vertex form: Complete the square, then graph based on transformations.

[Gijoefan1975]

Put the following functions into vertex form by completing the square, and then graph them based on transformations (as you did on HW 12).

. . .\(\displaystyle a)\mbox{ }\, f(x)\, =\, x^2\, +\, 6x\, +\, 9\)

. . .\(\displaystyle b)\mbox{ }\, y\, =\, x^2\, -\, 8x\, +\, 16\)

. . .\(\displaystyle c)\mbox{ }\, y\, =\, x^2\, +\, 6x\, +\,5\)

. . .\(\displaystyle d)\mbox{ }\, y\, =\, x^2\, -\, 10x\, -\, 10\)

. . .\(\displaystyle e)\mbox{ }\, y\, =\, x^2\, +\, 2x\, +\, 8\)

. . .\(\displaystyle f)\mbox{ }\, y\, =\, 2x^2\, +\, 4x\, +\, 1\)

. . .\(\displaystyle g)\mbox{ }\, y\, =\, -x^2\, +\, x\, +\, 4\)

. . .\(\displaystyle h)\mbox{ }\, g(x)\, =\, 2\, -\, x^2\, +\, x\)

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Would it be okay to see these as set to zero?

 

[ksdhart2]

Sorry, but I don't understand. What do you mean by "see these as set to zero?" Do you, perhaps, mean setting the function equal to 0 and solving, in order to find the roots (aka "solutions," aka "zeroes") of the function?

 

[Gijoefan1975]

Sorry, but I don't understand. What do you mean by "see these as set to zero?" Do you, perhaps, mean setting the function equal to 0 and solving, in order to find the roots (aka "solutions," aka "zeroes") of the function?

Hmm Yes I think so....Thanks for asking,.... So like the first one Would I see it as x^2 +6x+9=0? Since we then have to add negative 9 to both sides then divide the 6 by two and then Square the 3 so that we then would end up with (x+3)^2+9

Thanks!

 

[HallsofIvy]

You do not need to set them to 0, just complete the square as suggested. For example, if the problem were \(\displaystyle y= x^2- 8x+ 7\), then, because \(\displaystyle -8/2= -4\) and \(\displaystyle (-4)^2= 16\) we would write \(\displaystyle x^2- 8x+ 16- 16+ 7= (x- 4)^2- 9\) so that the vertex is at (4, 9).

 

[Gijoefan1975]

You do not need to set them to 0, just complete the square as suggested. For example, if the problem were \(\displaystyle y= x^2- 8x+ 7\), then, because \(\displaystyle -8/2= -4\) and \(\displaystyle (-4)^2= 16\) we would write \(\displaystyle x^2- 8x+ 16- 16+ 7= (x- 4)^2- 9\) so that the vertex is at (4, 9).

ahh...okay so setting them to zero is not the right wording, so I guess I could look at it is there just is nothing on the other side of the equal sign?
Thanks

 

[Mrspi]

You do not need to set them to 0, just complete the square as suggested. For example, if the problem were \(\displaystyle y= x^2- 8x+ 7\), then, because \(\displaystyle -8/2= -4\) and \(\displaystyle (-4)^2= 16\) we would write \(\displaystyle x^2- 8x+ 16- 16+ 7= (x- 4)^2- 9\) so that the vertex is at (4, 9).

Or, you might look at it this way......

y = x² - 8x + 7

Let's subtract 7 from both sides so that we have just the terms containing x on the right:

y - 7 = x² - 8x

Go about the processing of completing the square: -8/2 = -4. (-4)² = 16. So, we need a +16 on the right side to make that a perfect square. We also need to add 16 to the left side!

y - 7 + 16 = x² -8x + 16
y + 9 = (x - 4)²
y = (x - 4)² - 9

Of course, that's exactly the same result you get with Halls' method, but perhaps it may be helpful to view what happens with that "y" on the left side. You can't really just ignore it......

 

[mmm4444bot]

...we would write \(\displaystyle [ y ] = (x- 4)^2- 9\) so that the vertex is at (4, 9).

There's a typo. The vertex is at (4, -9).

 

[stapel]

ahh...okay so setting them to zero is not the right wording, so I guess I could look at it is there just is nothing on the other side of the equal sign?

It sounds like you were in class for the lesson on solving quadratic equations by completing the square, but missed the lessons on graphing quadratic functions by completing the square. To learn the basics (so the replies make sense), try here.

 

[HallsofIvy]

There's a typo. The vertex is at (4, -9).

Yes, thank you.