Function g is a transformation of the parent quadratic function f. Which statements are true about this function?
g(x) = -(x - 4)2 + 6
Select all the correct answers.
Function g reflected function facross the x-axis.
Function g translated the vertex from (0,0) to (-4,6).
For function g, as x approaches negative infinity, 8(x) approaches negative infinity.
Function g is always decreasing,
Function g is always negative.
Function g is symmetrical about the point (4,6).

Answers

Answer 1

Answer: Function g reflected function f across the x-axis.

For function g, as x approaches negative infinity, g(x) approaches negative infinity.

Function f is symmetrical about the point (4,6).

Answer 2

The correct options are Option A and Option C : Function g is reflected of function f across the x-axis. and as x approaches to the -∞, g approaches to the -∞.

Here given that the function g(x)= -(x-4)²+6

vertex of the graph of the function g is

dg/dx=0

⇒2(x-4)=0

⇒x-4=0

⇒x=4

putting x=4 in g(x) , g(4)= -(0-0)²+6=6

Hence the vertex of the graph of g is (4,6).

The parent function is f(x)= x²

here the vertex of the function is (0,0).

Option A: This option is true as the graph of function g is the inverse image of f. So,  Function g is reflected of function f across the x-axis.

Option B: This option is incorrect. as the vertex of the function f is transformed to (4,6) which is the vertex of g(x).

Option C:  This option is true because

 [tex]\lim_{x \to- \infty} g(x)[/tex]

= [tex]\lim_{x \to- \infty} -(x-4)^2+6\\[/tex]

= -(∞-4)²+6

=-∞²+6

=-∞

Hence as x approaches to the -∞, g approaches to the -∞

Option D. This option is incorrect because

if the function is always decreasing its derivative is negative for all x∈R

dg/dx<0

⇒2(x-4)<0

⇒x-4<0

⇒x<4

so the function is decreasing for x<4 not for all x.

Option E: This option is incorrect. because g is not always rather it is positive for 4-√6 < x < 4+√6

g(x)>0

⇒-(x-4)²+6 >0

⇒-(x-4)²>-6

⇒(x-4)²<6  

⇒|x-4|<√6

⇒4-√6<x<4+√6

so, g(x) is positive for 4-√6 < x < 4+√6

Option E:
This option is incorrect as g(x) is symmetrical about line x=4.

Therefore the correct options are Option A and Option C : Function g is reflected of function f across the x-axis. and as x approaches to the -∞, g approaches to the -∞.

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Answer

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Answers

Answer:

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Answers

Answer:

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Step-by-step explanation:

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Answers

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Step-by-step explanation:

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Answers

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Step-by-step explanation:

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Answers

Answer:

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Answer:

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Step-by-step explanation:

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Which statement describes the inverse of m(x) = x2 – 17x?

Answers

Answer:

The correct option is;

[tex]The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}[/tex]

Step-by-step explanation:

The given information is that m(x) = x² - 17·x

The above equation can be written in the form;

y = x² - 17·x

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0 = x² - 17·x - y

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[tex]x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}[/tex]

By comparison to the equation,0 = x² - 17·x - y, we have;

a = 1, b = -17, and c = -y

Substituting the values of a, b and c into the formula for the general solution of a quadratic equation, we have;

[tex]x = \dfrac{-(-17)\pm \sqrt{(-17)^{2}-4\times (1) \times (-y)}}{2\times (1)} = \dfrac{17\pm \sqrt{289+4\cdot y}}{2}[/tex]

Which can be simplified as follows;

[tex]x = \dfrac{17\pm \sqrt{289+4\cdot y}}{2}= \dfrac{17}{2} \pm \dfrac{1}{2} \times \sqrt{289+4\cdot y}} = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +\dfrac{4\cdot y}{4} }}[/tex]

And further simplified as follows;

[tex]x = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +y }} = \dfrac{17}{2} \pm \sqrt{y + \dfrac{289}{4} }}[/tex]

Interchanging x and y in the function of the inverse, m⁻¹(x), we have;

[tex]m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}[/tex]

We note that the maximum or minimum point of the function, m(x) = x² - 17·x found by differentiating the function and equating the result to zero, gives;

m'(x) = 2·x - 17 = 0

x = 17/2

Similarly, the second derivative is taken to determine if the given point is a maximum or minimum point as follows;

m''(x) = 2 > 0, therefore, the point is a minimum point on the graph

Therefore, as x increases past the minimum point of 17/2, m⁻¹(x) increases to give;

[tex]The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}[/tex] to increase m⁻¹(x) above the minimum.

Answer:

b

Step-by-step explanation:

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Answers

Answer:  (x - 3)² + (y + 0.5)² = 3.25    →    [tex](x-3)^2+\bigg(y+\dfrac{1}{2}\bigg)^2=\dfrac{13}{4}[/tex]

Step-by-step explanation:

Concentric means they have the same center but different radii.

The equation of a circle is:  (x - h)² + (y - k)² = r²    where

(h, k) is the center of the circler is the radius of the circle

Complete the square to find the center of the circle

x² - 6x + _____  + y² + y + _____ = 1 + ____ + ____

    ↓                              ↓

-6/2 = -3                    1/2 = 1/2

Equation: (x - 3)² + (y + 1/2)² = r²

Since the circle passes through point (x, y) = (4, -2), input that into the equation to find r².

(x - 3)² +  (y + 1/2)² = r²

(4 - 3)² + (-2 + 1/2)² = r²

  (1)²    +  (-3/2)²    = r²

   1      +     9/4      = r²

                 13/4     = r²

Equation: (x - 3)² + (y + 1/2)² = 13/4    →   [tex](x-3)^2+\bigg(y-\dfrac{1}{2}\bigg)^2=\dfrac{13}{4}[/tex]

In decimal form: (x - 3)² + (y + 0.5)² = 3.25

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i- i just want points so sure
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