find an equation for the tangent to the curve at the given point. F(x) (8/x^2), (2,2)?

Answer 1

First, use differentiation to determine the slope of the curve at the x-value 2. Then use the point-slope form of the equation of a line to get your answer.

y = 8/(x^2) = 8x^(-2) ==> y' = -2*8x^(-3) = -16/(x^3)

When x = 2, y' = -16/(2^3) = -16/8 = -2. So the slope at this point is -2, and point-slope form using slope of -2 and the point (2, 2) gives us y - 2 = -2(x - 2), which you may convert into any different form of the equation that you'd like.

Answer 2

f (x) = 8.x^(-2)
f `(x) = - 16 x^(-3)
f `(2) = - 16 / 2³ = - 2 = m
f (2) = 8 / 4 = 2
Tangent passes thro` (2 ,2) and m = - 2
y - 2 = - 2.(x - 2)
y - 2 = - 2x + 4
y = - 2x + 6 is equation of tangent at (2,2)

Answer 3

F(x)=8x^(-2)
F'(x)=-16x^(-3)
F'(2)=-2
You want a line with slope of -2 through (2,2)
y-2=-2(x-2)
y=-2x+6

Answer 4

diffrentiate it using rule:ax^m-1
=8x^-2
=-16x^-3
=-16/x^3
subsitute x=2 into the equation
dy/dx=f'x=m=-2
y-2=-2(x-2) then expand it.
y=-2x+6(ans)