# I have to sketch a graph of f(x)=2(1/2)^(x) +1. I do not know how to put the x as an exponent of (1/2). Then I have to state the function's domain, its range, and the equation of its asymptote.

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`f(x) =2(1/2)^x + 1` is an exponential graph with restrictions.

Note the asymptotes: these are the horizontal (y = ... ) and sometimes (although not in this case) vertical (x = ...) lines representing the restrictions. In other words, the graph cannot touch or cross these lines.

To show that something is an exponent, a universal practice is to use the ^ and then people understand to make it an exponent in a question. So just use the ^ and follow it by (in this case x).

In this graph, `f(x)=2(1/2)^x + 1` we can see 2 things. The graph has shifted - which is why it has a `+1` . And it has a fraction with x as the exponent(power, indice). Think about the rules of exponents so as to understand this graph better. The equivalent for `2(1/2)^x +1= 2(2^(-x)) +1` because the rules of exponents tell us that if the expression is divided, we must minus the exponents. So we know that this graph is negative and that it has shifted.

In exponential graphs, the x axis would normally be the asymptote (where y=0) but in this case, we know that the asymptote is y=1 as indicated by the `+1` because the graph has moved (up) one place in an upward direction. ` `

Now to find the values to allow you to draw it, use a table (or any preferred method). For example, if x=1 we would have:

`f(x) = 2(1/2)^1) + 1` which we can simplify to `f(x)= 1+1` . `therefore y=2 ` So the co-ordinate is `(1;2)` . Now find another co-ordinate, say when x=2:

`therefore f(x) = 2(1/2)^(x) +1 = 2(1/2)^2 +1 = 3/2` . So we have `(2;3/2)`

Now we have enough information to draw. Note that the line y=1 is drawn in red and this graph will never touch it although it will get closer and closer to it:

Domain: x is unlimited so `therefore x in RR`

Range: y has a restriction (1,`oo`).

For the equation, `f(x) = 2(1/2)^x +1` and to make a T-table for the numbers 1,2,3,4, you plug in the values of x as x=1, x=2 and so on and calculate the corresponding y values:

x / `( f(x) = 2(1/2)^x +1 ) ` / (x,y)

1 / `2(1/2)^1 +1 = 2 ` / (1; 2)

2 / `2(1/2)^2 + 1 = 3/2 ` / ( 2; 3/2)

3 / `2(1/2)^3 +1 = 5/4 ` / (3; 5/4)

4 / `2(1/2)^4 +1 = 9/8 ` / (4; 9/8)

Note how the x values have been substituted with 1,2,3 or 4 and the last column is the co-ordinate (when x = 1, for example, y = 2).

In terms of domain and range, answers can actually be written in terms of infinity (`oo` ) or as an element (`in` ) of real numbers (`RR` ). Your teacher/ instructor/ educator probably has a preferred one.

The domain would be (-`oo`,`oo`). The range would be (1,`oo`). The graph in the second answer is correct.