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A.) Find the slope of the graph of f(x) = 1 - e^x at the point where it crosses the x -axis B.) Find the Equation of the tangent line to the curve at this point
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So we need to find the x-intercepts if any then find f' and do f'(x-intercept)
i think the answer to a would be 1. Am i correct with this? I set 1 - e^x = 0 and worked from there
1=e^x => x=0 right? thats the x-intercept
so now find f' f'(x)=-e^x and do f'(0)
so that would be -1 then
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yep yep thats the slope :)
then plug in original to find the y point?
so y=mx+b y=-1x+b we know a point on the line(x=0,y=1-e^{0})=(0,0) the y-intercept is b=0 y=-x is the tangent line
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