SOLUTION: Find the equation of the tangent to the curve y = e^x at the point where it crosses the Y-axis. I differentiated y = e^x (which is e^x), which gives you your gradient for the ta

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Find the equation of the tangent to the curve y = e^x at the point where it crosses the Y-axis. I differentiated y = e^x (which is e^x), which gives you your gradient for the ta      Log On


   

Question 842109: Find the equation of the tangent to the curve y = e^x at the point where it crosses the Y-axis.
I differentiated y = e^x (which is e^x), which gives you your gradient for the tangent..
Subbing in x= 0 into the original equation;
y = e^(0) = 1
Therefore if y=1, x=0 (exponential function)
Sub into the formula:
y-y1 = m(x-x1)
y-1 = e^x(x-0)
y= xe^x + 1
However the answers say y = x+1. Whilst in theory, i completely understand whyh it would be x+1, i can't seem to resolve the question quantatively. Thank you for your help!
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

"I differentiated y = e^x (which is e^x), which gives you your gradient for the tangent"(line)

highlight_green%28m%29 = e^x and when x = 0 ⇒ highlight_green%28m%29 = 1

The "tangent to the curve is a LINE:  y-1 = highlight_green%281%29(x-0)  0r y = x + 1