# Find the equation of the tangent line to the curve y=x^4+2 e^x at the point (0,2).?

Jan 25, 2018

$y = 2 x + 2$

#### Explanation:

First, find the gradient of the line by using the derivative:

$y = {x}^{4} + 2 {e}^{x} \to \frac{\mathrm{dy}}{\mathrm{dx}} = 4 {x}^{3} + 2 {e}^{x}$

To get the gradient of the tangent, evaluate this at $x = 0$

$4 {\left(0\right)}^{3} + 2 {e}^{0} = 2$

So ${m}_{\tan} = 2$

We know the line must pass through the point: $\left(0 , 2\right)$ so with:

$\left(a , b\right) = \left(0 , 2\right)$

Use:

$y - b = m \left(x - a\right)$

$y - 2 = 2 \left(x - 0\right)$

$\to y = 2 x + 2$

Jan 25, 2018

$y = 2 x + 2$

#### Explanation:

The slope of the tangent line is the first derivative evaluated at the the given x coordinate; this compels us compute the first derivative:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 4 {x}^{3} + 2 {e}^{x}$

The slope, m, is the first derivative evaluated at $x = 0$

$m = 4 {\left(0\right)}^{3} + 2 {e}^{0}$

$m = 2$

Use the point slope form of the equation of the line:

$y = m \left(x - {x}_{0}\right) + {y}_{0}$

$y = 2 \left(x - 0\right) + 2$

$y = 2 x + 2$