Ask your own question, for FREE!
Mathematics 17 Online
OpenStudy (princesssleelee):

Given the function f(x) = x2 and k = 2, which of the following represents a horizontal shift? f(x) + k kf(x) f(x + k) f(kx)

OpenStudy (misty1212):

HI!!

OpenStudy (misty1212):

it has nothing to do with the particular function you are given compared to the graph of \(y=f(x)\) the graph of \(y=f(x+k)\) is always a horizontal shift

OpenStudy (princesssleelee):

Would you go with C as well? Thats what I am going for!

OpenStudy (misty1212):

lol it is always C when in doubt, charlie out

OpenStudy (princesssleelee):

Given the functions f(x) = 3x2, g(x) = x2 − 4x + 5, and h(x) = –2x2 + 4x + 1, rank them from least to greatest based on their axis of symmetry. f(x), g(x), h(x) f(x), h(x), g(x) g(x), h(x), f(x) g(x), f(x), h(x)

OpenStudy (misty1212):

the axis of symmetry of \(y=ax^2+bx+c\) is always \(x=-\frac{b}{2a}\)

OpenStudy (misty1212):

in the first one \(f(x)=3x^2\) you have \(b=0\) the the axis of symmetry is \(x=0\)

OpenStudy (misty1212):

can you do the rest?

OpenStudy (princesssleelee):

In that case, Im going with C again! lol

OpenStudy (misty1212):

i would not

OpenStudy (princesssleelee):

Nnevermind! they want least to greatest! not greatest to least

OpenStudy (misty1212):

it says "least to greatest" right? pretty sure \(x=0\) is the least

OpenStudy (princesssleelee):

Yes! my misunderstanding

OpenStudy (misty1212):

but you got it right except that you put it backwards

OpenStudy (princesssleelee):

I see that.

OpenStudy (princesssleelee):

i cant figure out what h(X) is...

OpenStudy (misty1212):

you mean this one\[ h(x) = –2x^2 + 4x + 1\]

OpenStudy (misty1212):

in this case \(a=-2,b=4\) the axis of symmetry is \(x=-\frac{4}{2\times (-2)}=1\)

OpenStudy (princesssleelee):

therefore it is second. ok, thank you very much. making it B, overall.

OpenStudy (misty1212):

yes, it is B\[\color\magenta\heartsuit\]

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!