Question

Okay, guys, I need confirmation, very important:
I need to find an equation of the tangent line to the parabola y=x^3-5x+1. Now, I know the slop at f'(1) = -2. Using y2-y1=m(x-x1),

Answer 1

I am not sure what it gives me. I need to be 100% sure before I submit anything. And I don't think my answer to y = ? is going to be correct.

Answer 2

The points are (1,-3)

Answer 3

Are you finding the tangent line with the delta-x limit definition, or deriving it? With what you are saying im guessing that you are using delta-x limit.

Answer 4

I derived then plugged in f'(1) like they asked, they then asked me to use it to find what I had written up there.

Answer 5

I had gotten some answers which turned out to be wrong, so I can't afford to be wrong again due to the way I submit my HW.

Answer 6

Never mind you supplied the slope at f(1). So now you have the right slope, and the 2 points, you need to plug it into:
\[y-y _{1}=m(x-x _{1})\]
where y sub 1 is the y value at f(1)
and x sub 1 is the x value at f(1)

Answer 7

That is the point slope formula. You take a slope, and a point, and it makes the line for you. So, I'll set it up for you:
\[y - (-3) = -2 (x - 1)\]

Answer 8

Now just simplify for the equation for the line.

Answer 9

y = -2x+2-3 = -2x-1?

Answer 10

Correct

Answer 11

I see where I made my mistake. I didn't account for the negative sign already included in the formula. Thank you very much.