Throughout this question, use exact values or at least 5 decimal place accuracy.

Question

Throughout this question, use exact values or at least 5 decimal place accuracy. Consider f(x)=9-e^x.

A. Find the slope of the graph of f(x) at the point where the graph crosses the x-axis. slope = -9
B. Find the equation of the tangent line to the curve at this point. y =
C. Find the equation of the line perpendicular to the tangent line at this point. (This is the normal line.) y = (1/9)x - (1/9)ln(9)

Unable to figure out B!
The answer i originally got for B was -9x - 9ln(9) but this is apparently incorrect.
coulr some please explain how to got to the answer please :)

Explanation / Answer

f(x)=y=9-e^x.

crosses x axis =>y =0

9-e^x.=0

=>x =ln9

y=9-e^x

for slope differentiate with respect to x

dy/dx=0-e^x

at x =ln9

dy/dx=-e^ln9

dy/dx =-9

slope of tangent m=-9

point on x axis here curve intersects is (ln9,0)

by point slope form equation of tangent is y-0=-9(x -ln9)

y =-9x +9ln9 is equation of tangent line

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