Here's the question you clicked on:
cos00155079
can someone check my answer? find the vertex, just not sure about the negatives
usually, to check which value is the x-coordinate of the vertex, just check what value of x will make the bracket zero. in this case it's (-4 + 4)^2 - 33 so x-coordinate = -4
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
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|
yeah, well I have all the steps I'm just confused with the negatives
yeah, what cambpell said. so it will be (-33,-4)
no the parabola is concave right.... its x = y^2 + 8x - 17
the methodology is correct... its problem was identifying the type of parabola you have...
let me show u my steps ill upload a pic
I only need the vertex, but i dont have to graph it... wait a sec please
I've seen your method.... but the parabola is not the traditional concave up parabola...
so it would be (-4,-33) ?
yes, @slaaibak but its (h,k) so (-4,-33) right?
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
ok got it! thanks a lot :)