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anonymous

  • 5 years ago

can someone help me!!??? please i would really appreciate it. question: If y=-4x²+kx-1determine the value(s) of k for which the maximum value of the function is an integer. Explain your reasoning using pictures, numbers, and words.

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  1. amistre64
    • 5 years ago
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    lets try this from the quad formula perspective... sqrt(k^2 -4(-1)(-4)) would have to be equal to or greater than 0

  2. amistre64
    • 5 years ago
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    k^2 -16 >= 0 k^2 >= 16 k >= +-sqrt(16) k>= -4, or 4

  3. amistre64
    • 5 years ago
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    or rather k>= |4 ro -4|...might be a better scenario :)

  4. amistre64
    • 5 years ago
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    since -4x^2 gives us an upsode down "U" for a graph, we can determine the "largest value"... at the vertex: -k/2(-4)

  5. amistre64
    • 5 years ago
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    -k/-8 = k/8 we have some value for "x" which is equal to k/8 can we use that in our equation?

  6. amistre64
    • 5 years ago
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    y=-4(k/8)²+k(k/8)-1 y = -4k^2/61 + k^2/8 - 1

  7. amistre64
    • 5 years ago
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    61 = 64 in my world lol

  8. amistre64
    • 5 years ago
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    y = -4k^2/64 + 8k^2/64 -1 y = 4k^2/64 - 1 y = 4k^2/64 - 1 y = k^2/16 - 1 can we do anything with this "value"?

  9. amistre64
    • 5 years ago
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    or am I making this to hard on myself....

  10. amistre64
    • 5 years ago
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    -4x^2 +kx -1 = 0 factors to.. 4x^2 -kx +1 = 0 1,4 = 4 2,2 = 4 (x- 1)(4x- 1) or (2x-1) (2x-1)

  11. amistre64
    • 5 years ago
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    k=5.... i beieve

  12. anonymous
    • 5 years ago
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    how??

  13. amistre64
    • 5 years ago
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    the values we get for the product of "ac" are: 4(1) = 4 1+4 = 5, k=5 in this instance 2+2 = 4, k=4 in this instance. but I might have a "sign" out of place so I would have to recheck it all :) but does that make sense?

  14. anonymous
    • 5 years ago
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    can we do this from the beginning again??

  15. anonymous
    • 5 years ago
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    if y=-4x^2 +k -1 dont we plug it into the discriminant?

  16. anonymous
    • 5 years ago
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    a=-4 b=k c=-1

  17. amistre64
    • 5 years ago
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    which beginning :) I was trying to determine a good manner in which to appraoch the problem...the discriminate there was not an "easy" way.... that I could see.

  18. anonymous
    • 5 years ago
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    i mean from the beggining of the question...so far thats what i know

  19. anonymous
    • 5 years ago
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    but the question states tht the value of the function is an integer so does tht mean that the value of k <0

  20. amistre64
    • 5 years ago
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    its easier to approach from the "meaing" that is given to a,b, and c the "c" term is a product of 2 numbers; the "b" term is a sum of 2 numbers. for "b" to have the greatest value, the sum of the factors of "c" would have to combine to the biggest value right?

  21. anonymous
    • 5 years ago
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    i have no idea what u just said?? lol

  22. amistre64
    • 5 years ago
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    an integer is any number between -infinity and +infinity... :)

  23. anonymous
    • 5 years ago
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    ooo okay...

  24. amistre64
    • 5 years ago
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    the bigest integer value can p[ossibly be a negative value as long as it is the biggest integer we can get :)

  25. amistre64
    • 5 years ago
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    but lets try this approach... first we factor out that -1 to make our lives easier: y = -1(4x^2 -kx +1) we good here?

  26. anonymous
    • 5 years ago
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    okay..

  27. amistre64
    • 5 years ago
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    we need to gather a "pool" of options for the middle term. that "pool" comes from the number we get when we multiply "a" and "c". 4 times 1 = 4. we good so far?

  28. anonymous
    • 5 years ago
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    wait when u r doing this r u plugging it into the discriminant??? or just doing it...?

  29. amistre64
    • 5 years ago
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    just doing it.... the discrimant is only valueable if you "know" what k is.... so lets continue on this path and see where it leads :)

  30. amistre64
    • 5 years ago
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    we need two number: call them "m" and "n", that multiply together to get "4" and ADD together to get our value for "k". does that make sense?

  31. anonymous
    • 5 years ago
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    yes ur doing product n sum

  32. amistre64
    • 5 years ago
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    correct: our options are: 4(1) = 4; 4+1 = 5 2(2) = 4; 2+2 = 4 which number is greater.... 4 or 5?

  33. anonymous
    • 5 years ago
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    the 2(2) seems reasonable..cuz u need greates common factor right??

  34. amistre64
    • 5 years ago
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    we couldnt care less about "common" factors here :) what we care about is the greatest value that we can get for "k". the value for "k" is obtained by "adding together" the factors that make up "4". That is our only concern :)

  35. anonymous
    • 5 years ago
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    so if we r looking for the greatest common value then we use the numbers 1 n 4...

  36. anonymous
    • 5 years ago
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    i mean the greatest value for k

  37. amistre64
    • 5 years ago
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    thats right.... so lets see if 5 is a good answer by plugging it into our original formula and seeing what we get:) y = -4x^2 +(5)x - 1 how would we TEST this situation?

  38. anonymous
    • 5 years ago
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    if y =0??

  39. amistre64
    • 5 years ago
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    our "highest point" is at our vertex: -b/2a -5/2(-4) = -5/-8 = 5/8 5/8...our "vertex is at 5/8. what is the value of y at x = 5/8?

  40. anonymous
    • 5 years ago
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    where did u get -b/2a?

  41. amistre64
    • 5 years ago
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    that is from the quadratic formula as well.... do you know the quadratic formula?

  42. anonymous
    • 5 years ago
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    yes

  43. anonymous
    • 5 years ago
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    i didnt tht it gives u the highest point at the vertex..

  44. anonymous
    • 5 years ago
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    i dint know

  45. amistre64
    • 5 years ago
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    the quad formula says that the roots of a quadratic expression are equal distances from a certain value of x.... right? -b sqrt(b^2-4ac) --- + or - ------------- 2a 2a right?

  46. anonymous
    • 5 years ago
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    yes

  47. amistre64
    • 5 years ago
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    so the cernter, or vertex, of our equation lies between these 2 points..... the x value of our "vertex" then must be -b/2a.... am I right?

  48. amistre64
    • 5 years ago
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    if you move 5 feet to your left, or 5 feet to your right.... where is the middle of that?.... exactly where you were standing to begin with right?

  49. amistre64
    • 5 years ago
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    -sqrt(b^2-4ac).......-b.......+sqrt(b^2-4ac) ------------ --- ------------- 2a 2a 2a I hope this displays correctly lol

  50. amistre64
    • 5 years ago
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    does that make sense?

  51. anonymous
    • 5 years ago
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    yupp

  52. anonymous
    • 5 years ago
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    i think i got it thanks i think i can handle it from here

  53. anonymous
    • 5 years ago
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    thanks for ur help

  54. amistre64
    • 5 years ago
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    ok :) but if this might clarify things a bit

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