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KengaRu [80]
3 years ago
12

PLEASE ANSWER! Given the functions f(x) = x2 + 6x - 1, g(x) = -x2 + 2, and h(x) = 2x2 - 4x + 3, rank them from least to greatest

based on their axis of symmetry.
a. f(x), g(x), h(x)
b. h(x), g(x), f(x)
c. g(x), h(x), f(x)
d. h(x), f(x), g(x)
Mathematics
1 answer:
Ratling [72]3 years ago
7 0

Answer:

<em>he rank from least to great based on their axis of symmetry: </em>

0, 1, -3 ⇒ g(x), h(x), f(x)

So, <em>option C</em> is correct.

Step-by-step explanation:

A quadratic equation is given by:

ax^2+bx+c =0

Here, a, b and c are termed as coefficients and x being the variable.

<em>Axis of symmetry can be obtained using the formula</em>

x = \frac{-b}{2a}

Identification of a, b and c in f(x), g(x) and h(x) can be obtained as follows:

f(x) = x^2 + 6x - 1

⇒ a = 1, b = 6 and c = -1

g(x) = -x^2 + 2

⇒ a = -1, b = 0 and c = 2

h(x) = 2^2 - 4x + 3

⇒ a = 2, b = -4 and c = 3

So, axis of symmetry in f(x) = x^2 + 6x - 1 will be:

x = \frac{-b}{2a}

x = -6/2(1) = -3

and axis of symmetry in g(x) = -x^2 + 2 will be:

x = \frac{-b}{2a}

x = -(0)/2(-1) = 0

and axis of symmetry in h(x) = 2^2 - 4x + 3 will be:

x = \frac{-b}{2a}

x = -(-4)/2(2) = 1

<em>So, the rank from least to great based on their axis of symmetry: </em>

0, 1, -3 ⇒ g(x), h(x), f(x)

So, <em>option C</em> is correct.

<em>Keywords: axis of symmetry, functions</em>

<em>Learn more about axis of symmetry from brainly.com/question/11800108</em>

<em>#learnwithBrainly</em>

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For the hyperbola with an equation of the form (x - h)²/a² - (y - k)²/b² = 1, the foci in on the points (h + c, k) and (h - c, k), where c = √(a² + b²).

For the hyperbola with an equation of the form (y - k)²/a² - (x - h)²/b² = 1, the foci in on the points (h, k + c) and (h, k - c), where c = √(a² + b²).

<u>Among the options</u>:

(a) (x - 24)²/24² - (y -1)²/7² = 1.

The equation is of the form (x - h)²/a² - (y - k)²/b².

h = 24, k = 1, a = 24, b = 7.

c = √(a² + b²) = √(24² + 7²) = √(576 + 49) = √625 = 25.

Thus, foci are at the points (h + c, k), (h - c, k) = (24 + 25, 1), (24 - 25, 1) = (49, 1), and (-1, 1), which are in different quadrants.

(b) (y - 12)²/5² - (x - 6)²/12² = 1.

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c = √(a² + b²) = √(5² + 12²) = √(25 + 144) = √169 = 13.

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(c) (y - 16)²/15² - (x - 2)²/8² = 1.

The equation is of the form (y - k)²/a² - (x - h)²/b².

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The equation is of the form (y - k)²/a² - (x - h)²/b².

h = -1, k = 16, a = 9, b = 12.

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Thus, foci are at the points (h, k + c), (h, k - c) = (-1, 16 + 15), (-1, 16 - 15) = (-1, 31), and (-1, 1), which are in the same quadrants (QUADRANT II).

Thus, the hyperbola having both foci lying in the same quadrant is <u>(y - 16)²/9² - (x + 1)²/12² = 1</u>, making the <u>4th option</u> a right choice.

Learn more about the foci of a hyperbola at

brainly.com/question/9384729

#SPJ4

For the options, refer the image.

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