Tangent and normal to eˣ

Statement

Find the tangent and the normal to the graph of $y=e^x$ at the point $x=0$.

Solution

At the point: $f(0)=e^0=1$, so $A(0;1)$. The derivative is $f'(x)=e^x$, from which $f'(0)=1$: slope of the tangent $m=1$.

Tangent: $y-1 = 1\cdot(x-0) \Rightarrow \boxed{y = x+1}$.

Normal: slope $-1/m = -1$, so $y-1 = -1\cdot(x-0) \Rightarrow \boxed{y = -x+1}$.

Links

Topics: Derivatives
Concepts: Line · Normal line · Tangent line
Skills: Analytic geometry
Exercise type: Derivative computation