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

Question

15130 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-25T20:05:56

#y=2x+2#

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

#y=x^4+2e^x->dy/dx=4x^3+2e^x#

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

#4(0)^3+2e^(0)=2#

So #m_(tan)=2#

We know the line must pass through the point: #(0,2)# so with:

#(a,b)=(0,2)#

Use:

#y-b=m(x-a)#

#y-2=2(x-0)#

#->y=2x+2#

Answer 2 · 2018-01-25T20:06:58

#y = 2x+2#

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

#dy/dx = 4x^3+2e^x#

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

#m = 4(0)^3+2e^0#

#m = 2#

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

#y = m(x-x_0)+y_0#

#y = 2(x-0)+2#

#y = 2x+2#