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3 years ago

Which best describes the range of the function f(x) = 2(one-fourth) Superscript x after it has been reflected over the y-axis?

2 answers:

9 0

"All real numbers greater than 0" best describes the range of the function after it has been reflected over the y-axis ⇒ 3rd answer

Step-by-step explanation:

The form of the exponential function is $f(x)=a(b)^{x}$ , where

a is the initial value (value f(x) at x = 0)
b is the growth/decay factor
The domain of the function is {x : x ∈ R} (all real numbers)
The range of the function is {y : y > 0} (all positive real numbers)

∵ $f(x)=2(\frac{1}{4})^{x}$

- Compare it with the form above

∴ a = 2 ⇒ initial value

∴ b = $\frac{1}{4}$ ⇒ decay factor because its between 0 and 1

∵ The domain of the function is the values of x

∴ The domain of the function is {x : x ∈ R} ⇒ all real numbers

∵ The range of the function is values of y

∴ The range of the function is {y : y > 0}

∵ Reflection over the y-axis change the sign of x

∴ The image of f(x) is $y=2(\frac{1}{4})^{-x}$

- That means, all x-coordinates of the points on the graph

opposite in signs, but that does not effect the domain

because the domain is all real numbers

∵ There is no change of the sign of y

- That means reflection over y-axis does not effect the range

∴ The range of the function after reflection is the same as the

range of f(x)

∴ The range of the image of f(x) is {y : y > 0}

"All real numbers greater than 0" best describes the range of the function after it has been reflected over the y-axis

Learn more:

You can learn more about the reflection in brainly.com/question/4057490

#LearnwithBrainly

Assoli18 [71]3 years ago

6 0

Answer:

Answer is C

Step-by-step explanation:

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boyakko [2]

Step-by-step explanation:

18.

the last one.

-4.9t² + 24.5t + 117.6 = 0

the ball starts at 117.6 meters height. then, during the first seconds the thrust upwards is adding height, until finally, with more and more seconds passing, gravity will win with a vengeance and pulls the ball down faster and faster than any other force in any other direction.

19.

the general solution for a quadratic equation

y = ax² + bx + c

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

x = t

a = -4.9

b = 24.5

c = 117.6

t = (-24.5 ± sqrt(600.25 - -4×4.9×117.6))/-9.8 =

= (-24.5 ± sqrt(600.25 + 2304.96))/-9.8 =

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t1 = (-24.5 + 53.9)/-9.8 = -3

t2 = (-24.5 - 53.9)/-9.8 = 8

a negative time does not make any sense, so the real solution is : the ball reached the ground after 8 seconds.

20.

now the equation is

-4.9t² + 24.5t + 117.6 = 49

but this is quickly transformed again into a "= 0" equation :

-4.9t² + 24.5t + 68.6 = 0

t = (-24.5 ± sqrt(600.25 - -4×4.9×68.6))/-9.8 =

= (-24.5 ± sqrt(600.25 + 1344.56))/-9.8 =

= (-24.5 ± sqrt(1944.81))/-9.8 = (-24.5 ± 44.1)/-9.8

t1 = (-24.5 + 44.1)/-9.8 = -2

t2 = (-24.5 - 44.1)/-9.8 = 7

again, negative time does not make sense.

the ball will be at 49 m height after 7 seconds.

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