cos00155079 2 years ago can someone check my answer? find the vertex, just not sure about the negatives

1. cos00155079

2. slaaibak

correct

3. slaaibak

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

4. slaaibak

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

5. campbell_st

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|

6. cos00155079

yeah, well I have all the steps I'm just confused with the negatives

7. slaaibak

yeah, what cambpell said. so it will be (-33,-4)

8. campbell_st

no the parabola is concave right.... its x = y^2 + 8x - 17

9. campbell_st

the methodology is correct... its problem was identifying the type of parabola you have...

10. cos00155079

let me show u my steps ill upload a pic

11. cos00155079

I only need the vertex, but i dont have to graph it... wait a sec please

12. campbell_st

I've seen your method.... but the parabola is not the traditional concave up parabola...

13. cos00155079

14. cos00155079

so it would be (-4,-33) ?

15. cos00155079

yes, @slaaibak but its (h,k) so (-4,-33) right?

16. campbell_st

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

17. cos00155079

ok got it! thanks a lot :)