Question

Finding an Equation of a Tangent Line In Exercise, find an equation of the tangent line to the graph of the function at the given point.

y = e^2 - x, (2, 1)

Answer 1

Equation of tangent to the line will be
`2e^2x-4e^2-2x+5`

Explanation:

We have given the equation of the line
`y=e^2-x`

Now slope of tangent will be equal to
`(dy)/(dx)=e^2x-(x^2)/(2)`

Slope at point (2,1)

So
`(dy)/(dx)=e^2* 2-(2^2)/(2)=2e^2-2`

Now equation of line is given as
`y-y_1=m(x-x_!)`

So
`y-1=(2e^2-2)(x-2)`

`y=(2e^2-2)(x-2)+1=2e^2x-4e^2-2x+4+1=2e^2x-4e^2-2x+5`