Ask your own question, for FREE!
Mathematics 52 Online
OpenStudy (study_buddy99):

Graphing functions practice

Directrix (directrix):

Let's look at the last equation and see what happens to its graph if a negative x + 1 is under the radical. y = - (-x+1) ^ (1/3) + 5 You can copy and paste that into the Desmos.com calculator function slot. I'll be over at Desmos.com seeing what I get. Ready?

OpenStudy (study_buddy99):

yep

Directrix (directrix):

If the copy and paste option will not work, use Control + V or you can type it in.

OpenStudy (study_buddy99):

okay I got it

Directrix (directrix):

I added a second function with the lead negative removed.

Directrix (directrix):

What next? I have a very small collection in my head of what I call "Great Graphs" which are very basic equations and the associated graphs. (parent graphs) I remember those to have some idea of how the graph of a more complicated equation might appear.

OpenStudy (study_buddy99):

what about this one?\[f(x)=-(x)^\frac{ 1 }{ 2 }+5\]

Directrix (directrix):

>okay I got it Do you have radical equation questions to answer that do not involve simply drawing the graphs.

OpenStudy (study_buddy99):

like what we were doing last night?

Directrix (directrix):

I was thinking about questions about the functions. What is its x-intercept. What are the end behaviors of the graph?

OpenStudy (study_buddy99):

yeah, my worksheet asks me what the y-int and the:\[as x \rightarrow \infty, f(x)\rightarrow?\]

Directrix (directrix):

I don't know how to interpret that. asx --> Is that something to do with the x-intercept or the vertical asymptote?

OpenStudy (study_buddy99):

as x*

Directrix (directrix):

That is end behavior talk.

Directrix (directrix):

What is the equation of the function that goes with that?

OpenStudy (study_buddy99):

\[f(x)= -\sqrt[3]{x+1}+5\]

Directrix (directrix):

Think about an equation like y = 2*x + 3 As x increases without bound, what are the values of y doing? Are they increasing without bound? Are they approaching 0? So, looking at the graph. |dw:1456210048228:dw|

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!