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`