How do you find the equation of the tangent line to the curve #y=x^2+2e^x# at (0,2)?

Question

Scarlett Allison · Accepted Answer

#y=2x+2# #"we use the formula "y-y_1=m(x-x_1)" where:" # #(x_1,y_1) " is a known co-ordinate on the line"# #"& "m=" the gradient of the tangent"# #y=x^2+2e^x " at "(0,2)# #m=[(dy)/(dx)]_(x=0)# #y=x^2+2e^x=>(dy)/(dx)=2x+2e^x # #m=[(dy)/(dx)]_(x=0)=2xx0+2e^0# #m=2# #:."eqn of tgt. "y-y_1=m(x-x_1)# #y-y_1=m(x-x_1)# #y-2=2(x-0)# #y-2=2x=>y=2x+2#

Ezra Adams · Answer

undefined To find the equation of the tangent line to the curve y=x^2+2e^x at (0,2), we need to find the slope of the tangent line at that point.

First, we find the derivative of the curve y=x^2+2e^x with respect to x.

The derivative of x^2 is 2x, and the derivative of 2e^x is 2e^x.

So, the derivative of y=x^2+2e^x is dy/dx = 2x + 2e^x.

Next, we substitute x=0 into the derivative to find the slope at (0,2).

dy/dx = 2(0) + 2e^0 = 2(1) = 2.

Therefore, the slope of the tangent line at (0,2) is 2.

Using the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, we can substitute the values to find the equation of the tangent line.

Plugging in the values, we have y - 2 = 2(x - 0).

Simplifying, we get y - 2 = 2x.

Rearranging the equation, we have y = 2x + 2.

Therefore, the equation of the tangent line to the curve y=x^2+2e^x at (0,2) is y = 2x + 2.