Answer :
Answer:
[tex] y= -2e^{-2} (x-2) = -2e^{-2} x +(4e^{-2}+1) [/tex]
Where the slope is [tex] m = -2e^{-2}[/tex] and the intercept [tex] b=4e^{-2}+1[/tex]
Step-by-step explanation:
For this case w ehave the following function [tex] y = e^{-2x}[/tex] and a point givn P(2,1)
In order to find an equation of the tangent line to the graph of the function given at the point P first we need to find the derivat eof our original equation.
[tex] \frac{dy}{dx} =\frac{d}{dx} (e^{-2x}) [/tex]
[tex] \frac{dy}{dx} = -2 e^{-2x}[/tex]
And this value represent the slope for our tangent line at the given point, we can replace the value of x and we got:
[tex] m = \frac{dy}{dx} = -2 e^{-2(1)}= -2e^{-2}=-0.2708[/tex]
Now the general equation for the tangent line is given by:
[tex] y -y_0 = m (x-x_0)[/tex]
And for this case [tex] x_0 = 2, y_0 = 1[/tex] and if we replace we got:
[tex] y -1 = -2e^{-2} (x-2) = -2e^{-2} x +4e^{-2}[/tex]
And if we simplify we got:
[tex] y= -2e^{-2} (x-2) = -2e^{-2} x +(4e^{-2}+1) [/tex]
Where the slope is [tex] m = -2e^{-2}[/tex] and the intercept [tex] b=4e^{-2}+1[/tex]