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anonymous
 one year ago
What is the maximum height that the projectile will reach? Show your work
H(t) = 16t2 + 60t + 100
anonymous
 one year ago
What is the maximum height that the projectile will reach? Show your work H(t) = 16t2 + 60t + 100

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SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Oh, don't use calculus. Find the vertex.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0many people start doing a derivative on this problem, when all it need is algebra:)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Do you know how to rewrite this in a vertex form ? (Note: since the leading coefficient of this quadratic is negative your parabola is opening down.)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I do not, But thanks because you just answered my other question xD

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0I answered your another question ?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0In any case. I will show you the steps in general

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0\(\large\color{black}{ f(x)=\displaystyle \color{red}{\rm a}x^2+ \color{green}{\rm b}x+ \color{blue}{\rm c} }\) \(\large\color{black}{ f(x)=\displaystyle \color{red}{\rm a}\left(x^2+ \frac{\color{green}{\rm b}}{\color{red}{\rm a}}x\right)+ \color{blue}{\rm c} }\) the number you need inside the parenthesis to complete the square is (b/(2a))^2 so we will do a trick. \(\large\color{black}{ f(x)=\displaystyle \color{red}{\rm a}\left(x^2+ \frac{\color{green}{\rm b}}{\color{red}{\rm a}}x+\left[\left(\frac{\color{green}{\rm b}}{\color{red}{\rm a}}\div 2\right)^2\right]\left[\left(\frac{\color{green}{\rm b}}{\color{red}{\rm a}}\div 2\right)^2\right] \right)+ \color{blue}{\rm c} }\) I add a magic zero so to speak. \(\large\color{black}{ f(x)=\displaystyle \color{red}{\rm a}\left(x^2+ \frac{\color{green}{\rm b}}{\color{red}{\rm a}}x+\left[\frac{\color{green}{\rm b^2}}{\color{red}{\rm 4a^2}}\right] \left[\frac{\color{green}{\rm b^2}}{\color{red}{\rm 4a^2}}\right] \right)+ \color{blue}{\rm c} }\) \(\large\color{black}{ f(x)=\displaystyle \color{red}{\rm a}\left(x^2+ \frac{\color{green}{\rm b}}{\color{red}{\rm a}}x+\frac{\color{green}{\rm b^2}}{\color{red}{\rm 4a^2}} \right)\color{red}{\rm a}\left[\frac{\color{green}{\rm b^2}}{\color{red}{\rm 4a^2}}\right]+ \color{blue}{\rm c} }\) \(\large\color{black}{ f(x)=\displaystyle \color{red}{\rm a}\left(x+ \frac{\color{green}{\rm b}}{\color{red}{\rm 2a}} \right)^2\color{red}{\rm a}\left[\frac{\color{green}{\rm b^2}}{\color{red}{\rm 4a^2}}\right]+ \color{blue}{\rm c} }\) \(\large\color{black}{ f(x)=\displaystyle \color{red}{\rm a}\left(x+ \frac{\color{green}{\rm b}}{\color{red}{\rm 2a}} \right)^2\frac{\color{green}{\rm b^2}}{\color{red}{\rm 4a}}+ \color{blue}{\rm c} }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0this is just me doing abstract thing..... you can look examples while I am finishing.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks man, I appreciate it

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0\(\large\color{black}{ f(x)=\displaystyle \color{red}{\rm a}\left(x+ \frac{\color{green}{\rm b}}{\color{red}{\rm 2a}} \right)^2\frac{\color{green}{\rm b^2}}{\color{red}{\rm 4a}}+\frac{\color{green}{\rm 4a\color{blue}{\rm c}}}{\color{red}{\rm 4a}} }\) \(\large\color{black}{ f(x)=\displaystyle \color{red}{\rm a}\left(x+ \frac{\color{green}{\rm b}}{\color{red}{\rm 2a}} \right)^2+\frac{\color{green}{\rm 4a\color{blue}{\rm c}}\color{green}{\rm b}^2}{\color{red}{\rm 4a}} }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0in this case the vertex is \(\large\color{black}{ \displaystyle \left(\frac{\color{green}{\rm b}}{2\color{red}{\rm a}}~,~\frac{4\color{red}{\rm a}\color{blue}{\rm c} \color{green}{\rm b}^2}{4\color{red}{\rm a}}\right) }\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0:/ Whats the highest though, Im confused.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0don't use my abstract as a formula. if you try to perform the steps and don't know how to proceed then look at this. it is kind of outline. here are some help links: https://mathway.com/examples/Algebra/ConicSections/FindingtheVertexFormofaCircle?id=817 https://www.youtube.com/watch?v=XyDMsotfJhE

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But you cant give me answer :( @SolomonZelman
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