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KATRIN_1 [288]
2 years ago
9

Could i have some super super quick help? :(( (part iv is not necessary)

Mathematics
2 answers:
ladessa [460]2 years ago
7 0

i)

<u>Factor using completing square method</u>

\dashrightarrow \sf y = -x^2+16x-64

\dashrightarrow \sf y = -(x^2-16x+64)

\dashrightarrow \sf y = -(x^2-8x-8x+64)

\dashrightarrow \sf y = -(x(x-8)-8(x-8))

\dashrightarrow \sf y = -((x-8)(x-8))

ii)

<u>Find zeros of a function, f(x) = 0</u>

\dashrightarrow \sf -((x-8)(x-8))=0

\dashrightarrow \sf (x-8)=0,  \  (x-8)=0

\dashrightarrow \sf x = 8

iii)

<u>In order to find vertex use the formulae : x =  -b/2a</u>

\dashrightarrow \sf x = \dfrac{-(16)}{2(-1)}

\dashrightarrow \sf x = 8

Then find y:

\dashrightarrow \sf y = -(8)^2+16(8)-64

\dashrightarrow \sf y = 0

coordinates: (8, 0)

iv) Sketched Below:

Bezzdna [24]2 years ago
5 0
  • y=-x²+16x-64

#1

  • y=-x²+8x+8x-64
  • y=-x(x+8)+8(x+8)
  • y=(x+8)(-x+8)
  • y=-(x-8)(x-8)

#2

Zeros are the x intercepts

Here they are

  • 8,8

#3

  • y=-x²+16x-64
  • y=-[x²-16x+64]
  • y=-(x-8)²+0

Vertex form of parabola:-y=a(x-h)²+k

  • Vertex=(h,k)=(8,0)

#4

Attached

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The taylor series for the f(x)=8/x centered at the given value of a=-4 is -2+2(x+4)/1!-24/16 (x+4)^{2}/2!+...........

Given a function f(x)=9/x,a=-4.

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Hence the taylor series for the f(x)=8/x centered at the given value of a=-4 is -2+2(x+4)/1!-24/16 (x+4)^{2}/2!+...........

Learn more about taylor series at brainly.com/question/23334489

#SPJ4

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