Completing the square / Changing format of equation

[danielle36]

I'm supposed to use the method of completeing the square to write an equation in the form [tex] y = a(x - p)^{2} + q [/tex].

Here's one of the equations: [tex] y = x^{2} + 6x [/tex]
and another: [tex] y = -2x^{2} + 8x + 5 [/tex]

I really don't know where to start, which is why I can't include any work... The main problem here is I don't know how to get the equation in the [tex] (x - p)^{2} [/tex] format. I can complete the square but its the - sign that's throwing me off.
If someone could just explain how to go about doing this it would be greatly appreciated.

 

[ice109]

I'm supposed to use the method of completeing the square to write an equation in the form [tex] y = a(x - p)^{2} + q [/tex].

Here's one of the equations: [tex] y = x^{2} + 6x [/tex]
and another: [tex] y = -2x^{2} + 8x + 5 [/tex]

I really don't know where to start, which is why I can't include any work... The main problem here is I don't know how to get the equation in the [tex] (x - p)^{2} [/tex] format. I can complete the square but its the - sign that's throwing me off.
If someone could just explain how to go about doing this it would be greatly appreciated.

theres a completeing the square formula but the first one is very obvious so i'll do it for you

it'll be easier to see if you rewrite this
[tex] y = (ax^2-2xp +p^2) -p^2 [/tex]
then your first problem is

[tex] y = x^{2} + 6x + 9 - 9 = (x^2 +3)^2 - 9 [/tex]

 

[symbolipoint]

danielle36,

Other messages have mentioned this; but if you could find a graphical representation of Completing The Square, you will probably understand the process very well. The idea of converting the equation from general form into standard form will require subtracting the "piece" that you add in order for the expression to be equivalent to the original expression.

 

[olgranpappy]

[tex] y = x^{2} + 6x + 9 - 9 = (x^2 +3)^2 - 9 [/tex]

...just fixing a typo. the last x should not be squared on the far RHS
[tex]
(x+3)^2-9
[/tex]

 

[HallsofIvy]

(x-a)²= x²- 2ax+ a²

If you want to change x²- 6x to a perfect square what must "a" be? You know that -6x= -2ax.

-2x²+ 8x+ 5 is a little harder. Write it as -2(x²- 4x)+ 5. Now look just at the x²- 4x. Remembering that -2ax= -4x, what is a? What is a²? Add and subtract a² inside the parentheses and then take part outside the parentheses.