How do you find the equation of a line tangent to #y=4/x# at (2,2)?

Question

Luke Alvarez · Accepted Answer

We know that the tangent passes through the point #(2,2)#, so we only need to find out the slope of the tangent at that point.

But the slope of the tangent of #y# at a given point is the value of the first derivative at that point. Now, for this function: #y'=-4/x^2#, and so evaluating at #x=2# we get #y'(2)=-4/(2^2)=-1# Then we now know that the tangent is #y=(-1)*x+b#, and since the point #(2, 2)# is on the line we also know that: #2=(-1)*2+b#. Then #b=4# and the equation of the tangent is: #y=-x+4#

Emily Ammerman · Answer

undefined To find the equation of a line tangent to the curve y = 4/x at the point (2,2), we can use the concept of differentiation.

First, we need to find the derivative of the function y = 4/x. The derivative of 4/x can be found using the power rule of differentiation, which states that the derivative of x^n is n*x^(n-1). Applying this rule, we get:

dy/dx = d(4/x)/dx = -4/x^2

Now, we can find the slope of the tangent line at the point (2,2) by substituting x = 2 into the derivative:

m = dy/dx = -4/(2^2) = -1

The slope of the tangent line is -1.

Next, we can use the point-slope form of a linear equation to find the equation of the tangent line. The point-slope form is given by:

y - y1 = m(x - x1)

Substituting the values of the point (2,2) and the slope (-1), we have:

y - 2 = -1(x - 2)

Simplifying, we get:

y - 2 = -x + 2

Rearranging the equation, we obtain the equation of the tangent line:

y = -x + 4