Question

Find an equation for the tangent to the curve at the given point: y=x^2+e^x−e; at (1,1).

Answer 1

Answer:

$y = (2 + e)x - e - 1$

Step-by-step explanation:

$\frac{dy}{dx} = 2x + {e}^{x} \\ \\ \frac{dy}{dx} = 2 + e \\ \\ equation \: of \: tangent \: is \: \\ \\ \\ y - 1 = (2 + e)(x - 1) \\ \\ y = 1 + (2 + e)x - 2 - e \\ \\ y = (2 + e)x - e - 1$