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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

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  1. anonymous
    • one year ago
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    @SolomonZelman

  2. SolomonZelman
    • one year ago
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    Oh, don't use calculus. Find the vertex.

  3. SolomonZelman
    • one year ago
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    many people start doing a derivative on this problem, when all it need is algebra:)

  4. SolomonZelman
    • one year ago
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    Do you know how to re-write this in a vertex form ? (Note: since the leading coefficient of this quadratic is negative your parabola is opening down.)

  5. anonymous
    • one year ago
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    I do not, But thanks because you just answered my other question xD

  6. SolomonZelman
    • one year ago
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    I answered your another question ?

  7. SolomonZelman
    • one year ago
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    In any case. I will show you the steps in general

  8. SolomonZelman
    • one year ago
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    \(\large\color{black}{ f(x)=\displaystyle \color{red}{\rm a}x^2+ \color{green}{\rm b}x+ \color{blue}{\rm c} }\) \(\large\color{black}{ f(x)=\displaystyle \color{red}{\rm a}\left(x^2+ \frac{\color{green}{\rm b}}{\color{red}{\rm a}}x\right)+ \color{blue}{\rm c} }\) 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. \(\large\color{black}{ f(x)=\displaystyle \color{red}{\rm a}\left(x^2+ \frac{\color{green}{\rm b}}{\color{red}{\rm a}}x+\left[\frac{\color{green}{\rm b^2}}{\color{red}{\rm 4a^2}}\right] -\left[\frac{\color{green}{\rm b^2}}{\color{red}{\rm 4a^2}}\right] \right)+ \color{blue}{\rm c} }\) \(\large\color{black}{ f(x)=\displaystyle \color{red}{\rm a}\left(x^2+ \frac{\color{green}{\rm b}}{\color{red}{\rm a}}x+\frac{\color{green}{\rm b^2}}{\color{red}{\rm 4a^2}} \right)-\color{red}{\rm a}\left[\frac{\color{green}{\rm b^2}}{\color{red}{\rm 4a^2}}\right]+ \color{blue}{\rm c} }\) \(\large\color{black}{ f(x)=\displaystyle \color{red}{\rm a}\left(x+ \frac{\color{green}{\rm b}}{\color{red}{\rm 2a}} \right)^2-\color{red}{\rm a}\left[\frac{\color{green}{\rm b^2}}{\color{red}{\rm 4a^2}}\right]+ \color{blue}{\rm c} }\) \(\large\color{black}{ f(x)=\displaystyle \color{red}{\rm a}\left(x+ \frac{\color{green}{\rm b}}{\color{red}{\rm 2a}} \right)^2-\frac{\color{green}{\rm b^2}}{\color{red}{\rm 4a}}+ \color{blue}{\rm c} }\)

  9. SolomonZelman
    • one year ago
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    this is just me doing abstract thing..... you can look examples while I am finishing.

  10. anonymous
    • one year ago
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    Thanks man, I appreciate it

  11. SolomonZelman
    • one year ago
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    \(\large\color{black}{ f(x)=\displaystyle \color{red}{\rm a}\left(x+ \frac{\color{green}{\rm b}}{\color{red}{\rm 2a}} \right)^2-\frac{\color{green}{\rm b^2}}{\color{red}{\rm 4a}}+\frac{\color{green}{\rm 4a\color{blue}{\rm c}}}{\color{red}{\rm 4a}} }\) \(\large\color{black}{ f(x)=\displaystyle \color{red}{\rm a}\left(x+ \frac{\color{green}{\rm b}}{\color{red}{\rm 2a}} \right)^2+\frac{\color{green}{\rm 4a\color{blue}{\rm c}}-\color{green}{\rm b}^2}{\color{red}{\rm 4a}} }\)

  12. SolomonZelman
    • one year ago
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    in this case the vertex is \(\large\color{black}{ \displaystyle \left(-\frac{\color{green}{\rm b}}{2\color{red}{\rm a}}~,~\frac{4\color{red}{\rm a}\color{blue}{\rm c}- \color{green}{\rm b}^2}{4\color{red}{\rm a}}\right) }\)

  13. anonymous
    • one year ago
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    :/ Whats the highest though, Im confused.

  14. SolomonZelman
    • one year ago
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    don'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/Conic-Sections/Finding-the-Vertex-Form-of-a-Circle?id=817 https://www.youtube.com/watch?v=XyDMsotfJhE

  15. anonymous
    • one year ago
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    But you cant give me answer :( @SolomonZelman

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