Question

How do you find an equation of a line tangent to the graph of E when given a point?

Answer 1

since u already have a `point`,
if u could find the `slope` somehow, you can write the tangent line ins `point-slope` form

Answer 2

taking the derivative of a function at a point gives u the slope at that point :

slope = derivative

Answer 3

Okay, so if I have the point (0,1), how do I find the slope (derivative) at that point?

Answer 4

where is the graph/function ?

Answer 5

It's called "E", or the basic graph of an exponential function. It intersects the X-axis at (0,1).

Answer 6

ahh ok :)

so your function is : \(\large y = e^x\)

Answer 7

exactly

Answer 8

you dont want to use the derivative formula ?

derivative of \(e^x\) is \(e^x\) itself :

\(\dfrac{d}{dx}(e^x) = e^x\)

Answer 9

since you want the slope at (0,1),

plugin x = 0

derivative of \(e^x\) at x = 0 is :
\(e^0 = 1\)

Answer 10

so your slope at (0, 1) is `1`

Answer 11

Huh, well that's easy. I have another problem that says that the X value of a point is 2, could you walk me through that real quick?
Wouldn't it be \[e ^{2}=2e ?\]

Answer 12

\(e^2 = e^2 \ne 2e\)

Answer 13

Is this the qyestion ?
Given the function : \(y = e^x\)
find the tangent line at \(x = 2\)

?

Answer 14

It gives me 2 as the abscissa, not the point which makes it trickier. Otherwise, yes.

Answer 15

you can get the y-coordinate of point by plugging in x = 2 into ur equation

Answer 16

Right.

Answer 17

y = e^x

plugin x = 2 :
y = e^2

so ur point is (2, e^2)

Answer 18

since the derivative of \(e^x\) is \(e^x\) itself,
slope = derivative = \(e^2\)

Answer 19

you have hte point and slope,
you can write the equation of line

Answer 20

point : \((2, e^2)\)

slope : \(e^2\)

Answer 21

And yes, \(e\) is just a number.

Answer 22

Oooh, I got it! Thank you very much for your help my friend.

Haha, yeah, my team drilled that all class XD

Answer 23

u wlc :)