Ask your own question, for FREE!
Mathematics 18 Online
OpenStudy (anonymous):

What transformation has changed the parent function f(x) = log2 x to its new appearance shown in the graph below? @rm152

OpenStudy (anonymous):

@ranga

OpenStudy (inkyvoyd):

we need pictures lol

OpenStudy (anonymous):

sorry i forgot @inkyvoyd

OpenStudy (anonymous):

@rm152

OpenStudy (anonymous):

It has been shifted upwards 3 units.

OpenStudy (anonymous):

f(x + 2) + 3 like that

OpenStudy (ranga):

Compare it to how \[\Large y = \log_{2} x\]would look like and you will see what shifts have taken place. The function has been shifted both along the x-axis and the y-axis. If you compare the graph you will know the amount by which it has been shifted along each axis and you can come up with the function.

OpenStudy (anonymous):

I have a feeling it is also compressed or stretched. What are the options.

OpenStudy (anonymous):

f(x + 3) + 2 f(x − 3) − 2 f(x + 2) + 3 f(x − 2) − 3

OpenStudy (anonymous):

what do you think @ranga @rm152

OpenStudy (anonymous):

The first answer f(x+3) + 2

OpenStudy (ranga):

The function y = log(x) has a vertical asymptote at x = 0. This function has a vertical asymptote at x = -2. Therefore, the function has been shifted to the left by 2 units. Shifting to the left by 2 units is same as replacing x by (x + 2). \[\Large y = \log_{2} (x+2) + k\] When x = 0, the log function is 1. The y-intercept on the graph is 4. So k must be 3. Therefore, \[\Large y = \log_{2} (x+2) + 3\] or y = f(x+2) + 3

OpenStudy (anonymous):

theres a part b to the question which is equivalent to \[\log_{125}25 \] 1.076 3 / 2 2 / 3 2/5

OpenStudy (ranga):

Let \[\log_{125}25 = \log_{5}25 / \log_{5}(125)= ?\]

OpenStudy (anonymous):

2/3

OpenStudy (ranga):

Yes.

OpenStudy (anonymous):

woohooo!

OpenStudy (ranga):

Alright.

OpenStudy (anonymous):

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!