Respuesta :

rosarioaaliyah79

Look at the graph below carefully

Observe the results of shifting ={2}^{x}f(x)=2​x

​​  vertically:

The domain, (−∞,∞) remains unchanged.

When the function is shifted up 3 units to ={2}^{x}+3g(x)=2​x +3:

The y-intercept shifts up 3 units to (0,4).

The asymptote shifts up 3 units to y=3y=3.

The range becomes (3,∞).

When the function is shifted down 3 units to ={2}^{x}-3h(x)=2 ​x​​ −3:

The y-intercept shifts down 3 units to (0,−2).

The asymptote also shifts down 3 units to y=-3y=−3.

The range becomes (−3,∞).

Ver imagen rosarioaaliyah79
sqdancefan

9514 1404 393

Answer:

  • reflection over the x-axis
  • translation up 'e' units and right 3 units

Step-by-step explanation:

The transformation ...

  g(x) = a×f(x-h) +k

performs a vertical stretch by the factor 'a', and a translation by (h, k), which is h units right and k units up. When 'a' is negative, there is a reflection over the x-axis.

Here, we have 'a' = -1, h = 3, k = 'e'. So, the graph of f(x) has been ...

  reflected over the x-axis

  translated right by 3 units

  translated up by 'e' units

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Additional comment

We don't know if your 'e' is a typo. When talking about exponential functions, 'e' often refers to the base of natural logarithms, an irrational number approximately equal to 2.718281828459045...

Here's a graph of an exponential function transformed as you have indicated.

Ver imagen sqdancefan

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