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mafiozo
4 days ago
9

HELP!!! Determine if the graph is symmetric about the x-axis, the y-axis, or the origin. r = 4 cos 3θ

Mathematics
2 answers:
Zina [9.1K]4 days ago
6 0

Response:

                The graph exhibits symmetry concerning the x-axis.

Step-by-step outline:

  • Symmetry about the polar axis (x-axis) is observed when replacing (r,θ) with (r,-θ), resulting in an equation identical to the original.
  • For the y-axis, replacing (r,θ) with (-r,-θ) gives the same equation as the original.
  • For symmetry around the pole (origin), a switch of (r,θ) to (-r,θ) must yield the same original equation.

We have the following equation:

           r=4\cos 3\theta

  • When substituting (r,θ) with (r,-θ), the result is:

r=4\cos 3(-\theta)\\\\i.e.\\\\r=4\cos (-3\theta)\\\\i.e.\\\\r=4\cos 3\theta

(  Since, we are aware that:

\cos (-\tehta)=\cos \theta  )

Therefore, the graph is symmetric with respect to the x-axis.

  • For when we substitute (r,θ) with (-r,-θ)

-r=4\cos 3(-\theta)\\\\i.e.\\\\-r=4\cos (-3\theta)\\\\i.e.\\\\-r=4\cos 3\theta\\\\i.e.\\\\r=-4\cos 3\theta

In this case, we see that the graphs are not equivalent upon performing this substitution.

Consequently, the graph lacks symmetry concerning the y-axis.

  • Upon substituting (r,θ) with (-r,θ), we find:

-r=4\cos 3\theta\\\\i.e.\\\\r=-4\cos 3\theta

Again, it's clear that the resulting graphs do not equate.

Thus, the graph does not exhibit symmetry with respect to the pole (origin).[TAG_63]]

Leona [9.2K]4 days ago
3 0
To evaluate y-symmetry, let’s define r = 5 cos 3θ and set <span>θ=π−θ</span> which implies that <span><span><span>r=5cos3θ</span><span>θ=(π−θ)</span><span>r=5cos(3(π−θ))⟹r=5cos(3π−3θ)</span></span></span>
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<pthus our="" equation="" is:="">

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<pexamining the="" table:="">

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<pthus:>

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<pthe symmetry="" of="" the="" quadratic="" function="" implies="" that="" axis="" lies="" between="" and="" located="" at="" x="3s.&lt;/p"><pin a="" standard="" quadratic="" function:="">

a*x^2 + b*x + c

The symmetry line is given by:

x = -b/2a

<pin this="" instance:="">

b = v0

a = a

<ptherefore we="" derive:="">

3s  = -v0/(2*a)

v0 = -3s*(2a)

<phaving gathered="" all="" necessary="" data="" for="" our="" equation="" we="" can="" express="" it="" as:="">

h(x) = a*x^2 - 6s*a*x + 6ft

<pnext focusing="" on="" just="" one="" variable="" we="" know="" that="" at="" x="2s," h=""><pso:>

h(2s) = 22ft = a*(2s)^2 - 6s*a*2s + 6ft

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</pthus></p></pso:></pnext></phaving></ptherefore></pin></pin></pthe></pthus:></pexamining></pthus>
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