Need Help with a Calculus equation fast!!!!!!!?

Question

Consider the graph Y=e^x

a) Find the equation of the tangent line to the graph at (a, e^a)

b) Find the X- and y- intercepts of the line in part (a)

Answer 1

y=(e^a)∙x - ae^a + e^a
The slope of the line tangent to e^x at a is e^a.
So the equation is y=(e^a)x + b where b is the y-intercept
Since (a,e^a) is on the line, e^a = (e^a)a + b,
or b = e^a - ae^a

x-intercept = a-1
y-intercept = e^a(1-a)

Answer 2

The slope of the line will be the derivative of e^x at x = a, which is e^a. The line's equation is then y = e^a*x + b. At x = a, y = e^a, and that point must be on the line, so e^a = a*e^a + b, then b = e^a*(1-a) and the equation is y = e^a*x + (1-a)*e^a.

The y intercept is the value of y when x = 0, or (1-a)*e^a
the x-intercept is the value of x when y = 0:

0 = e^a*x + (1-a)*e^a

e^-a = x + 1-a

x = e^-a - 1 + a

Answer 3

a)tangent line = derivative of function. derivative of e^x is e^x. (a,e^a)= (1,e^a)
b) x intercept is always one and and y intercept is a value.

Answer 4

eqn of tangent line requires point-slope
point is given (a,e^a)
slope is y ' = e^a

Line is y - e^a = e^a(x - e^a)

(b) to find x-intercepts, let y=o
- e^a = e^a(x - e^a) ===>x = e^a - 1

to find y-intercept, let x=o
y - e^a = e^a(0 - e^a)
y = e^(2a) + e^a