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cos00155079
Group Title
can someone check my answer? find the vertex, just not sure about the negatives
 one year ago
 one year ago
cos00155079 Group Title
can someone check my answer? find the vertex, just not sure about the negatives
 one year ago
 one year ago

This Question is Closed

slaaibak Group TitleBest ResponseYou've already chosen the best response.1
usually, to check which value is the xcoordinate of the vertex, just check what value of x will make the bracket zero. in this case it's (4 + 4)^2  33 so xcoordinate = 4
 one year ago

slaaibak Group TitleBest ResponseYou've already chosen the best response.1
oh, sorry, I never saw the equation was x = y^2 + 8y  17 it's gonna be a bit different now, I'll post the solution now
 one year ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
the vertex is at (h, k) and the general form is y=(x h )^2 + k but since the parabola x = y^2 + 8y  17 dw:1354919627767:dw
 one year ago

cos00155079 Group TitleBest ResponseYou've already chosen the best response.0
yeah, well I have all the steps I'm just confused with the negatives
 one year ago

slaaibak Group TitleBest ResponseYou've already chosen the best response.1
yeah, what cambpell said. so it will be (33,4)
 one year ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
no the parabola is concave right.... its x = y^2 + 8x  17
 one year ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
the methodology is correct... its problem was identifying the type of parabola you have...
 one year ago

cos00155079 Group TitleBest ResponseYou've already chosen the best response.0
let me show u my steps ill upload a pic
 one year ago

cos00155079 Group TitleBest ResponseYou've already chosen the best response.0
I only need the vertex, but i dont have to graph it... wait a sec please
 one year ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
I've seen your method.... but the parabola is not the traditional concave up parabola...
 one year ago

cos00155079 Group TitleBest ResponseYou've already chosen the best response.0
so it would be (4,33) ?
 one year ago

cos00155079 Group TitleBest ResponseYou've already chosen the best response.0
yes, @slaaibak but its (h,k) so (4,33) right?
 one year ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
the methodolgy is correct... here is the graph attached the error is in the general form that you have used. for parabola that are in the form x = ay^2 + by + c the vertex form is x = (y  k)^2 + h
 one year ago

cos00155079 Group TitleBest ResponseYou've already chosen the best response.0
ok got it! thanks a lot :)
 one year ago
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