## livelaughlilz 2 years ago what is the minimum value of 7^(x^2-4x+7)

1. saifoo.khan

$\Large 7^{x^2-4x+7}$Like this?

2. livelaughlilz

yes

3. saifoo.khan

@mathmate @UnkleRhaukus

4. saifoo.khan

Minimum value should be there where dy/dx = 0.

5. livelaughlilz

what do you mean by the "d"?

6. saifoo.khan

In other words minimum value occurs where the derivative of the function is equal to zero.

7. livelaughlilz

isn't derivatives calculus? i haven't learned that yet and i don't think we're expected to know it to solve this problem

8. saifoo.khan
9. livelaughlilz

haha i was thinking about it and wouldn't it be one over infinity? because if x^2-4x+7 was negative infinity than 7^(x^2-4x+7) would be 1/infinity

10. mathmate

Hint: 7^x is an increasing function. So if we minimize x, 7^x is also a minimum.

11. mathmate

x^2-4x+7 cannot be at -inf. It is a parabola concave upwards.!

12. saifoo.khan

And are we going to try that by trial and error? @mathmate

13. KingGeorge

No, we need to simply find the vertex of the parabola to find the minimum value of the parabola.

14. saifoo.khan

Howw??

15. mathmate

Good point!] @livelaughlilz can you take it from here?

16. mathmate

Completing the squares will do the job.

17. livelaughlilz

yeah the vertex of the parabola is (2,3)

18. livelaughlilz

(x-2)^2=y-3

19. mathmate

x^2-4x+7=(x-2)^2+3 so the vertex is at (2,3)

20. saifoo.khan
21. mathmate

Good job livelaughlilz! Speed+accuracy!

22. mathmate

I like his movies! :)

23. livelaughlilz

oh so the answer is 343. the lowest value of x is 2, and if you plug it in you get 7^3=343

24. saifoo.khan

Yes^ @livelaughlilz

25. mathmate

7^2 = ???

26. saifoo.khan

@mathmate : haha, nice. My brother just love his movies. i don't. :/

27. mathmate

Nevermind, I was out of my mind. Yes, 7^3=343

28. saifoo.khan

You were thinking about Jackie Chan. :D

29. KingGeorge

As a sidenote @mathmate, you can find the x-coordinate of the vertex using the formula $v_x=\frac{-b}{2a}$

30. mathmate

Thank you for the defence, even though it was a lame one! Good thinking! :)

31. saifoo.khan

I simply love this method. Satellite73 taught me this. -b/2a

32. livelaughlilz

to add onto @KingGeorge the y-coordinate of the vertex is -[(b^2-4ac)/(4a)]

33. mathmate

Thank you @KingGeorge, my memory is limited and sometimes defective. But thanks for the shortcut anyway!

34. livelaughlilz

you can figure all that out by completing the square of y=ax^2+bx+c

35. livelaughlilz

and making y=ax^2+bx+c into vertex form

36. mathmate

Actually I know the -b/2a part, that's how I get to do the completing the square. It's just I never managed to memorize the -[(b^2-4ac)/(4a)] part.

37. livelaughlilz

well b^2-4ac is also the discriminant...

38. saifoo.khan

@livelaughlilz is a boss at this.

39. mathmate

Perhaps from this time on I will remember. Practice makes perfect! :)

40. mathmate

Actually what I do is finding it by c-(b/2a)^2, which is exactly what we do when we complete the square.

41. livelaughlilz

oh

42. livelaughlilz

i'm going to close this question now. thanks everyone!

43. mathmate

See you all!