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Bookworm14
 one year ago
I am so confused, please help me! (I know HOW to write the equations, but just not when the info is given to me like this :'( )
Bookworm14
 one year ago
I am so confused, please help me! (I know HOW to write the equations, but just not when the info is given to me like this :'( )

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Vertex form: \(y=a(xh)^2+k\), where (h, k) is the vertex. Looking at the info they gave you, can you tell what either coordinate for the vertex is?

Bookworm14
 one year ago
Best ResponseYou've already chosen the best response.0(2, ? ) im sorry this is a new concept for me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no problem. The axis of symmetry gives the xcoordinate of the vertex. The maximum (or minimum when you get a problem like that) height gives the ycoordinate. So the vertex is (2, 2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Put that into the equation to get \(y = a(x 2)^2+2\)

Bookworm14
 one year ago
Best ResponseYou've already chosen the best response.0because it is maximum, wont that make it a negative cooridinate or does that not apply here?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no. The negative doesn't come into play here. Just take the number as they give it to you. The negative comes in as a check. Since it's a maximum \(a\) is supposed to be negative. But you have to plug in the point (3, 0) to find it.

Bookworm14
 one year ago
Best ResponseYou've already chosen the best response.0okay, i dont know how to plug in the points in runs through, im not really sure about that part

Bookworm14
 one year ago
Best ResponseYou've already chosen the best response.0and yes, i was looking at my chart wrong, i see now that the a is whats supposed to be negative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You have \(y=a(x2)^2+2\) You have to plug in 3 for x and 0 for y, then solve for a. \[0=a(32)^2+2\]

Bookworm14
 one year ago
Best ResponseYou've already chosen the best response.0a is our maximum value which is 2, but its negative? so 2?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a is not the maximum value. The ycoordinate of the vertex is the maximum value.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443061442308:dw

Bookworm14
 one year ago
Best ResponseYou've already chosen the best response.0so the final answer is \[y=2(x2)^22\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no, it's not 2 \(0=a(3−2)^2+2\) \(0=a + 2\) \(2=a\)

Bookworm14
 one year ago
Best ResponseYou've already chosen the best response.0oh ok, i see what you did there. do you think you could help me with one more? (its similar, but worded differently, i will definitely contribute more this time, because i think i understand slightly better now...)
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