# Let f(x) = \frac{2}{x^2}. a. Find the slope of the tangent line to the graph of this function at...

## Question:

Let {eq}\displaystyle f(x) = \frac{2}{x^2} {/eq}.

a. Find the slope of the tangent line to the graph of this function at {eq}x = 3 {/eq}.

b. Find the equation of the tangent to the graph of this function at {eq}x = 3 {/eq}.

## Tangent Line

We can use the first derivative to find the instantaneous rate of change of a function. Since the rate of change is a slope, we can use it to construct the line that has the same instantaneous slope.

## Answer and Explanation:

a. In order to find the slope of the tangent line, we need to take the first derivative. We can rewrite our function using a negative exponent so that we can easily use the Power Rule.

{eq}f(x) = 2x^{-2}\\ f'(x) = -4x^{-3} {/eq}

By evaluating this at x = 3, we will find the slope of the tangent line to that point.

{eq}f'(3) = -4(3)^{-3} = -\frac{4}{27} {/eq}

b. In order to use this slope to construct the tangent line, we need both the x and y coordinate of the point on the original function where we want this line to touch. We have the x value, so we can evaluate our function at this point to find the y value.

{eq}f(3) = \frac{2}{(3)^2} = \frac{2}{9} {/eq}

Now that we have both a point and a slope, we can construct the tangent line using point-slope form.

{eq}y - \frac{2}{9} = -\frac{4}{27} (x - 3)\\ y = -\frac{4}{27} x + \frac{4}{9} + \frac{2}{9}\\ y = -\frac{4}{27} x + \frac{2}{3} {/eq}

Thus, the tangent line to this function at the given point is {eq}y = -\frac{4}{27} x + \frac{2}{3} {/eq}.

#### Learn more about this topic:

Tangent Line: Definition & Equation

from NY Regents Exam - Geometry: Tutoring Solution

Chapter 1 / Lesson 11
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