Find equations of all lines having slope -1 that are tangent to the curve y = 1/x+5.

Answer 1

y = 1/x +5

y' = -1/x^-2 = -1

1/x^2 = 1

x^2 = 1

x = + or - 1

y =1/1 +5 = 6 and 1/-1 +5 = 4

y = 4 or 6

the points with slope = -1 are (1,6) and (-1,4)

equations for the tangent lines are

y -6 = -(x-1) in point slope form

or

x+y = 7 in standard form

and

y-4 = -(x+1) in point slope form

or

x+y = 3 in standard form

that assumes you meant y = (1/x) + 5

but if you meant y = 1/(x+5), the tangent lines and equations are different

f(x) is a hyperola, with two separate branches, either way, and two points with the slope -1

two lines no matter how you interpret the problem

f(x) = 1/(x+5)

f'(x) = y' = -1/(x+5)^2 =-1

1/(x+5)^2 = 1

x+5)^2 = 1

x+5 = + or -1

x = -6 or -4

the two points are then

(-4, 1) and (-6,-1)

the lines are

y-1 =-(x+4) in point slope form

or

x+y = -3 in standard form

and

y +1 = -(x+6) in point slope form

or

x+y = -7 in standard form

it might help to graph f(x) and look where the slope of the curve is at about a -45 degree angle with the x axis. That's what slope -1 means. try to locate, at least roughly what the points are. Use a graphing calculator, online or handheld

Answer 2

y' = - 1/x² = - 1; x² = 1; x = ±1

x = 1; y = 6; y = 6 - (x - 1) or y = 7 - x

x = - 1, y = 5, y = 5 - (x + 1); y = 4 - x.