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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^-2x + x2 , (2 , 1)

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Answer:


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

Where the slope is
m = -2e^(-2) and the intercept
b=4e^(-2)+1

Explanation:

For this case w ehave the following function
y = e^(-2x) 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.


(dy)/(dx) =(d)/(dx) (e^(-2x))


(dy)/(dx) = -2 e^(-2x)

And this value represent the slope for our tangent line at the given point, we can replace the value of x and we got:


m = (dy)/(dx) = -2 e^(-2(1))= -2e^(-2)=-0.2708

Now the general equation for the tangent line is given by:


y -y_0 = m (x-x_0)

And for this case
x_0 = 2, y_0 = 1 and if we replace we got:


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

And if we simplify we got:


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

Where the slope is
m = -2e^(-2) and the intercept
b=4e^(-2)+1

User DanielNolan
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