Remember to show work and explain.

12) Write the equation for the function. (Find y-intercept and common ration)​

f(x) =

Domain:

Range:

Remember to show work and explain12 Write the equation for the function Find yintercept and common rationfx DomainRange class=

Respuesta :

Using given,

f(0)=2=2/1=2/3°

f(1)=2/3=2/3¹

f(2)=2/9=2/3²

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Similarly

f(x)=2/(3^x)

[Explanation: All I did here is find a pattern or behaviour of function with the help of given values of function (for x=0,1,2). You can clearly see that numerator in f(0),f(1) and f(2) remains constant while denominator follows the pattern 3ⁿ . So given function has to be equal to 2/3ⁿ]

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Y-intercept means where does the given function cuts the graph at y-axis. It cuts y-axis at 2, so,

Y-intercept=2

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To find common ratio,

f(1)/f(0)=(2/3)/2=1/3

f(2)/f(1)=(2/9)/(2/3)=1/3

Therefore,

common ratio=1/3

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The equation for function will be same as function but f(x) will be replaced by y,

y=2/(3^x)➡️is the required equation

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Since the given function is exponential function, its domain will be always (-∞,+∞) no matter what shifts or translations.

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The graph seems to touch x-axis or the line y=0 but never reach or touch y=0 .So it doesn't quite reach y=0. Also the graph goes upward in positive direction which means it will increase infinitely in upward or positive direction. So its range will be (0,+∞) which means it will not touch y=0 line and will infinitely increase in positive or upward direction.

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Answer:  f(x) = 2(1/3)ˣ,   D: x is All Real Numbers,   R: y > 0

Step-by-step explanation:

f(x) = a(b)ˣ

  • a represents the y-intercept
  • b represents the common ratio

The graph shows the y-intercept as 2   ⇒     a = 2

[tex]f(2) \div f(1)\ \text{is the common ratio}\\\\\dfrac{2}{9}\div\dfrac{2}{3}=b\implies \dfrac{2}{9}\times\dfrac{3}{2}=b\implies \dfrac{1}{3}=b[/tex]

[tex]f(x)=2\bigg(\dfrac{1}{3}\bigg)^x[/tex]

Domain: there are no restrictions on the x-value ⇒ x is All Real Numbers

Range: there are no values of x that will result in the y-value being 0 or negative, so y > 0.

We can show this algebraically by taking the log of both sides of the equation.  

log y = log 2(1/3)ˣ    

You cannot take the log of 0 or a negative number so y has to be greater than 0.