Asked by Randy
g(x) = x^4-8x^3-16x^2+128x
the graph goes through the x through
(-4,0) , (0,0) (4,0), (8,0)
horizontal axis has a scale of 2,4,6,8,10 (each tick increases by 2)
Y-axis value that is the highest is (-4,30) and (8,30)
Assume
g(x) = k(x-p)(x-q)(x-r)(x-s) p<q<r<s
I need help finding K and P please.
k = ?
p = ?
q = 0
r = 4
s = 8
IF you need more information let me know please.
the graph goes through the x through
(-4,0) , (0,0) (4,0), (8,0)
horizontal axis has a scale of 2,4,6,8,10 (each tick increases by 2)
Y-axis value that is the highest is (-4,30) and (8,30)
Assume
g(x) = k(x-p)(x-q)(x-r)(x-s) p<q<r<s
I need help finding K and P please.
k = ?
p = ?
q = 0
r = 4
s = 8
IF you need more information let me know please.
Answers
Answered by
bobpursley
you have the zeroes.
the only thinking is to put them in order.
equation will be
g(x)=k(x+4)(x)((x-4)(x-8)
changing order to make s>r>q>p
g(x)=k(x+4)(x)(x-4)(x-8)
so that give you pqrs Now on k, I don't understand. You say it has zeros at 8, and -4, but then you staed it is Highest at those points...I dunno at this point on k.
the only thinking is to put them in order.
equation will be
g(x)=k(x+4)(x)((x-4)(x-8)
changing order to make s>r>q>p
g(x)=k(x+4)(x)(x-4)(x-8)
so that give you pqrs Now on k, I don't understand. You say it has zeros at 8, and -4, but then you staed it is Highest at those points...I dunno at this point on k.
Answered by
Randy
never mind forget how i put (-4,30) and and the (8,30) bit... Forget i said that. I just meant to say the highest endpoints you can see before the graph cuts off is those points
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