Ask your own question, for FREE!
Mathematics 16 Online
OpenStudy (anonymous):

Find the maximun y-value on the graph of y=f(x). f(x)=-x^2+2x+2

OpenStudy (tkhunny):

You must select words more carefully. Words mean things. You have defined y = f(x). Therefore, it makes no sense to sak "What is y?" You just defined it. If you copied this accurately, you should complain to the author. Anyway, how do you propose that we might find a solution? Perhaps a 1st Derivative? Maybe a 2nd Derivative? What say you?

OpenStudy (anonymous):

change -x^2 +2x+2 into completed square form

OpenStudy (anonymous):

can u?

OpenStudy (anonymous):

no how do I do that

OpenStudy (tkhunny):

So, algebra it is, then. You MUST know how to complete the square. Utilize this principle: \((a+b)^{2} =a^{2} +2ab+b^{2}\) How do you apply that to your right-hand expression?

OpenStudy (anonymous):

notsu re

OpenStudy (anonymous):

-x^2+2x+2 -(x^2-2x)+2 -{(x)^2 - 2*x*1 + 1^2 - 1^2 } +2 -(x-1)^2 +1+2 -(x-1)^2+3

OpenStudy (anonymous):

So, f(x)=-(x-1)^2+3 right?

OpenStudy (anonymous):

right

OpenStudy (anonymous):

Now, (x-1)^2>=0 u know it?

OpenStudy (anonymous):

is that y

OpenStudy (anonymous):

y=f(x)=-(x-1)^2+3

OpenStudy (anonymous):

so y=(x-1)^2+3

OpenStudy (anonymous):

NO

OpenStudy (anonymous):

U missed the -ve sign

OpenStudy (anonymous):

where

OpenStudy (anonymous):

y=-(x-1)^2+3

OpenStudy (anonymous):

oh gotsa

OpenStudy (anonymous):

So, (x-1)^2>=0 -(x-1)^2<=0 -(x-1)^2+3<=3 y<=3

OpenStudy (anonymous):

thanks

OpenStudy (anonymous):

So, the max value of y is 3 got it?

OpenStudy (anonymous):

yes

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!
Latest Questions
Arriyanalol: i need help
1 day ago 7 Replies 1 Medal
Arriyanalol: CONGRATS to me
1 day ago 8 Replies 2 Medals
Legomaster346: What colleges are good that are near Arab Alabama?
1 week ago 5 Replies 0 Medals
Arriyanalol: i need help
1 week ago 6 Replies 2 Medals
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!