## Janu16 one year ago The following graph of f(x)=x^2 has been shifted into the form f(x)=(x-h)^2+k

1. Janu16

This

2. Nnesha

in that equation (h,k) is the vertex point

3. Janu16

ya so im lookinf for k

4. Nnesha

y-coordinate of the vertex would be the k |dw:1443886284134:dw| so look at the graph find what's x and y

5. Janu16

2?

6. Nnesha

it should be (x,y) pair

7. Janu16

2,2? not sure

8. Nnesha

you can draw a vertical line like this |dw:1443886489382:dw| when x = 2 y =-1 so (2,-1)

9. Janu16

ohhyaa

10. Nnesha

so what would be the vertex in the given graph ?

11. Janu16

2?

12. Janu16

wait

13. Janu16

-1?

14. Nnesha

that was an example.

15. Janu16

ohh so the vertex is -2? cause that dot is on negative line

16. Nnesha

$\huge\rm f(x)=a(x-h)^2+k$ vertex form of parabola where (h,k) is the vertex point shape of the parabola depends on the sign of leading coefficient if a is positive then it will opens up|dw:1443886757241:dw| and when a is negative parabola opens down

17. Nnesha

|dw:1443886878265:dw| that's the vertex

18. Janu16

oookk

19. Nnesha

or in other words h = horizontal shift k= vertical shift to find k you should know how many units the graph going up or down from the origin

20. Nnesha

is*

21. anonymous

why cant i see the drawing

22. Janu16

so if you know that you would get the k?

23. Janu16

you cant see it if you are on the phone app.

24. Nnesha

yes. so what's the k ?

25. Nnesha

i'll give u an exxample |dw:1443887248429:dw|

26. Janu16

-2?

27. Janu16

thats what i got

28. Nnesha

|dw:1443887515749:dw| the point is in first quadrant where x and y both are positive

29. Janu16

ohh so positive 2

30. Nnesha

yes x =2 but we also need y -coordinate so when x = 2 y = what ?

31. Janu16

3

32. Nnesha

right (h,k)= (2,3)

33. Janu16

so h is 2 aznd k is 3?

34. Janu16

thx so much nnesha!!

35. Nnesha

np :=) and yes right k=3