# What is the equation of the line tangent to  f(x)=e^(x^2+x)  at  x=0 ?

Apr 27, 2018

$y - x - 1 = 0$

#### Explanation:

To find the point (0,y) which lies on the tangent We substitute in the function with the value $x = 0$

$f \left(0\right) = {e}^{0 + 0} = 1$

So the point is $\left(0 , 1\right)$

to find the slope$\left(m\right)$ of the tangent we find the first derivative $\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$ value at the point $\left(0 , 1\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{{x}^{2} + x} \left(2 x + 1\right)$ "rarrChain Rule

Substitute with the point $\left(0 , 1\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{0 + 0} \cdot \left(2 \times 0 + 1\right) = 1$

The Equation of a Straight Line

color(green)((y-y_1)=m(x-x_1)

Substitute

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

$y - x - 1 = 0$