#### Answer

$y=−2x+6$

#### Work Step by Step

The question asks us to find the equation of the tangent of the function
$f(x)=\frac{8}{x^2}$ at the point (2,2)
Given that we are looking for the equation of a tangent, the first reasonable step would be to find the gradient at the points given, which can be found using the derivative function.
Using the power rule:
$f(x)=\frac{8}{x^2}=8x^{-2}$
$f′(x)=(-2)8x^{-2-1}=-16x^{-3}=\frac{-16}{x^{3}}$
$f′(2)=−2$
Thus, the gradient of that function at the value x=2 is -2.
Using this information, we can find the intercept value of the tangent line.
The equation of the gradient is given by:
$\frac{y_{1}−y_{0}}{x_{1}−x_{0}}=m$
We can now use the points (2,2) and (0,c) (where c is the y-intercept).
$\frac{2−c}{2-0}=-2$
Solving for c gives us c=6
Thus, the equation of the tangent at (3,3) is $y=−2x+6$