SOLUTION: A person standing close to the edge on the top of a 170 foot building throws a baseball vertically upward. The quadratic function given below models the ball's height above the gro

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: A person standing close to the edge on the top of a 170 foot building throws a baseball vertically upward. The quadratic function given below models the ball's height above the gro      Log On


   

Question 756464: A person standing close to the edge on the top of a 170 foot building throws a baseball vertically upward. The quadratic function given below models the ball's height above the ground, s(t), in feet, t in seconds after it was thrown.
S(t)=-16t+64t+170
The ball reaches its maximum height of ___ feet after ____seconds.
Answer by KMST(5398) About Me  (Show Source):
You can put this solution on YOUR website!
The equation must be
S%28t%29=-16t%5E2%2B64t%2B170
A quadratic equation of the form y=ax%5E2%2Bbx%2Bc with a%3C%3E0 is the equation of a parabola in standard form. It has a minimum at the vertex if a%3E0, and it has a maximum at the vertex if a%3E0.
The equation can be written in vertex form as y=a%28x-h%29%5E2%2Bk based on the coordinates of the vertex: x=h and y=k

S%28t%29=-16t%5E2%2B64t%2B170 is the equation of a parabola, and the function S%28t%29 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%5E2%2Bbx%2Bc is the line x=-b%2F2a and contains the vertex. In other words, thw x-coordinate of the vertex is h=-b%2F2a.
The axis of symmetry of S%28t%29=-16t%5E2%2B64t%2B170 is the line t=-64%2F%282%2A%28-16%29%29a --> t=-64%2F%28-32%29 --> highlight%28t=2%29seconds.
That is the time-coordinate of the vertex of S%28t%29=-16t%5E2%2B64t%2B170, the time when S%28t%29 is maximum.
At that time, the height of the baseball is maximum, and that height is
S%282%29=-16%2A2%5E2%2B64%2A2%2B170 --> S%282%29=-16%2A4%2B128%2B170 --> S%282%29=-64%2B128%2B170 --> S%282%29=highlight%28234%29feet.

TRANSFORMING THE EQUATION INTO VERTEX FORM:
S=-16t%5E2%2B64t%2B170 --> S-170=-16t%5E2%2B64t --> %28S-170%29%2F%28-16%29=%28-16t%5E2%2B64t%29%2F%28-16%29 --> %28S-170%29%2F%28-16%29=t%5E2-4t
Now we can "complete the square."
%28S-170%29%2F%28-16%29=t%5E2-4t --> %28S-170%29%2F%28-16%29%2B4=t%5E2-4t%2B4 --> %28S-170%29%2F%28-16%29%2B%28-64%29%2F%28-16%29=%28t-2%29%5E2 --> %28S-170-64%29%2F%28-16%29=%28t-2%29%5E2 --> %28S-234%29%2F%28-16%29=%28t-2%29%5E2 --> S-234=-16%28t-2%29%5E2 --> S=-16%28t-2%29%5E2%2B234
S=-16%28t-2%29%5E2%2B234 is the vertex form of the equation
The equation S=-16%28t-2%29%5E2%2B234 tells you that the maximum shappens
for t=highlight%282%29seconds, when S=highlight%28234%29%7D%7Dfeet%2C%0D%0Abevause+for+any+other+value+of+%7B%7B%7Bt, S is 16%28t-2%29%5E2 less than that.