need to find equation for line tangent at the indicated point?

Question

f(X)= 64/x @(2,32)

Answer 1

f (x) = 64 x^(-1)
f `(x) = - 64 x^(-2)
f `(x) = - 64 / x²
f `(2) = - 64 / 4 = - 16
m = - 16
y - 32 = (-16) (x - 2)
y = (-16) x + 64

Answer 2

First you need to know the point-slope form of the equation of a straight line is y-y1 = m(x-x1) . You already know that (x1,y1) is the point (2,32). The slope at a point is given by the derivative at that point so find d(64/x)/dx = d(64x^-1)/dx = -64/x^2. Substitute 2 for x in this derivative to find the slope at the point: -64/2^2 =
-16. Now put all these values in the equation of the line to get

y-32 = -16(x-2) and simplify to y = -16x +64. Notice when you let x=2 in this equation you get y=32 so this line does go through the point (2,32) and it has the slope we calculated at the point.

Answer 3

The slope of the line=slope of the function at the tangent. So f'(x) = -64/x^2. When x=2, f'(x)= -16
So tangent line will have equation y= -16x + B
Since line also includes (2,32), B can be solved for, and equals 64. The line is y=-16x+64.

Answer 4

Look in a book!