plz help!! test tomorrow!

• Jun 11th 2006, 09:48 AM
alientype
plz help!! test tomorrow!
here is what i need to do_

''Rewrite the following functions in the form y=a(x-h)^2+k.Find the vertex.''

here is my problem_

y = x^2 + 4x - 9

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I tried to learn this but i just can't...my teacher is 2 lazy and not willing to spend any extra time with me. Also not a single stundent in my class knows this stuff. so please HELP! This is my last test and if i dont pass it i will have 2 go to summer school or even worse.

here some more problems_( i know these will be on a test)

y = x^2 - 6x + 16

y = x^2 - 8x - 1

y = x^2 + 5x + 2

thanks
• Jun 11th 2006, 10:36 AM
earboth
Quote:

Originally Posted by alientype
here is what i need to do_

''Rewrite the following functions in the form y=a(x-h)^2+k.Find the vertex.''

here is my problem_
y = x^2 + 4x - 9
----------
...
here some more problems_( i know these will be on a test)
y = x^2 - 6x + 16
y = x^2 - 8x - 1
y = x^2 + 5x + 2
thanks

Hello,

with all your problems you have to do the following steps:

If the factor before x^2 is 1 (as it is with your problems) you can calculate a complete square and an additional summand:
$y = x^2 + 4x - 9$ = $x^2 + 4x +\left({4\over 2}\right)^2-\left({4\over 2}\right)^2-9=(x+2)^2-13$
Coordinates of the Vertex are V(-2,-13).

1. step: Divide the coefficient of x (not x^2!!) by 2.
2. step:Square this quotient.
3. step: Add this square and subtract it at once
4. step: The first three summands are a complte square.
5. step: What you've subtracted and the constant together give the y-coordinate of the vertex.

$y = x^2 - 6x + 16$ = $x^2 - 6x +\left({6\over 2}\right)^2-\left({6\over 2}\right)^2+16=(x-3)^2+7$
Coordinates of the Vertex are V(3,7).

$y = x^2 - 8x - 1$ = $x^2 - 8x +\left({8\over 2}\right)^2-\left({8\over 2}\right)^2-1=(x-4)^2-17$
Coordinates of the Vertex are V(4,-17).

$y = x^2 + 5x + 2$ = $x^2 + 5x +\left({5\over 2}\right)^2-\left({5\over 2}\right)^2+2=(x+\frac{5}{2})^2-\frac{15}{4}$
Coordinates of the Vertex are $V\left(-\frac{5}{2},-\frac{15}{4}\right)$.

Greetings

EB
• Jun 11th 2006, 10:38 AM
Soroban
Hello, alientype!

Quote:

Rewrite the following functions in the form $y \:=\:a(x-h)^2+k.$
Find the vertex.

$1)\;y \,= \,x^2 + 4x - 9$
This requires "Completing the Square".
I hope you're famiiar with it . . .

We have: . $y\;=\;x^2 + 4x - 9$

Take one-half of the x-coefficient and square it:
. . $\frac{1}{2}(4) = 2\quad\Rightarrow\quad 2^2 = 4$

On the right side, add 4 and subtract 4:
. . . $y \;= \;x^2 + 4x$ + 4 $+\, 9$ - 4

And we have: . $y \;\:= \;\:(x + 2)^2 \,+ \,5$
. . . . . . . . . . . . . . . . . . . $\searrow\;\;\;\swarrow$
Therefore, the vertex is: . $(-2,\,5)$

Quote:

$2)\;y \:= \:x^2 - 6x + 16$
We have: . $\frac{1}{2}(-6) = -3\quad\Rightarrow\quad(-3)^2 = 9$

Add and subtract 9: . $y \;= \;x^2 - 6x$ + 9 $+ 16$ - 9

And we have: . $y \;= \;(x - 3)^2 \,+ \,7$
. . . . . . . . . . . . . . . . . . . $\searrow\;\;\,\swarrow$
Therefore, the vertex is: . . . $(3,7)$

Think you can do the rest of them now?
• Jun 11th 2006, 10:54 AM
alientype
thks alot guys i think i can do them now! 20 more probs. 2 go lol!
• Jun 11th 2006, 10:55 AM
earboth
Quote:

Originally Posted by Soroban
...

We have: . $y\;=\;x^2 + 4x - 9$

Take one-half of the x-coefficient and square it:
. . $\frac{1}{2}(4) = 2\quad\Rightarrow\quad 2^2 = 4$

On the right side, add 4 and subtract 4:
. . . $y \;= \;x^2 + 4x$ + 4 $+\, 9$ - 4

And we have: . $y \;\:= \;\:(x + 2)^2 \,+ \,5$
. . . . . . . . . . . . . . . . . . . $\searrow\;\;\;\swarrow$
Therefore, the vertex is: . $(-2,\,5)$
...

Hello, Soroban,

I don't want to be a know-it-all (I found this expression in my dictionary and I don't know if it is appropriate here), but in comparison with the problem and your solution I believe that you changed the sign before the 9.

Greetings

EB