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.
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
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.
To turn the equation g(x) = -e^(x+1) + 1 into exponential form, you need to isolate the exponential term, which in this case is e^(x+1).
Step 1: Start with the equation g(x) = -e^(x+1) + 1.
Step 2: Move the constant term (1 in this case) to the other side of the equation by subtracting it from both sides:
g(x) - 1 = -e^(x+1).
Step 3: By multiplying both sides of the equation by -1, you can change the sign to get:
1 - g(x) = e^(x+1).
Now, the equation is written in exponential form, where the base is e and the exponent is (x+1).
However, note that not all exponential equations can be easily rearranged into this form. In some cases, it may be more practical to work with the equation in its original form, such as when graphing a function.