Finding equation tangent to the line?

ok basically first you sub x=0 into the normal equation

im assuming y=x+3 as the normal derivate?

so y=0+3

therefore y = 3

coords = (0,3)

then we need to find m1.

sub x=0 into the differentiated form this will give you the gradient of m1.

then to find the tangent you need to sub the coords with m1 into the form y=mx+c

you will end up with an answer such as y=4x (not the answer just an example).

then you need to find the normal which is m1/-1 with changing the sign and turning the value upside down.

therefore if m1 was say 2

m2 would be 2/-1 = -1/2

you then substitute the new found gradient (m2) into y=mx+c again with the original coordinates.

after this you should end up with the equation of the tangent to the line.

this can be given in y=mx+c or ax+by+c=0

hope i helped.

Y=2+e^2 is indeed a line -- a horizontal line. At x = 0 it's still a horizontal line. You say the answer is y=x+3, but are unsure how to get there. WE need to know the question!