Ask your own question, for FREE!
Mathematics 21 Online
OpenStudy (bluebeta):

Will someone check my work please. Attaching pictures. Will medal.

OpenStudy (bluebeta):

1

OpenStudy (bluebeta):

Need help on 12-15

OpenStudy (bluebeta):

Need help on 15-17

OpenStudy (bluebeta):

Need help on 18-19

OpenStudy (bluebeta):

Topic is transformations subject is pre-calculus

OpenStudy (solomonzelman):

So far, everything in the first attachment is correct. I will check the other attachments now.

OpenStudy (solomonzelman):

For #8, yes it flips, and in mathematical terms it is called "reflection across the x-axis"

OpenStudy (solomonzelman):

#10, you left out the fact that it is streched by a scale factor of a>1 (you will get that when you factor out of 2 from the parenthesis - which you apparently did)

OpenStudy (solomonzelman):

Please redo #11.

OpenStudy (solomonzelman):

#12 is correct. (#13 & #14 are not yet completed.)

OpenStudy (bluebeta):

How do I do #11? & yes any of the ones not answered I need help with.

OpenStudy (solomonzelman):

\(\large\color{black}{ \displaystyle y=2\left(\frac{1}{2}(x+2)\right)^2-7 }\) you know that: \(\large\color{blue}{ \displaystyle \left(a\times b\right)^n=a^n\times b^n }\) and by that rule, you get: \(\large\color{black}{ \displaystyle y=2\left(\frac{1}{2}\right)^2\left(x+2\right)^2-7 }\) which simplies to: \(\large\color{black}{ \displaystyle y=2\left(\frac{1}{4}\right)\left(x+2\right)^2-7 }\) \(\large\color{black}{ \displaystyle y=\frac{1}{2}\left(x+2\right)^2-7 }\)

OpenStudy (solomonzelman):

then you tell me how to describe this function in terms of the parent function.

OpenStudy (bluebeta):

Ok one second @solomonzelman

OpenStudy (bluebeta):

Ok @solomonzelman I'm confused now? I thought what I put was good.

OpenStudy (solomonzelman):

No, it is not-:(

OpenStudy (bluebeta):

How do I fix it ): @solomonzelman

OpenStudy (bluebeta):

We're you able check anything else?

OpenStudy (bluebeta):

@SolomonZelman ??

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!