## miah 2 years ago ohkay so I REALLY need help.

1. miah

Using the direction, size and vertex of your parabola, think of a quadratic equation in general form, y = a(x - h)2 + k, that would closely match your parabola. Remember (h, k) represents the vertex of the parabola. Enter that equation in the Input bar.

2. miah

3. miah

@amistre64

4. miah

@KyrstanMiller24

5. KyrstanMiller24

what grade of math are you in?

6. miah

12th

7. miah

@AirPotta

8. miah

@Mertsj

9. miah

10. miah

Im doing a project thing. and I was going thru the steps then I go to that one and I stopped cause I didnt know what to do

11. miah

@Mertsj

12. miah

its the pic in the attatchment

13. miah

14. miah

Select the point tool , then place a point at the vertex of the parabola in your image. Using the direction, size and vertex of your parabola, think of a quadratic equation in general form, y = a(x - h)2 + k, that would closely match your parabola. Remember (h, k) represents the vertex of the parabola. Enter that equation in the Input bar. See how closely the graphed parabola matches your image parabola. You may need to try different values of a to make the parabola fit. To change the values of the equation, right-click on the parabola, choose Object Properties…, and change the equation under the Basic tab. Close the window. Once the graphed parabola matches as close as possible to the image, make note of the quadratic equation, both in general and standard form (y = ax2 + bx + c). When completed, the graphed parabola should lie on top of the parabola in your image. If necessary, right-click on the image and go to the Color tab to make the image more transparent so the graphed parabola can be seen over the image.

15. miah

16. amistre64

well, a parabola can be defined by its x intercepts, called zeroes or roots. For what x values does your parabola cross the x axis at?

17. miah

2,8 ?

18. amistre64

i cant tell that part, pictures not that detailed for me to see, |dw:1365103947488:dw|

19. amistre64

are those x values 2 and 8 on the picture you posted?

20. amistre64

you have 2 arches ... which one are you using?

21. amistre64

if its the front one, im guessing its a 2.5 and a 12.5 but im not sure what you are looking at

22. miah

the back 1

23. amistre64

...why? front one seems simpler; 3 and 13, while the back one might be iffier

24. miah

ohkay front

25. amistre64

lol, do you agree that 3 and 13 look about right for those x values?

26. miah

yes

27. amistre64

we will also want to know the value of x that is in the center of those, that will be our "line of symmetry".

28. amistre64

what value is the center between 3 and 13?

29. miah

uhm ?

30. amistre64

if you cant find the middle between 3 and 13, then you are definantly not prepared to do quadratics ....

31. amistre64

we want some value such that 3+n = 13-n can you solve for n?