How do you find the equation of the tangent and normal line to the curve
#y=9x^-1# at x-3?

Question

How do you find the equation of the tangent and normal line to the curve 
#y=9x^-1# at x-3?

Isaiah Allen · Accepted Answer

I'm not sure if you meant #x=3# or #x=-3#

If #x=3# Then
Tangent is # y=-x+6 # and Normal is # y=x #

If #x=-3# Then
Tangent is # y=-x-6 # and Normal is # y=x #

I'm not sure if you meant #x=3# or #x=-3# so let's do both:

First we need the derivative, which will give us the slope of the tangent at any point:

# y=9/x = 9x^-1 => dy/dx = 9(-x^-2) #

# :. dy/dx = -9/x^2 #

We will use the formula # y-y_1=m(x-x_1)# to find the tangents and normals, and use the fact that the tangents and normals are [perpendicular (so their product is -1)

When x=3
# x=3 => y=3 ; y'=-9/9=-1#

Tangent:
Passes through #(3,3)# and has slope #m=-1#
# :. y-3=-1(x-3) #
# :. y-3=-x+3 #
# :. y=-x+6 #

Normal:
Passes through #(3,3)# and has slope #m=1#
# :. y-3=1(x-3) #
# :. y-3=x-3 #
# :. y=x #

When x=-3
# x=-3 => y=-3 ; y'=-1#

Tangent:
Passes through #(-3,-3)# and has slope #m=-1#
# :. y-(-3)=-1(x-(-3)) #
# :. y+3=-x-3 #
# :. y=-x-6 #

Normal:
Passes through #(-3,-3)# and has slope #m=1#
# :. y-(-3)=1(x-(-3)) #
# :. y+3=x+3 #
# :. y=x #

Adam Adkins · Answer

undefined To find the equation of the tangent line to the curve y = 9x^(-1) at x = 3, we first find the derivative of the function y with respect to x, which is dy/dx = -9x^(-2).

Then, evaluate dy/dx at x = 3 to find the slope of the tangent line:

dy/dx = -9(3)^(-2) = -9/9 = -1.

So, the slope of the tangent line is -1 at x = 3.

Now, we use the point-slope form of a line to find the equation of the tangent line. Given the point (3, 9/3) on the curve, and the slope -1, the equation of the tangent line is:

y - y₁ = m(x - x₁), where (x₁, y₁) = (3, 9/3).

Substitute the values into the equation:

y - 3 = -1(x - 3).

Now, simplify the equation:

y - 3 = -x + 3.

y = -x + 6.

This is the equation of the tangent line.

To find the equation of the normal line, we use the fact that the slope of the normal line is the negative reciprocal of the slope of the tangent line. So, the slope of the normal line is 1.

Using the point-slope form again, with the point (3, 9/3) and slope 1, we get:

y - 3 = 1(x - 3).

Simplify:

y - 3 = x - 3.

y = x.

This is the equation of the normal line.

Charles Anctil · Answer

undefined To find the equation of the tangent line to the curve $ y = 9x^{-1} $ at $ x = 3 $, you first need to find the derivative of the function with respect to $ x $. Then, evaluate the derivative at $ x = 3 $ to find the slope of the tangent line. Next, use the point-slope form of a line with the point $ (3, 3) $ and the slope you found to write the equation of the tangent line. The normal line will have a slope that is the negative reciprocal of the slope of the tangent line, and you can use the same point to find the equation of the normal line.

How do you find the equation of the tangent and normal line to the curve #y=9x^-1# at x-3?