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Mathematics 28 Online
OpenStudy (anonymous):

please help i will fan!!! Given f(x) and g(x) = f(x) + k, look at the graph below and determine the value of k. k = ______ Graph of f of x and g of x. f of x equals 1 over 3 x minus 2 and g of x equals 1 over 3 x plus 3. Numerical Answers Expected! Answer for Blank 1:

OpenStudy (anonymous):

@magicremix123

OpenStudy (vickyisthesmartone123):

hahaha...Your face in the picture is so dramatic!;)

OpenStudy (vickyisthesmartone123):

Go to mathway.com

OpenStudy (anonymous):

right ;) @Vickyisthesmartone123

OpenStudy (trojanpoem):

f(x) = 1/(3x-2) , g(x) = 1/(3x+3) is it like that ?

OpenStudy (vickyisthesmartone123):

LOL

OpenStudy (trojanpoem):

or f(x) = 1/3x - 2 , g(x) = 1/3x + 3

OpenStudy (anonymous):

looks like you've gotten help.....

OpenStudy (vickyisthesmartone123):

hehe

OpenStudy (vickyisthesmartone123):

@TrojanPoem I like your picture so spooky and violent abstract LOL

OpenStudy (anonymous):

OpenStudy (vickyisthesmartone123):

Go to desmos.com

OpenStudy (trojanpoem):

This is a linear equation: y = mx + b ( m : slope , b : y-intercept) f(x) = mx -2 Point (6,0) is on f(x). 0 = -6m -2 , m = 1/3 f(x) = x/3 - 2 ( As you mentioned above) g(x) = x/3 + 3 ( same slope parallel lines , you found that too). g(x) = f(x) + k x/3 - 2 = x/3 + 3 + k (Slove for k) k = -5

OpenStudy (anonymous):

so -5??

OpenStudy (vickyisthesmartone123):

Oh! your so smart @TrojanPoem

OpenStudy (vickyisthesmartone123):

Yes the correct answer is -5!

OpenStudy (trojanpoem):

nah, 5 as I made a typo. x/3 + 3 = x/3 -2 + k k = 5

OpenStudy (anonymous):

ha......

OpenStudy (vickyisthesmartone123):

HA!

OpenStudy (vickyisthesmartone123):

The correct answer is 5 Then...LOL

OpenStudy (anonymous):

well since we all are giving medals miles give me one xD

OpenStudy (vickyisthesmartone123):

I gave you one!:)

OpenStudy (anonymous):

ITS 5 XD AND OKAY< sorry caps, i do this a lot -_-

OpenStudy (vickyisthesmartone123):

hehe! It's ok

OpenStudy (vickyisthesmartone123):

I guess...

OpenStudy (anonymous):

thank you guysssss

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