Ask your own question, for FREE!
Mathematics 13 Online
OpenStudy (anonymous):

22. (04.07) Use the graph of a translated function below to answer this question: Given the parent function of f(x) = x3, what is the value of h in the translated graph of f(x - h) + k? h = -3 h = -2 h = 2 h = 3

OpenStudy (anonymous):

OpenStudy (anonymous):

I would have said -3 because h is a horizontal shift and negatives cause a shift to the right where as a positive value causes a shift to the left, but appearantly the correct answer is h = 3, why??

OpenStudy (anonymous):

@Michele_Laino

OpenStudy (michele_laino):

hint: the equation of the graph, is: \[\huge y = {\left( {x - 3} \right)^3} + 2\]

OpenStudy (anonymous):

So h is -3 :/ which is what I thought but my teacher says thats wrong... is it -3?

OpenStudy (michele_laino):

please, we have to compare with this equation: \[\huge y = {\left( {x - h} \right)^3} + k\] so \(h=3,\;k=2\)

OpenStudy (anonymous):

Okay so you do not count to - in front of the h? .-.

OpenStudy (michele_laino):

no, since if we substitute, \(h=3,\;k=2\) we get my formula above

OpenStudy (anonymous):

Okay, but why does it shift to the right?

OpenStudy (michele_laino):

physically (in the sense of Physics) speaking, when I write: \(y=f(x)\) and \(g=f(x-3)\) I'm speaking about progressive waves, which are moving to the right

OpenStudy (anonymous):

Wait, so 3 is positive because the standard formula has (x-h) and if h were negative it would be (x-(-h)) and a negative value (x+h) actually goes left and a positive (x-h) actually goes right but a short cut would just to go right if you see (-) and left if you see (+)?

OpenStudy (michele_laino):

yes! That's right!

OpenStudy (anonymous):

Okay I get it now. C: My textbook did not explain anything! It just said (x+h) goes left and (x-h) goes right so I just assumed that whenever I saw (x-h) it meant h was negative and went right, never thought about it the other way but now it makes sense :D thanks!

OpenStudy (michele_laino):

:)

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!