OpenStudy (sophiesky):

Another question! These confuse me.. Evaluate g(x+4) if g(x)=(x^2+3)/(2x)+4

1 year ago
OpenStudy (loser66):

Where are you stuck?

1 year ago
OpenStudy (sophiesky):

Well, do I set the eqaution equal to 0?

1 year ago
OpenStudy (loser66):

Nope, just from g(x) , to get g(x+4) you need replace x by x+4. Dat sit

1 year ago
OpenStudy (sophiesky):

Ohhhhhhh wait wait wait

1 year ago
OpenStudy (loser66):

\(g(\color{red}{x})=\dfrac{\color{red}{x}^2+3}{2\color{red}{x}+4}\) \(g(\color{red}{(x+4)})=???\)

1 year ago
OpenStudy (sophiesky):

So basically it's \[g(x)=(x+4)^{2}+3/2x\]

1 year ago
OpenStudy (loser66):

why don't you replace x by x+4 on denominator? where is your +4 at last?

1 year ago
OpenStudy (sophiesky):

Oh geez. I'm totally not on my game right now. One sec, let me put it all together.

1 year ago
OpenStudy (sophiesky):

Okay, \[\frac{ (x+4)^{2} }{ 2(x+4) }+4\]

1 year ago
OpenStudy (sophiesky):

Is that the right format?

1 year ago
OpenStudy (loser66):

where is your +3 from numerator?

1 year ago
OpenStudy (jim_thompson5910):

Replace every copy of 'x' with 'x+4' \[\Large g(x) = \frac{x^2+3}{2x}+4\] \[\Large g(x) = \frac{(x)^2+3}{2(x)}+4\] \[\Large g({\color{red}{x}}) = \frac{({\color{red}{x}})^2+3}{2({\color{red}{x}})}+4\] \[\Large g({\color{red}{x+4}}) = \frac{({\color{red}{x+4}})^2+3}{2({\color{red}{x+4}})}+4\]

1 year ago
OpenStudy (jim_thompson5910):

I'll let you simplify it out

1 year ago
OpenStudy (sophiesky):

Alright I'm simplifying it out and I have \[\frac{ x^2+19 }{ 2x+8 }+4\]

1 year ago
OpenStudy (jim_thompson5910):

You made a mistake with \(\Large (x+4)^2\)

1 year ago
OpenStudy (jim_thompson5910):

(x+4)^2 is NOT equal to x^2+16

1 year ago
OpenStudy (sophiesky):

Oh... It's not?

1 year ago
OpenStudy (jim_thompson5910):

(x+4)^2 = (x+4)*(x+4) use foil or the box method https://www.youtube.com/watch?v=3OvHmKjVTGU

1 year ago
OpenStudy (jim_thompson5910):

hopefully that video helps show an example of the box method in action

1 year ago
OpenStudy (sophiesky):

Ah, alright. So it would be \[\frac{ 2x^2+8x+3 }{ 2x+8 }+4\]

1 year ago
OpenStudy (jim_thompson5910):

|dw:1473297020420:dw|

1 year ago
OpenStudy (jim_thompson5910):

|dw:1473297035921:dw|

1 year ago