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

1. anonymous

@SolomonZelman

2. SolomonZelman

Oh, don't use calculus. Find the vertex.

3. SolomonZelman

many people start doing a derivative on this problem, when all it need is algebra:)

4. SolomonZelman

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

I do not, But thanks because you just answered my other question xD

6. SolomonZelman

7. SolomonZelman

In any case. I will show you the steps in general

8. SolomonZelman

$$\large\color{black}{ f(x)=\displaystyle \color{red}{\rm a}x^2+ \color{green}{\rm b}x+ \color{blue}{\rm c} }$$ $$\large\color{black}{ f(x)=\displaystyle \color{red}{\rm a}\left(x^2+ \frac{\color{green}{\rm b}}{\color{red}{\rm a}}x\right)+ \color{blue}{\rm c} }$$ 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

this is just me doing abstract thing..... you can look examples while I am finishing.

10. anonymous

Thanks man, I appreciate it

11. SolomonZelman

$$\large\color{black}{ f(x)=\displaystyle \color{red}{\rm a}\left(x+ \frac{\color{green}{\rm b}}{\color{red}{\rm 2a}} \right)^2-\frac{\color{green}{\rm b^2}}{\color{red}{\rm 4a}}+\frac{\color{green}{\rm 4a\color{blue}{\rm c}}}{\color{red}{\rm 4a}} }$$ $$\large\color{black}{ f(x)=\displaystyle \color{red}{\rm a}\left(x+ \frac{\color{green}{\rm b}}{\color{red}{\rm 2a}} \right)^2+\frac{\color{green}{\rm 4a\color{blue}{\rm c}}-\color{green}{\rm b}^2}{\color{red}{\rm 4a}} }$$

12. SolomonZelman

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

:/ Whats the highest though, Im confused.

14. SolomonZelman

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

But you cant give me answer :( @SolomonZelman