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- Thread starter theredeemer
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eumyang

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The object wouldn't be in the air anymore when it hits the ground, no? So what should you set h(t) equal to?

Assuming that you know nothing about calculus, recall that the vertex of a quadratic equation is the maximum or minimum point. So rewrite the equation in vertex form.

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VietDao29

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There are a couple of ways of achieving this:

**The first way**is to use your imagination. Notice that, when you launch the object from the ground, its speed is greatest, right? And then, as it flies off, the speed decreases. And when its speed is 0, it's also the time that object reaches its max height (before it starts to falls off).

So, to find it's max height, we can calculate h(t_{0}), where**t**._{0}is the time that its speed is 0

**The second way**is to try to*Complete the Square*to find the maximum value for*h*(*t*). Have you studied how to Complete the Square?

In case you haven't, I'll explain it briefly so you can get the idea. Completing the Square is the manipulation you do to change your expression into something like this:

(*a*+*b*)^{2}+*c*, or*c*- (*a*+*b*)^{2}

The identity we should use here is the Square of Sum: (*a*+*b*)^{2}=*a*^{2}+ 2*ab*+*b*^{2}

**Example**: Complete the Square for:- [tex]x ^ 2 + 3x + 1[/tex]
- [tex]-x ^ 2 + 6x - 3[/tex]

------------------------------

**Problem 1:**[tex]x ^ 2 + 3x + 1[/tex].

First we'll split it so that it contains the first 2 terms*a*^{2}+ 2*ab*, like this:

[tex]x ^ 2 + 3x + 1 = x ^ 2 + 2.(x)\left(\frac{3}{2}\right) + 1[/tex]

Then, we'll add, and subtract*b*^{2}, note that, in this example: [tex]b = \frac{3}{2}[/tex] like this:

[tex]... = x ^ 2 + 2.x\frac{3}{2} + 1 = x ^ 2 + 2.x\frac{3}{2} + \frac{9}{4} - \frac{9}{4} + 1 = \left(x ^ 2 + 2.x\frac{3}{2} + \frac{9}{4}\right) - \frac{5}{4}[/tex]

[tex]= \left( x + \frac{3}{2} \right) ^ 2 - \frac{5}{4}[/tex]

**Problem 2:**[tex]-x ^ 2 + 6x - 3[/tex].

It's the same, except that we have a little minus sign "-", in front of*x*^{2}, right? So we can factor -1 out, like this, and do the same as what we've done above:

[tex]-x ^ 2 + 6x - 3 = -(x ^ 2 - 6x + 3) = - (x ^ 2 + 2.x.3 + 3) = - (x ^ 2 + 2.x.3 + 9 - 9 + 3)[/tex]

[tex]= - [(x ^ 2 + 2.x.3 + 9) - 6] = -[(x + 3) ^ 2 - 6] = 6 - (x + 3) ^ 2[/tex]

After Completing the Square for*h*(*t*), you can find its maximum value by noting that: [tex]\alpha ^ 2 \ge 0 , \forall \alpha \in \mathbb{R}[/tex]

Let's see if you can tackle this problem on your own. :)

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Moderator's note:

Since this thread is a duplicate, I am closing it.

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