# precalculus

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When asked to sketch the graph of
g(x)= -e^(x+1) +1, how do I turn this equation into exponential form?
Or can I? I always get confused on what to do especially when I need to the x part of the equation out of the exponent.

• precalculus -

g-1 = -e^(x+1)
1-g = e^(x+1)
x + 1 = ln (1 - g)
x = ln(1-g) -1

g will always be less than 1.

Is that what you want? I don't know what you mean by "exponential form"

You can always sketch the graph using the original equation. Just assume different x values, compute g for each, and plot the result

• precalculus -

My book says a way for me to graph/plot points for my function is to write it in exponential form. For example. "Graph (fx)= log base 2(x-2)
y=log base 2 (x-2)
x-2=2^y
x=2^y + 2
So now I can just plug any y value in to get my x value to graph my translated graph. So I was wondering on how to be able to put the above function in exponential form as well? I know when you're done with the graph it's supposed to go through the point (-1,0) and the asymptote is y=1, but I don't know how to get there.

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