another calculus?

Question

The tangent line to the graph y=e^(2-x) at the point (1,e) intersects both coordinate axes. What is the area of the triangle formed by this tangent line and the coordinate axes?
a. 2e
b. e^2-1
c. e^2
d. 2e (squareroot of e)
e. 4e

Answer 1

I like to split off the e^2, since it's a constant:

y = e^(2 - x) = e^2e^(-x)

Then the derivative is obvious:

y' = -e^2e^(-x)

Find the value of the slope at x = 1:

y'(1) = -e^2e^(-1) = -e

So the tangent line would be (using point-slope equation):

y - e = -e(x - 1) = -ex + e
y = -ex + 2e

So one side of the triangle is the y-intercept, 2e. The other side is found by looking for the x-intercept:

0 = -ex + 2e
ex = 2e
x = 2

So the area would be 1/2(2e)(2) = 2e, choice a.

Answer 2

Should we find the equation of the tangent line? Let's do.

The slope of the line is given by the first derivative of the function.

There's no sense me trying to answer these questions when Jim is online at the same time.

Answer 3

y' = e^(2-x)(-1)
y'(1) = -e

Equation of the tangent,
y-e = -e(x-1)

x-int = 2
y-int = 2e

area = (1/2)(x-int)(y-int) = 2e