I need help with one problem. Automatic Medal and Fan if you help meee!!!!!
A ball is thrown vertically upward. After t seconds its height h (in feet) is given by the function h(t)=68t-16t^2. What is the maximum height that the ball will reach? Do not round your Answer.
http://www.algebra.com/algebra/homework/word/mixtures/Mixture_Word_Problems.faq.question.359046.html
the answer is on that website
look it is a different question
@alibaby
put your equation into standard form (for a parabola) \[ h(t)=-16t^2 +68t + 0\] (we normally don't show the +0 ) now match that up with ax^2 + bx +c and use the formula for the vertex (the peak of the parabola) which is at x= -b/(2a) or in this case, at t= fill in the numbers for a and b)
oh
once you find t, you can use it in the formula to find the actual height
what do you get for t?
How exactly would I get t?
first look at your equation \[ h(t)=-16t^2 +68t + 0 \] which number corresponds to "a" ? any idea ?
The idea is "a" stands for the coefficient on the "highest order term" i.e. the t^2 term
no idea
in other words, what is the number in front of the \(t^2\)?
16
almost
-16
yes and to "talk about" that number we call it "a" and we call the number in front of the \( t\) term (the 2nd term) "b" what is b?
68?
yes. and the formula for the vertex of the \( \cap\) is \[ t= \frac{-b}{2a}\] we replace b with the number we found and we replace the a with the number you found for a can you do that ?
so \[\frac{ 16 }{ 136 }\]
?
@phi
b is 68 -b means put a minus sign (multiply by -1) in front of b: so -b is -68 -68 goes in as the "top" in -b/(2a) in other words we have (so far) -68/(2a) now you try the bottom: you want to put in 2 times a as the bottom
that's what I did \[\frac{ 16 }{ 136 }\]
-16/136
can you do 2 * a (remember a is -16)
you switched a and b. right idea, but wrong numbers
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