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
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) (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) = e^x and when x = 0 ⇒ = 1
The "tangent to the curve is a LINE: y-1 = (x-0) 0r y = x + 1