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scoundrel [369]
2 years ago
9

Y=-4(x-2)^2+4 what is the vertex and axis of symmetry of this equation?

Mathematics
1 answer:
alexira [117]2 years ago
6 0

Answer:

Step-by-step explanation:

for vertex

x-2=0 gives x=2

when x=2,y=4

vertex (2,4)

axis of symetry

x-2=0

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Identify the coefficients, terms, like terms, and constants in
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\huge\boxed{\sf{Hello. :D}}

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2 years ago
How to solve this ? I am not sure pls if you know the answer answer it I really neeed it for marks
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Answer:

L.S = R.S ⇒ Proved down

Step-by-step explanation:

Let us revise some rules in trigonometry

  1. sin²α + cos²α = 1
  2. sin2α = 2 sin α cosα
  3. cscα = 1/sinα

To solve the question let us find the simplest form of the right side and the left side, then show that they are equal

∵ L.S = csc2α + 1

→ By using the 3rd rule above

∴ L.S = \frac{1}{sin2\alpha} + 1

→ Change 1 to \frac{sin2\alpha}{sin2\alpha}

∴ L.S = \frac{1}{sin2\alpha} + \frac{sin2\alpha}{sin2\alpha}

→ The denominators are equal, then add the numerators

∴ L.S = \frac{1+sin2\alpha}{sin2\alpha}

∵ R. S = \frac{(sin\alpha+cos\alpha)^{2} }{sin2\alpha}

∵ (sinα + cosα)² = sin²α + 2 sinα cosα + cos²α

∴ (sinα + cosα)² = sin²α + cos²α + 2 sinα cosα

→ By using the 1st rule above, equate sin²α + cos²α by 1

∴ (sinα + cosα)² = 1 + 2 sinα cosα

→ By using the 2nd rule above, equate 2 sinα cosα by sin2α

∴ (sinα + cosα)² = 1 + sin2α

→ Substitute it in the R.S above

∴ R. S = \frac{1+sin2\alpha}{sin2\alpha}

∵ L.S = R.S

∴ csc 2α + 1 = \frac{(sin\alpha+cos\alpha)^{2} }{sin2\alpha}

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