Question

How do you sketch `y=e^absx`?

Answer 1

See explanation...

Explanation:

When `x = 0` the given formula gives `y = e^abs(color(blue)(0)) = 1`

Also note that since `y` is a function of `abs(x)`, its graph will be symmetrical about the `y` axis.

The function `f(x) = e^x` is well known for being equal to its own derivative. That is, the instantaneous slope at any point is itself `e^x`.

In symbols we can write:

`d/(dx) e^x = e^x`

So we find:

`lim_(x->0) d/(dx) e^x = lim_(x->0) e^x = e^0 = 1`

Hence the slope of our given function approaches `1` as `x->0^+`, that is as `x->0` from the right.

As a result, the graph of `y = e^abs(x)` forms a V shape at `(0, 1)` with slope `+-1`

We can also evaluate `y=e^abs(x)` for a few values of `x` to provide some points through which the graph passes:

`e^abs(1) ~~ 2.718`

`e^abs(2) ~~ 7.389`

Putting this information together we can plot the graph of `y = e^abs(x)` as a kind of steep U shape with a pointed vertex:
graph{(y-e^abs(x))(x^2+(y-1)^2-0.02)((x-1)^2+(y-e)^2-0.02)((x-2)^2+(y-e^2)^2-0.02) = 0 [-10.375, 9.625, -1.32, 8.68]}