Question

Write an equation for the tangent line to f(x) = (2x) - (3/x) at x=-2

I have around 9 more of these kinds of questions I need to solve. Please show some work.

Thank you!

Answer 1

You will have to take the derivative of f, and then calculate f'(-2).
Do you know how to do this?

Answer 2

Would the derivative be 2 - (3/x^2)
?

Answer 3

Yes, it would!

The equation you are looking for, has the form y = ax + b.
Here, the slope a of the line is also f'(-2), so plug in this value.

Answer 4

11/4 ?

Answer 5

f'(-2) = 2 - 3/4 = 1 1/4, oh I see now what you mean. (1.25)

So you have y = 1.25x + b.
To find b, you will have to know a point that is on the line. That point is (-2, f(-2)), because that is the point where the tangent line "touches" the graph of f.
Calculate f(-2). Then put the x and y in y=-2x + b to find b. Then you're done!

Answer 6

I mean y = 1.25x + b

Answer 7

f(-2) would be -2.5

Answer 8

So (-2, -2.5) is on the line. y = 1.25x + b, so b = y - 1.25x.
Set x = -2 and y = -2.5 to get b.

Answer 9

I got b = -5

Answer 10

answer choices are -10x-5y=-13, -11-4y = -12 11x-4y=-12 10x-5y=-13

Answer 11

any idea which one?

Answer 12

Oops. Error in f'(x).
It must be 2 + 3/x^2 (see why?).
So f'(-2)=2+3/4 = 2.75
y = 2.75x + b,
b = y - 2.75x = -2.5 - 2.75*-2 = -2.5 + 5.5 = 3, so

y = 2.75x + 3 or ( multiply by 4)
4y = 11x + 12, or

11x -4y = -12

Answer 13

ah! thanks!