document.write( "Question 378049: Please advise on how to sketch the graph of f(x)=(2)^x-2 + 1. Show at least 2 points and dash in any asymptotes. \n" ); document.write( "

Algebra.Com's Answer #268629 by jsmallt9(3758) $\"\"$ $\"About$

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$\"f%28x%29=%282%29%5E%28x-2%29+%2B+1\"$
\n" ); document.write( "First let's look at the base function:
\n" ); document.write( " $\"g%28x%29+=+2%5Ex\"$
\n" ); document.write( "As all base exponential functions do, this exponential function passes through (0, 1) and, since 2 to any power can never be zero or negative, this graph is entirely above the x-axis. It never touches or crosses the x-axis. But for very large negative values of x, g(x) will be very, very close to zero (i.e. the x-axis). So the x-axis (or y = 0) is a horizontal asymptote of g(x).

\n" ); document.write( "The equation for f(x) shows that its graph will be two units to the right (because of the x-2) of and one unit above (because of the +1) the graph of g(x). Moving the point (0, 1) two to the right and 1 up we get (2, 1). And the asymptote is also moved up one. So y = 1 is the horizontal asymptote for f(x).

\n" ); document.write( "You were asked to find at least two points. We have just one so far, (2, 1). I will leave it up to you to find the other point(s). Just pick a number (other than 2 since we already have that point) for x, insert it into f(x) and find the value of the function for that x. Each time you do this, the x and the value of f(x) are the coordinates of a point on the graph of f(x). \n" ); document.write( "

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