## JacksonJRB one year ago Algebraically determine the domain and range: y=x^2-8x+7

1. JacksonJRB

i already know the domain just need help with the range

2. jim_thompson5910

have you learned about completing the square?

3. JacksonJRB

Yes

4. anonymous

wait you got x you said right?

5. JacksonJRB

$x=y^2-8y+7$

6. JacksonJRB

And yes

7. jim_thompson5910

complete the square to get the equation into vertex form y = a(x-h)^2 + k the range will be $$\Large y \ge k$$ (replace k with a numeric value though) because 'a' is positive which means the parabola opens upward

8. JacksonJRB

I was taught to substitute the x for the y and vice vera

9. anonymous

@jim_thompson5910 can't he use substitution?

10. jim_thompson5910

@JacksonJRB you're thinking of the inverse

11. JacksonJRB

Ah

12. jim_thompson5910

first compute h = -b/(2a) in this case, a = 1, b = -8

13. freckles

there might be an easier way x^2-8x+7 is factorabole y=x^2-8x+7 you can find the average of the 0's so the 0's are a and b the average of a and b is (a+b)/2 so you can find the range by doing: $[f(\frac{a+b}{2}),\infty) \text{ since } a=1>0$

14. jim_thompson5910

once you know the value of h, plug it into y=x^2-8x+7 to find k

15. SolomonZelman

y-value of the vertex (the k of the vertex) is the minimum range. (and there is no maximum limit for range) no range limits, since it is a polynomial

16. jim_thompson5910

can't he use substitution? I'm not sure what you mean @GTA_Hunter35

17. anonymous

he said he has x

18. jim_thompson5910

He said he had the domain. Not just a single value of x.

19. anonymous

nvm it won't work

20. JacksonJRB

^^

21. anonymous

^_^

22. anonymous

23. JacksonJRB

I'm still very confused...

24. jim_thompson5910

were you able to compute h = -b/(2a) ?

25. freckles

do you want to complete the square or find the zeros or use the vertex formula?

26. JacksonJRB

Yes @jim_thompson5910

27. jim_thompson5910

what value did you get for h

28. anonymous

wait wats the vertex formula?

29. JacksonJRB

And either way @freckles

30. JacksonJRB

h=4

31. freckles

well if you find the average of the zeros you will find the x-coordinate of the vertex

32. jim_thompson5910

plug x = 4 into y=x^2-8x+7 and you get y = ??

33. freckles

nevermind you guys got the x-coordinate of the vertex

34. JacksonJRB

-9

35. jim_thompson5910

so k = -9 making the range $$\Large y \ge -9$$ we have a parabola that has the lowest point at (4,-9). All other points will have larger y values.

36. JacksonJRB

Ah. That makes sense now. Thank you so much!

37. jim_thompson5910

no problem

38. freckles

$f(x)=x^2-8x+7 \\ f(x)=(x-7)(x-1) \\ f(x)=0 \text{ when } x=1 \text{ or } x=7 \\ \text{ the average of the zeros is } \frac{7+1}{2}=\frac{8}{2}=4 \\ \text{ so since } a=1>0 \text{ then the range is } [f(4),\infty) \\ f(4)=4^2-8(4)+7=-9 \\ \text{ so the range is } [-9,\infty)$ just wanted to type what I was going for

39. anonymous

well ur right too but u used the function way

40. freckles

used the function way?

41. anonymous

u used f of x

42. anonymous

dat was the only difference

43. freckles

k