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algol [13]
3 years ago
7

Which function forms a geometric sequence when x = 1, 2, 3, . . . ?

Mathematics
1 answer:
irinina [24]3 years ago
6 0

Answer:

a) f(x)=3(4)^{x}

Step-by-step explanation:

The correct options are:

a) f(x)=3(4)^{x}

b) f(x)=3(x)^{2}

c) f(x)=2x+4

d) f(x)=x+(2)^{4}

We have to identify which of these options will form a geometric sequence. A geometric sequence is defined as the sequence in which the ratio of two consecutive terms remain the same. This ratio is known as common ratio of the sequence. General term of a geometric sequence is defined as:

a_{n}=a_{1}(r)^{n-1}

Here,

a_{n} is the nth term of the sequence. Replacing n by different values will give us the term of the sequence at that values.

The option resembling the general term of the geometric sequence is option a, with only difference that in place on "n", the variable x is used in the option. So option a should be our answer. Lets verify this.

f(x)=3(4)^{x}

For x = 1, the value will be:

f(1) = 12

For x = 2, the value will be:

f(2) = 48

For x =3, the value will be:

f(3) = 192

The ratio of these terms is:

Ratio of f(2) and f(1) = 4

Ratio of f(3) and f(2) = 4

Therefore, the function given in option a represents a geometric sequence.

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F(x)=2x^2-x-10 part a: what are the X intercepts of the graph of f(x)? Show your work
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A. x intercepts are where the graph hits the x axis or where f(x)=0
0=2x^2-x-10
solve
hmm
we can use the ac method
multiply 2 and -10 to get -20
what 2 numbers muliply to get -20 and add to get -1 (the coefint of the x term)
-5 and 4
split the midd into that
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group
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factor
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undistribute
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set each to 0
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0=-2

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B. ok, so for f(x)=ax^2+bx+c
if a>0, then the parabola opens up and the vertex is a minimum
if a<0 then the parabola opens down and the vertex is a max

f(x)=2x^2-x-10
2>0
opoens up
vertex is minimum

ok, the vertex

the x value of  the vertex in f(x)=ax^2+bx+c= is -b/(2a)
the y value of the vertex is f(-b/(2a)) so
given
f(x)=2x^2-x-10
a=2
b=-1
-b/2a=-(-1)/(2*2)=1/4
f(1/4)=2(1/4)^2-(1/4)-10
f(1/4)=2(1/16)-1/4-10
f(1/4)=1/8-1/4-10
f(1/4)=1/8-2/8-80/8
f(1/4)=-81/8
so the vertex is (1/4,-81/8) or (0.25,-10.125)


C. graph the x intercepts and the vertex
the vertex is min and the graph goes through the x intercepts
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