Question

Consider the graph of the function f(x)=2^x.

Answer 1

Answer:

Step-by-step explanation:

1 -     f(x) = f (x + 2) = 2^x+2

      { when x = 0, y = 2² = 4 }

     so the function { y - intercept at (0, 4) }

2 -   m(x) = -f(x) = - 2^x

      2^x increases as x increases

     So -2^x {decreases as x increases}

3 -  g(x) = 2f(x) = 2, 2^x = 2^x+1

    { when x = 0, y = 2^1 = 2 }

   So the function { y-intercept at (0, 2)

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

f(x)'s asymptote of y = 0

so {f(x) + 2}'s asymptote of y =2