# What is the equation of the normal line of #f(x)=(x^2-4)-e^(x+2)# at #x=2#?

First, we take the derivative of the function:

You can manipulate this however you want to get whatever form you need.

Final Answer

By signing up, you agree to our Terms of Service and Privacy Policy

To find the equation of the normal line at a given point, we need to determine the slope of the tangent line at that point and then find the negative reciprocal of that slope.

First, we find the derivative of the function f(x) with respect to x.

f'(x) = 2x - e^(x+2)

Next, we substitute x=2 into the derivative to find the slope of the tangent line at x=2.

f'(2) = 2(2) - e^(2+2) = 4 - e^4

The slope of the tangent line at x=2 is 4 - e^4.

To find the slope of the normal line, we take the negative reciprocal of the slope of the tangent line.

Slope of the normal line = -1 / (4 - e^4)

Now, we have the slope of the normal line and the point (2, f(2)) = (2, (2^2-4)-e^(2+2)) = (2, 0 - e^4).

Using the point-slope form of a line, we can write the equation of the normal line.

y - y1 = m(x - x1)

Substituting the values, we get:

y - 0 + e^4 = (-1 / (4 - e^4))(x - 2)

Simplifying further, we have:

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

This is the equation of the normal line of f(x) = (x^2-4) - e^(x+2) at x=2.

By signing up, you agree to our Terms of Service and Privacy Policy

- What is the equation of the line normal to # f(x)=lnx-x# at # x=2#?
- What is the equation of the tangent line of #f(x) =(e^x-x)(e^x-x^2)# at #x=4#?
- How do you find the average rate of change of #f(t)= 2t + 7# from [1,2]?
- What is the equation of the line tangent to #f(x)=(3x-1)(2x+4)# at #x=0#?
- What is the slope of the line normal to the tangent line of #f(x) = xsecx-cos(2x-pi/6) # at # x= (15pi)/8 #?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7