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JacksonJRB

  • one year ago

Algebraically determine the domain and range: y=x^2-8x+7

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  1. JacksonJRB
    • one year ago
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    i already know the domain just need help with the range

  2. jim_thompson5910
    • one year ago
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    have you learned about completing the square?

  3. JacksonJRB
    • one year ago
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    Yes

  4. anonymous
    • one year ago
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    wait you got x you said right?

  5. JacksonJRB
    • one year ago
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    \[x=y^2-8y+7\]

  6. JacksonJRB
    • one year ago
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    And yes

  7. jim_thompson5910
    • one year ago
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    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
    • one year ago
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    I was taught to substitute the x for the y and vice vera

  9. anonymous
    • one year ago
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    @jim_thompson5910 can't he use substitution?

  10. jim_thompson5910
    • one year ago
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    @JacksonJRB you're thinking of the inverse

  11. JacksonJRB
    • one year ago
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    Ah

  12. jim_thompson5910
    • one year ago
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    first compute h = -b/(2a) in this case, a = 1, b = -8

  13. freckles
    • one year ago
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    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
    • one year ago
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    once you know the value of h, plug it into y=x^2-8x+7 to find k

  15. SolomonZelman
    • one year ago
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    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
    • one year ago
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    `can't he use substitution?` I'm not sure what you mean @GTA_Hunter35

  17. anonymous
    • one year ago
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    he said he has x

  18. jim_thompson5910
    • one year ago
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    He said he had the domain. Not just a single value of x.

  19. anonymous
    • one year ago
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    nvm it won't work

  20. JacksonJRB
    • one year ago
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    ^^

  21. anonymous
    • one year ago
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    ^_^

  22. anonymous
    • one year ago
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    its your call @jim_thompson5910

  23. JacksonJRB
    • one year ago
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    I'm still very confused...

  24. jim_thompson5910
    • one year ago
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    were you able to compute h = -b/(2a) ?

  25. freckles
    • one year ago
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    do you want to complete the square or find the zeros or use the vertex formula?

  26. JacksonJRB
    • one year ago
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    Yes @jim_thompson5910

  27. jim_thompson5910
    • one year ago
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    what value did you get for h

  28. anonymous
    • one year ago
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    wait wats the vertex formula?

  29. JacksonJRB
    • one year ago
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    And either way @freckles

  30. JacksonJRB
    • one year ago
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    h=4

  31. freckles
    • one year ago
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    well if you find the average of the zeros you will find the x-coordinate of the vertex

  32. jim_thompson5910
    • one year ago
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    plug x = 4 into y=x^2-8x+7 and you get y = ??

  33. freckles
    • one year ago
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    nevermind you guys got the x-coordinate of the vertex

  34. JacksonJRB
    • one year ago
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    -9

  35. jim_thompson5910
    • one year ago
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    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
    • one year ago
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    Ah. That makes sense now. Thank you so much!

  37. jim_thompson5910
    • one year ago
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    no problem

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

  39. anonymous
    • one year ago
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    well ur right too but u used the function way

  40. freckles
    • one year ago
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    used the function way?

  41. anonymous
    • one year ago
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    u used f of x

  42. anonymous
    • one year ago
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    dat was the only difference

  43. freckles
    • one year ago
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    k

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