Calculus Help Please -Calculus 1-

Consider the function f(x) = 3(1 − e^x).

(a) Find the exact slope of the graph of f(x) at the point where it crosses the x-axis.
slope =

(b) Find the exact equation of the tangent line to the graph of f(x) at the point in part (a).
y(x) =

(c) Find the exact equation of the line which is perpendicular to the tangent line in part (b). This line is called a normal line.
y(x) =

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I've graphed it and looked at the point where it crosses the x access and it appears to be 0. Am I doing something wrong? Please help and thank you in advanced.

(a) dy/dx at (0,0) is -3
Use the chain rule to find the derivative and then plug in 0 for x to find the slope.

(b) tangent line: y = -3x
Use the point slope formula and plug in 0 for x and 0 for y.

(c) y(x) = 1/3x
The slope of the perpendicular line to the tangent line is the opposite reciprocal of the slope of the tangent line.

If you have any further questions, you can message me.

f(x)=3(1 − e^x)

f'(x) = −3e^x

(a) graph of f(x) at the point where it crosses the x-axis,
f(x) = 0 = 3(1 − e^x) => y = 0

slope = f'(x) = f'(0) = -3

(b) y - 0 = (-3) (x - 0)

y(x) = -3x

(c) slope = 1/3 (since we need -3 * 1/3 = -1)

y = 1/3 x

(Hope these are correct!)

You are correct. When x=0, e^x = 1 and f(x) = 0

f '(x) = -3e^x, so slope at x=0 is -3

Tangent line is y= -3x

normal line is y = 1/3 x (slope of normal line is negative inverse of original line)