Question

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

Answer 1

@SolomonZelman

Answer 2

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

Answer 3

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

Answer 4

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.)

Answer 5

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

Answer 6

I answered your another question ?

Answer 7

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

Answer 8

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

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

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.

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

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

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

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

Answer 9

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

Answer 10

Thanks man, I appreciate it

Answer 11

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

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

Answer 12

in this case the vertex is

\(\large\color{black}{ \displaystyle \left(-\frac{\color{green}{\rm b}}{2\color{red}{\rm a}}~,~\frac{4\color{red}{\rm a}\color{blue}{\rm c}- \color{green}{\rm b}^2}{4\color{red}{\rm a}}\right) }\)

Answer 13

:/ Whats the highest though, Im confused.

Answer 14

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

Answer 15

But you cant give me answer :( @SolomonZelman