Question

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

Answer 1

$y=4e^x+x\implies y'=4e^x+1$

At the point (0,1), the derivative has the value of $y'=4e^0+1=5$, so this is the slope of the tangent line. The equation for the line is then

$y-4=5(x-0)\implies y=5x+4$