetry of {{{S(t)=-16t^2+64t+170}}} is the line {{{t=-64/(2*(-16))a}}} --> {{{t=-64/(-32)}}} --> {{{highlight(t=2)}}}seconds.
That is the time-coordinate of the vertex of {{{S(t)=-16t^2+64t+170}}}, the time when {{{S(t)}}} is maximum.
At that time, the height of the baseball is maximum, and that height is
{{{S(2)=-16*2^2+64*2+170}}} --> {{{S(2)=-16*4+128+170}}} --> {{{S(2)=-64+128+170}}} --> {{{S(2)=highlight(234)}}}feet.
 
TRANSFORMING THE EQUATION INTO VERTEX FORM:
{{{S=-16t^2+64t+170}}} --> {{{S-170=-16t^2+64t}}} --> {{{(S-170)/(-16)=(-16t^2+64t)/(-16)}}} --> {{{(S-170)/(-16)=t^2-4t}}}
Now we can "complete the square."
{{{(S-170)/(-16)=t^2-4t}}} --> {{{(S-170)/(-16)+4=t^2-4t+4}}} --> {{{(S-170)/(-16)+(-64)/(-16)=(t-2)^2}}} --> {{{(S-170-64)/(-16)=(t-2)^2}}} --> {{{(S-234)/(-16)=(t-2)^2}}} --> {{{S-234=-16(t-2)^2}}} --> {{{S=-16(t-2)^2+234}}}
{{{S=-16(t-2)^2+234}}} is the vertex form of the equation
The equation {{{S=-16(t-2)^2+234}}} tells you that the maximum {{{s}}}happens
for {{{t=highlight(2)}}}seconds, when {{{S=highlight(234)}}feet,
bevause for any other value of {{{t}}}, {{{S}}} is {{{16(t-2)^2}}} less than that.