Question

Find the equation of a tangent line to the curve y=e^(x^2) that also passes through the point (1,0).
Thanks in advance !

Answer 1

y=e^(x^2)
dy/dx = 2x(e^(x^2))
when x = 1, dy/dx = 2(e^1) = 2e

so you now have the slope, m = 2e and a point (1,1)

use your usual way to find the equation of the straight line.

Answer 2

Thanks !

Answer 3

y = m x + b
when y = 0, x=1
so b = -m
that m is the slope where they hit
slope of curve = dy/dx = 2x e^(x^2) = m = -b at crash
at that point
y = m x - m
and y = e^(x^2) = m x - m
and
m = 2 x e^(x^2)
so
e^(x^2) = 2 x^2 e^(x^2) - 2 x e^(x^2)
1 = 2 x^2 - 2 x
2 x^2 - 2 x - 1 =0
x = 1.366 for + root
m = 2 x e^(x^2) = 2.732 * 6.46 = 17.65
b = -17.65

Answer 4

They do not hit at (1,0)

Answer 5

or at (1,1)

Answer 6

as above, we want a line through (h,k) on the curve, and
2h^2-2h-1=0
h = (1+√3)/2
So now we have a slope m = e^h^2/(h-1) (x-1)
That gives us the line
y = e^((1+√3)/2)^2/( (1+√3)/2-1) (x-1)

See the graphs at

https://www.wolframalpha.com/input/?i=plot+y%3De%5E%28x%5E2%29%2C+y%3De%5E%28%281%2B%E2%88%9A3%29%2F2%29%5E2%2F%28+%281%2B%E2%88%9A3%29%2F2-1%29%28x-1%29

Answer 7

argggh
For these type of questions, I always told my students that the first thing they should do is test if the given point actually lies on the function.
I did not do that, and the point (1,0) was not on y = e^(x^2).

Looks like oobleck, our dependable tutor saw that.
My bad!

Answer 8

I'm a bit stuck on the same problem. I'm not sure how to solve it bc (1,0) isn't on the graph of y.

Answer 9

wow this was all so useless...