Ask your own question, for FREE!
Mathematics 4 Online
OpenStudy (anonymous):

Please help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Rewrite y = x2 + 14x + 29 in general form. Rewrite y = 3x2 - 24x + 10 in general form.

OpenStudy (slaaibak):

Is the general form y = a(x-b)2 + c ?

OpenStudy (anonymous):

I think it's y=a(x-h)^2+k but im not sure

OpenStudy (anonymous):

im just not sure what the variable equal?? @slaaibak

OpenStudy (slaaibak):

Cool, i'll walk you through it

OpenStudy (anonymous):

I just need to know what to plug in. like what does k and h equal??

OpenStudy (anonymous):

@slaaibak????????

OpenStudy (slaaibak):

Let's start with the first one: y = x2 + 14x + 29 The key to doing this, is by completing the square for x: y = x2 + 14x + 29 Now add 49 and substract 49, because 49 = (14/2)^2 y = x2 + 14x + 49 - 49 + 29 but, x2 + 14x + 49 = (x+7)^2 so y = (x+7)^2 - 30 so h=-7, k = 30

OpenStudy (slaaibak):

sorry, k = -30

OpenStudy (anonymous):

thank you!!!

OpenStudy (slaaibak):

Sorry, I made a mistake haha. y = (x+7)^2 - 20 so h = -7 and k = -20

OpenStudy (anonymous):

so is the whole equations supposed to have exact numbers? where do i plug in x^2 14 x and 29?

OpenStudy (slaaibak):

read what I did. It explains everything. You just transform the equation.

OpenStudy (anonymous):

im so confused i meant i needed the form with the correct numbers but are you solving for the actual value?

OpenStudy (slaaibak):

I'm not solving anything.... I have the correct form. y = (x+7)^2 - 20

OpenStudy (anonymous):

You said h=-7?

OpenStudy (anonymous):

^^

OpenStudy (slaaibak):

yes, but I'm just showing how the general form relates to what I got. Nevermind. Just look at the answer: y = (x+7)^2 - 20

OpenStudy (anonymous):

could you show me how do the second one? i am still so confused. Sorry, thank you.

OpenStudy (slaaibak):

y = 3x2 - 24x + 10 First factor out the 3: y = 3(x2 - 8x + 10/3) add and subtract: (8/2)^2 = 16 y= 3(x2 - 8x + 16 - 16 + 10/3) Try to go further from here

OpenStudy (anonymous):

nvmd thanks anyway

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!