Find the vertex form of f(x) = 16 - x^2

unregistered

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Find the vertex form of f(x) = 16 - x^2

Is that vertex form of:

f(x) = 16 - x<sup>2</sup>

f(x) = -1(x - 16)<sup>2</sup> + 0

Is there a good rule of thumb when trying to put into vertex form that I can adhere to?

Thanks.
 

stapel

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Since -(x - 16)<sup>2</sup> = -x<sup>2</sup> + 32x - 256, not 16 - x<sup>2</sup>, your last function cannot be equivalent, and thus cannot be the completed square.

The following is the usual process for completing the square to find the vertex, as illustrated using a different function:

. . . . .Given, a function:
. . . . .y = 3x<sup>2</sup> + 6x + 5

. . . . .Separate the constant term:
. . . . .y = (3x<sup>2</sup> + 6x) + 5

. . . . .Factor out whatever is multiplied on the square:
. . . . .y = 3(x<sup>2</sup> + 2x) + 5

. . . . .Find half of the linear coefficient:
. . . . .(1/2)(2) = 1

. . . . .Square, and add inside and subtract outside,
. . . . .remembering to account for the factoring:
. . . . .y = 3(x<sup>2</sup> + 2x + 1) + 5 - 3(1)

. . . . .Simplify:
. . . . .y = 3(x<sup>2</sup> + 2x + 1) + 2

. . . . .Convert to squared form:
. . . . .y = 3(x + 1)<sup>2</sup> + 2
. . . . .y = 3(x - (-1))<sup>2</sup> + 2

. . . . .Read off the vertex:
. . . . .(h, k) = (-1, 2)

Follow this process with your exercise. If you get stuck, please reply showing all of your work and reasoning. Thank you.

Eliz.
 

unregistered

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Whoa, thank-you for such a great detailed response but are we working on the same equation? Where did you get the leading term 3 and the constant of 5? I wanted to try and get this into the vertex form:



Thanks.
 

galactus

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Completing the square is standard(vertex) format.

\(\displaystyle \L\\y=a(x-h)^{2}+k\)

h is the x-coordinate of the vertex and k is the y-coordinate of the vertex.

Your function is already in 'vertex' format.

The vertex of this parabola is at (0,16)

\(\displaystyle y=a(x-0)^{2}+16\)

Where a=-1

The parabola passes through (4,0)

\(\displaystyle 0=a(4-0)^{2}+16\)

\(\displaystyle 0=16a+16\)

\(\displaystyle a=-1\)

So, you have:

\(\displaystyle y=-1x^{2}+16\Rightarrow{y=16-x^{2}}\)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let's try another so you can see:

Let's say we have a parabola with vertex (2,3) that passes through (5,1):

\(\displaystyle y=a(x-h)^{2}+k\)

h=2 and k=3

\(\displaystyle y=a(x-2)^{2}+3\)

Use the x and y points:

\(\displaystyle 1=a(5-2)^{2}+3\)

\(\displaystyle a=\frac{-2}{9}\)

Therefore, \(\displaystyle \L\\y=\frac{-2}{9}(x-2)^{2}+3\)

Then, \(\displaystyle \frac{-2}{9}x^{2}+\frac{8}{9}x+\frac{19}{9}\)

Now, complete the square to get back to standard form:

\(\displaystyle \L\\y=\underbrace{\frac{-2}{9}}_{\text{a}}x^{2}+\underbrace{\frac{8}{9}}_{\text{b}}x+\underbrace{\frac{19}{9}}_{\text{c}}\)

\(\displaystyle \L\\a(x+\frac{b}{2a})^{2}+\underbrace{c-\frac{b^{2}}{4a}}_{\text{constant}}=(\frac{-2}{9})\left(x+\frac{\frac{8}{9}}{2(\frac{-2}{9})}\right)^{2}+\underbrace{\frac{19}{9}-\frac{(\frac{8}{9})^{2}}{4(\frac{-2}{9})}}_{\text{constant=3}}=\frac{-2}{9}(x-2)^{2}+3\)

Use the above 'completing the square' formula on your problem:

\(\displaystyle \L\\a(x+\frac{b}{2a})^{2}+c-\frac{b^{2}}{4a}\)

\(\displaystyle \L\\-1(x+0)^{2}+16-0\Rightarrow{-x^{2}+16}\)...Voila!
 

unregistered

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Great, thank for the explanation and the bonus example. The only thing that puzzles me now is in the original equation x was negative and when you wrote it out in standard/vertex form it wasn't. I know that you came out with the result for a as -1 which totally makes sense but is that a rule that I should follow? that is, if x is negative in the original equation than drop the negative and make it positive?

Thanks again for that awesome tutorial.
 

galactus

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unregistered said:
The only thing that puzzles me now is in the original equation x was negative and when you wrote it out in standard/vertex form it wasn't.
I don't follow. Where was x^2 positive?. Your function was already in standard form. Which was \(\displaystyle {-}x^{2}+16\)

Use the completing the square formula I showed you the next time you have one.

It factors to \(\displaystyle (4-x)(x+4)\), but that's not what you want.
 

stapel

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stapel said:
...as illustrated using a different function....
unregistered said:
...are we working on the same equation?
No. I provided a complete worked explanation using a different function, meaning that the function used in the mini-lesson was not the same.

I apologize for any confusion.

Eliz.
 

unregistered

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galactus said:
Completing the square is standard(vertex) format.

\(\displaystyle \L\\y=a(x-h)^{2}+k\)

h is the x-coordinate of the vertex and k is the y-coordinate of the vertex.

Your function is already in 'vertex' format.

The vertex of this parabola is at (0,16)

\(\displaystyle y=a(x-0)^{2}+16\)

In your answer above x became positive no? Sorry for any frustration, I wanted to answer your question. In the original equation it started out as \(\displaystyle f(x)=16-x^{2}\).
 

galactus

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That's the standard eqaution of a parabola format. The x in the parentheses is not your -x^2.
 

unregistered

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oh excellent, thank you for clearing that up and most of all for your patience.
 
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