At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
lets try this from the quad formula perspective... sqrt(k^2 -4(-1)(-4)) would have to be equal to or greater than 0
k^2 -16 >= 0 k^2 >= 16 k >= +-sqrt(16) k>= -4, or 4
or rather k>= |4 ro -4|...might be a better scenario :)
since -4x^2 gives us an upsode down "U" for a graph, we can determine the "largest value"... at the vertex: -k/2(-4)
-k/-8 = k/8 we have some value for "x" which is equal to k/8 can we use that in our equation?
y=-4(k/8)²+k(k/8)-1 y = -4k^2/61 + k^2/8 - 1
61 = 64 in my world lol
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"?
or am I making this to hard on myself....
-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)
k=5.... i beieve
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?
can we do this from the beginning again??
if y=-4x^2 +k -1 dont we plug it into the discriminant?
a=-4 b=k c=-1
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.
i mean from the beggining of the question...so far thats what i know
but the question states tht the value of the function is an integer so does tht mean that the value of k <0
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?
i have no idea what u just said?? lol
an integer is any number between -infinity and +infinity... :)
the bigest integer value can p[ossibly be a negative value as long as it is the biggest integer we can get :)
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?
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?
wait when u r doing this r u plugging it into the discriminant??? or just doing it...?
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 :)
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?
yes ur doing product n sum
correct: our options are: 4(1) = 4; 4+1 = 5 2(2) = 4; 2+2 = 4 which number is greater.... 4 or 5?
the 2(2) seems reasonable..cuz u need greates common factor right??
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 :)
so if we r looking for the greatest common value then we use the numbers 1 n 4...
i mean the greatest value for k
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?
if y =0??
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?
where did u get -b/2a?
that is from the quadratic formula as well.... do you know the quadratic formula?
i didnt tht it gives u the highest point at the vertex..
i dint know
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?
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?
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?
-sqrt(b^2-4ac).......-b.......+sqrt(b^2-4ac) ------------ --- ------------- 2a 2a 2a I hope this displays correctly lol
does that make sense?
i think i got it thanks i think i can handle it from here
thanks for ur help