# How do you find the line to the tangent to the curve #y=9x^-2# at the point (3,1)?

The gradient of the tangent to a curve at any particular point is give by the derivative of the curve at that point.

so If

#dy/dx = -18x^-3#

When

and

So the tangent we seek passes through

# \ \ \ \ \ y-1=-2/3(x-3) #

# :. y-1=-2/3x+2#

# :. \ \ \ \ \ \ \ y=-2/3x+3 #

We can confirm this solution is correct graphically:

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To find the equation of the tangent line to the curve y=9x^-2 at the point (3,1), we need to determine the slope of the tangent line at that point.

First, we find the derivative of the function y=9x^-2 with respect to x. The derivative of y=9x^-2 is dy/dx = -18x^-3.

Next, we substitute x=3 into the derivative to find the slope at the point (3,1). dy/dx = -18(3)^-3 = -18/27 = -2/3.

Therefore, the slope of the tangent line at the point (3,1) is -2/3.

Using the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, we substitute the values into the equation.

y - 1 = (-2/3)(x - 3)

Simplifying the equation, we get:

y - 1 = (-2/3)x + 2

Rearranging the equation, we find the equation of the tangent line to be:

y = (-2/3)x + 3

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