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September 19th, 2015, 05:19 PM   #1
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Help converting standard form of quadratic function to vertex form

Hello everyone,

New here. Anyways, I am trying to convert the following function from standard to vertex form.

$\displaystyle f(x)=x^2+3x+1/4$

Besides that, does anybody have a very good explanation for noobs like me to understand. I'm really confused and need some help here.

Thanks.
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September 19th, 2015, 06:01 PM   #2
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The technique for converting from standard form to vertex form is call "completing the square". That means that we look for a linear term in $x$ that we can square and then just add (or subtract) a constant from to get the same equation.

That is, we want $$(x + a)^2 + b = x^2 + 3x + \tfrac14$$
Expanding the left hand side we get
$$x^2 + 2ax + (a^2 + b) = x^2 + 3x + \tfrac14$$
We need to find the values of $a$ and $b$ which we can do by comparing the coefficients of the different powers of $x$.

The $x^2$ terms are equal (which is important for the answer to be correct, but doesn't tell us much). But for the terms in $x$ we have
$$2ax = 3x \implies \boxed{a = \tfrac32}$$
which is half of our job done. The other half is done by comparing the constant terms
$$a^2 + b = \tfrac14$$
But we know that $a = \tfrac32$, so we can put that into the equation to get
$$\tfrac94 + b = \tfrac14 \implies \boxed{b = -\tfrac84 = -2}$$

Thus we have a vertex form $$f(x) = (x-\tfrac32)^2 - 2$$

In practice we don't do all of this formally. After doing a few of these (or perhaps from studying the above) you will notice that $a$ will always be half of the coefficient of the $x$ term (at least, it is when the coefficient of the $x^2$ term is $1$).

For example where the coefficient of the $x$ term is not equal to $1$, the easiest approach is to factor that coefficient out, writing $$f(x) = px^2 + qx + r = p\left(x^2 + \tfrac{q}{p}x + \tfrac{r}{p}\right)$$ and then "completing the square" for the quadratic inside the brackets.
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September 19th, 2015, 06:10 PM   #3
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1. Take out the coefficient of $\displaystyle x^2$ (in this case, 1) from $\displaystyle ax^2 + bx$, but don't touch the constant term:
$\displaystyle f(x) = 1(x^2 + 3x) + \frac{1}{4}$

2. Turn the coefficient of the second term to the product of 1, 2 and a third number (let's call it $\displaystyle h$, which in our case is $\displaystyle \frac{3}{2}$):
$\displaystyle f(x) = 1(x^2 + 2(1)(\frac{3}{2})x) + \frac{1}{4}$

3. Add $\displaystyle h^2$ to the expression inside the bracket, and add $\displaystyle -ah^2$ to the constant term:
$\displaystyle f(x) = 1(x^2 + 2(1)(\frac{3}{2})x + (\frac{3}{2})^2) + \frac{1}{4} - 1(\frac{3}{2})^2 $

4. Simplify time:
$\displaystyle f(x) = (x + \frac{3}{2})^2 - 2$

Thus the vertex is $\displaystyle (-\frac{3}{2}, -2)$

Edit: v8archie beat me to it XD
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Last edited by 123qwerty; September 19th, 2015 at 06:16 PM.
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September 19th, 2015, 06:16 PM   #4
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Alternatively,

$\displaystyle ax^2+bx+c=a\left(x+\dfrac{b}{2a}\right)^2+c-\dfrac{b^2}{4a}$
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Last edited by greg1313; September 20th, 2015 at 02:56 PM.
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September 19th, 2015, 08:38 PM   #5
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Thanks guys for the help. Your all's method worked like a charm

I finally understand it...
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September 20th, 2015, 12:18 PM   #6
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Okay. Another question now.

I have a similar equation here.

$\displaystyle f(x)=(1/4(x))^2-2x-12$

When I'm converting from standard to vertex
I'm using $\displaystyle x^2+2ax+(a^2+b)=(1/4(x))^2-2x-12$

Now $\displaystyle x^2$ doesn't equal $\displaystyle (1/4(x))^2$

I tried to equal them to each other and work them out but I got a weird answer.
At the end of it all
I got $\displaystyle 1/2(x-1)^2-13$

I'm putting both the equations in the calculator and their not giving me the same points.

Any help is appreciated.
Thanks

Last edited by abc98; September 20th, 2015 at 12:23 PM.
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September 20th, 2015, 03:12 PM   #7
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I'd rewrite it as $\displaystyle \dfrac{1}{16}x^2-2x-12$, then apply one of the formulae
given above. The method your using only applies if the coefficient of $\displaystyle x^2$ is $\displaystyle 1$
(scroll down to the bottom of v8archie's post to see what to do if the
coefficient of $\displaystyle x^2$ is not $\displaystyle 1$).

By the way, I made an error in my post above (which I have since corrected).
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September 20th, 2015, 05:08 PM   #8
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Quote:
Originally Posted by greg1313 View Post
I'd rewrite it as $\displaystyle \dfrac{1}{16}x^2-2x-12$, then apply one of the formulae
given above. The method your using only applies if the coefficient of $\displaystyle x^2$ is $\displaystyle 1$
(scroll down to the bottom of v8archie's post to see what to do if the
coefficient of $\displaystyle x^2$ is not $\displaystyle 1$).

By the way, I made an error in my post above (which I have since corrected).
Thanks for the input. I'm still confused though with the last part of v8archie's post. I don't understand what the values of r and q represent.
Can you give me a detailed explanation like visual steps on how to attempt the problem with v8archie's formula that he posted.
Sorry if it's stupid question. I'm not good at math at all

Thanks for the help
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September 20th, 2015, 05:16 PM   #9
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$\displaystyle f(x)=(1/4(x))^2-2x-12 = \tfrac{1}{16}x^2 - 2x - 12 = \tfrac1{16}\left(x^2 - 32x - 192\right)$

Now convert $\displaystyle x^2 - 32x - 192$ into vertex form as in the first question.

Does the question really have $\displaystyle \left(\tfrac14 x\right)^2$? Or is it $\displaystyle \tfrac14 x^2$?
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September 20th, 2015, 05:18 PM   #10
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Quote:
Originally Posted by v8archie View Post
$\displaystyle f(x)=(1/4(x))^2-2x-12 = \tfrac{1}{16}x^2 - 2x - 12 = \tfrac1{16}\left(x^2 - 32x - 192\right)$

Now convert $\displaystyle x^2 - 32x - 192$ into vertex form as in the first question.

Does the question really have $\displaystyle \left(\tfrac14 x\right)^2$? Or is it $\displaystyle \tfrac14 x^2$?
Thanks for the help

It's $\displaystyle \tfrac14 x^2$
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