# How do you find the equation of the tangent and normal line to the curve y=e^x at x=0?

Jan 29, 2017

Tangent Equation: $y = x + 1$
Normal Equation: $\setminus y = - x + 1$

#### Explanation:

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. The normal is perpendicular to the tangent and so the product of their gradients is $- 1$

so If $y = {e}^{x}$ then differentiating wrt $x$ gives us $\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x}$.

When $x = 0 \implies y = {e}^{0} = 1$, and $\frac{\mathrm{dy}}{\mathrm{dx}} = 1$

So the tangent passes through $\left(0 , 1\right)$ and has gradient $1$, and the normal passes through the same point and has gradient $- 1$

Using the point/slope form $y - {y}_{1} = m \left(x - {x}_{1}\right)$ the equation of the tangent is:

$y - 1 = 1 \left(x - 0\right)$
$\setminus \setminus \therefore y = x + 1$

And for the normal:

$y - 1 = - 1 \left(x - 0\right)$
$\setminus \setminus \therefore y = - x + 1$

We can confirm these result graphically:
graph{(y-e^x)(y-x-1)(y+x-1)=0 [-6, 6, -5, 5]}