Ask your own question, for FREE!
Mathematics 11 Online
OpenStudy (jazzyfa30):

Write each function in vertex form, then match each quadratic function with it vertex axis of symmetry and direction of opening. y= \frac{1}{3} (x-1)^2+2

OpenStudy (rsadhvika):

\(y= \frac{1}{3} (x-1)2+2 \)

OpenStudy (jhannybean):

\[\large y=\cfrac12 (x-1)^2 +2\]

OpenStudy (rsadhvika):

compare it with \(y = a(x-h)^2 + k\)

OpenStudy (rsadhvika):

vertex = \(( h, k)\)

OpenStudy (rsadhvika):

since \(\large a = \frac{1}{3}\) is positive, the function will rise and will open UP

OpenStudy (jhannybean):

\[\large y=\color{green}a(x-\color{blue}h)^2+\color{blue}k\]\[\large y=\color{green}{\cfrac13}(x-\color{blue}1)^2+\color{blue}2\]

OpenStudy (jazzyfa30):

ok um what do u mean by vertex

OpenStudy (jhannybean):

\[\large \text{vertex} = \color{blue}{(h,k)}\]

OpenStudy (jazzyfa30):

so (1,2)

OpenStudy (jhannybean):

Yep, Vertex = axis of symmetry on a parabola.

OpenStudy (jazzyfa30):

so the answer (1,2)x=1 and up

OpenStudy (jhannybean):

Yep.

OpenStudy (jazzyfa30):

van you help me with a few more

OpenStudy (jazzyfa30):

*Can

OpenStudy (jhannybean):

http://fooplot.com/plot/acn192saks your graph.

OpenStudy (jazzyfa30):

idk how to use that

OpenStudy (jhannybean):

Just click the link and look at the graph posted there. It shows you your function and the axis of symmetry.

OpenStudy (jazzyfa30):

ok

OpenStudy (jhannybean):

Alrighty.

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!