Question

consider f(x)= 9-e^x
Find the slope of the graph of f(x) at the point where the graph crosses the x-axis.
1-slope=?
Find the equation of the tangent line to the curve at this point.
2-y=?
Find the equation of the line perpendicular to the tangent line at this point. (This is the normal line.)
3-y=?

Answer 1

the graph crosses the x-axis at the point e^x=9
x=ln9

1- in order to find the slope take the derivative:
f'(x) = -e^x
now plug in the ln9 :
f'(ln9) = -9
2-tangent line
well we have the slope which is -9 and we have the point : (ln9,0)
now the general equation for a line is : y = y1 + m(x-x1)
plug in the values: y = -9x +9ln9
3- perpendicular line will be opposite and reciprocal to the slope of the tangent line
so m(perpendicular) = 1/9
note ( m(perpendicular) * m(tangent) = -1 )
now as we did in 2 :
y = x/9 - ln9/9