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)

Question

asked Feb 22, 2021 195k views

1 Answer

Arun Yokesh · Answer 1 · 2021-02-28T02:41:40+0000

Answer:

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