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.

- anonymous

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- amistre64

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

- amistre64

k^2 -16 >= 0
k^2 >= 16
k >= +-sqrt(16)
k>= -4, or 4

- amistre64

or rather k>= |4 ro -4|...might be a better scenario :)

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## More answers

- amistre64

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

- amistre64

-k/-8 = k/8
we have some value for "x" which is equal to k/8
can we use that in our equation?

- amistre64

y=-4(k/8)²+k(k/8)-1
y = -4k^2/61 + k^2/8 - 1

- amistre64

61 = 64 in my world lol

- amistre64

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"?

- amistre64

or am I making this to hard on myself....

- amistre64

-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)

- amistre64

k=5.... i beieve

- anonymous

how??

- amistre64

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?

- anonymous

can we do this from the beginning again??

- anonymous

if y=-4x^2 +k -1 dont we plug it into the discriminant?

- anonymous

a=-4 b=k c=-1

- amistre64

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.

- anonymous

i mean from the beggining of the question...so far thats what i know

- anonymous

but the question states tht the value of the function is an integer so does tht mean that the value of k <0

- amistre64

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?

- anonymous

i have no idea what u just said?? lol

- amistre64

an integer is any number between -infinity and +infinity... :)

- anonymous

ooo okay...

- amistre64

the bigest integer value can p[ossibly be a negative value as long as it is the biggest integer we can get :)

- amistre64

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?

- anonymous

okay..

- amistre64

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?

- anonymous

wait when u r doing this r u plugging it into the discriminant??? or just doing it...?

- amistre64

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 :)

- amistre64

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?

- anonymous

yes ur doing product n sum

- amistre64

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

- anonymous

the 2(2) seems reasonable..cuz u need greates common factor right??

- amistre64

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 :)

- anonymous

so if we r looking for the greatest common value then we use the numbers 1 n 4...

- anonymous

i mean the greatest value for k

- amistre64

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?

- anonymous

if y =0??

- amistre64

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?

- anonymous

where did u get -b/2a?

- amistre64

that is from the quadratic formula as well.... do you know the quadratic formula?

- anonymous

yes

- anonymous

i didnt tht it gives u the highest point at the vertex..

- anonymous

i dint know

- amistre64

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?

- anonymous

yes

- amistre64

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?

- amistre64

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?

- amistre64

-sqrt(b^2-4ac).......-b.......+sqrt(b^2-4ac)
------------ --- -------------
2a 2a 2a
I hope this displays correctly lol

- amistre64

does that make sense?

- anonymous

yupp

- anonymous

i think i got it thanks i think i can handle it from here

- anonymous

thanks for ur help

- amistre64

ok :) but if this might clarify things a bit

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