# Hi. I don't quite get this: At the point where the function f(x) = e^x intersects the y-axis the tangent is drawn. Find the x-coordinate of the intercept of this tangent with the x-axis. ?

Mar 24, 2018

$x = - 1$

#### Explanation:

The given function is $f \left(x\right) = {e}^{x}$.

A tangent is said to be drawn at the point where $f \left(x\right)$ intersects the $y$-axis.

Let's find this $y$-intercept.

The $y$-intercept occurs when $x = 0$:

$R i g h t a r r o w f \left(0\right) = {e}^{0} = 1$

So, a tangent is drawn at the point where $x = 0$ and $y = 1$, i.e. $\left(0 , 1\right)$.

Now, we need to find an equation for the tangent drawn at this point.

Let's differentiate $f \left(x\right)$.

The derivative of ${e}^{x}$ is a standard derivative: it's ${e}^{x}$ as well:

$R i g h t a r r o w f ' \left(x\right) = {e}^{x}$

Now, we need to find the gradient of the tangent at $\left(0 , 1\right)$.

Let's substitute $0$ in place of $x$ in the derivative:

$R i g h t a r r o w f ' \left(0\right) = {e}^{0} = 1$

So, the gradient of the tangent is $1$.

We can now write an equation for the tangent drawn at $\left(0 , 1\right)$:

$R i g h t a r r o w y - 1 = 1 \left(x - 0\right) = x - 0 = x$

$R i g h t a r r o w y = x + 1$

Finally, we are required to find the $x$-coordinate of the intercept of the tangent with the $x$-axis.

So let's the equation of the tangent equal to zero:

$R i g h t a r r o w y = 0$

$R i g h t a r r o w x + 1 = 0$

$\therefore x = - 1$

Therefore, the tangent intersects the $x$-axis at $x = - 1$.