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AysviL [449]
3 years ago
8

What is the vertex of the parabola? y+4=12(x−7)2

Mathematics
2 answers:
IceJOKER [234]3 years ago
5 0
<span>The answer to this question is (7,-4).</span>
The parabola <span>y+4=12(x−7)^2 can be moved so that the y term is isolated like so
                                    y = 12(x-7)^2 - 4
A parabola of the form y = a(x-h)^2 + k has its vertex at (h,k). So it follows that we take h = 7 and k = - 4.
The answer to this question is (7,-4).
</span>
Alborosie3 years ago
3 0
Vertex is (7,-4) for answer.

Hope this helps. :)
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What are the potential zeros of f(x)=6x^4+ 2x^3 - 4x^2 +2?
Dmitry_Shevchenko [17]

The potential zeros of f(x)=6x^4+ 2x^3 - 4x^2 +2 are ±(1, 1/2, 1/3, 1/6, 2, 2/3)

<h3>How to determine the potential zeros of the function f(x)?</h3>

The function is given as:

f(x)=6x^4+ 2x^3 - 4x^2 +2

For a function P(x) such that

P(x) = ax^n +...... + b

The rational roots of the function p(x) are

Rational roots = ± Possible factors of b/Possible factors of a

In the function f(x), we have:

a = 6

b = 2

The factors of 6 and 2 are

a = 1, 2, 3 and 6

b = 1 and 2

So, we have:

Rational roots = ±(1, 2)/(1, 2, 3, 6)

Split the expression

Rational roots = ±1/(1, 2, 3, 6)/ and ±2/(1, 2, 3, 6)

Evaluate the quotient

Rational roots = ±(1, 1/2, 1/3, 1/6, 2, 1, 2/3, 1/3)

Remove the repetition

Rational roots = ±(1, 1/2, 1/3, 1/6, 2, 2/3)

Hence, the potential zeros of f(x)=6x^4+ 2x^3 - 4x^2 +2 are ±(1, 1/2, 1/3, 1/6, 2, 2/3)

The complete parameters are:

The function is given as:

f(x) = 3x^3 + 2x^2 + 3x + 6

The potential zeros of f(x)=6x^4+ 2x^3 - 4x^2 +2 are ±(1, 1/2, 1/3, 1/6, 2, 2/3)

Read more about rational roots at

brainly.com/question/17754398

#SPJ1

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