Finding Perpendicular Lines to y=0.25x and Tangent to f(x)=1/x

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Homework Help Overview

The discussion revolves around finding two straight lines that are perpendicular to the line defined by y=0.25x and tangent to the curve f(x)=1/x. Participants explore the relationships between slopes and tangents in this context.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the need to find slopes of tangent lines and their relationship to the given line's slope. There are attempts to equate the function and the line to find points of intersection, as well as questions about the implications of the slopes being perpendicular.

Discussion Status

There is ongoing exploration of the relationship between the slopes of the tangent lines and the original line. Some participants have provided equations based on their calculations, while others are questioning the accuracy of those equations and the assumptions made in the process.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can use or the methods they can apply. There is some uncertainty regarding the calculations of y-coordinates and the derivation of line equations.

fghtffyrdmns
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Homework Statement



Find two straight lines that are perpendicular to y=0.25x and tangent to the curve f(x) = 1/x.

Homework Equations



y=0.25x
f(x) = 1/x.

The Attempt at a Solution



What I did was equate y and f(x) and determined when they equal which is 2 and -2. The points are (2, 1/2) and (-2,-1/2).

Now, I took the derivative of f(x) and got -1/x^2. Would I use this to find the slope at the two points? Then I could make the equation.
 
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If the lines are perpendicular to y=(1/4)*x then what should the slopes of the tangent lines be? The slopes of the tangent lines should also be f'(x), right?
 
If the lines are perpendicular to y=0.25x, what is their slope?
 
Dick said:
If the lines are perpendicular to y=(1/4)*x then what should the slopes of the tangent lines be? The slopes of the tangent lines should also be f'(x), right?

The slope would be -4, no?

I've made the two equations of y = -4x + 17/2 and y = -4x -17/2.
 
The -4x is right, but your y-intercept is wrong. Now that you have the slope of the lines, what does the relationship have to be between that slope and f'(x) in order for the lines to be tangent to f(x)?
 
fghtffyrdmns said:
The slope would be -4, no?

I've made the two equations of y = -4x + 17/2 and y = -4x -17/2.

Yes, the slope should be -4. But how did you get those two line equations?
 
Dick said:
Yes, the slope should be -4. But how did you get those two line equations?

I equated y and f(x) to see when they intercept. I got 2 and -2. I put these two into f(x) to get the y cordinate so I can make the equation as I have the slope.
 
fghtffyrdmns said:
I equated y and f(x) to see when they intercept. I got 2 and -2. I put these two into f(x) to get the y cordinate so I can make the equation as I have the slope.

I'm still not totally clear what you are doing. But if you know the slope is -4, then f'(x) should be -4, right? What are the possibilities for x?
 
Dick said:
I'm still not totally clear what you are doing. But if you know the slope is -4, then f'(x) should be -4, right? What are the possibilities for x?

Ahhh, I equate the slope to -4. The possibilities of x are 1/2 and -1/2?
 
  • #10
Yes! If you know x=1/2 or -1/2, then you know y. So now you know x and y and the slope. Pretty easy, right?
 
  • #11
Dick said:
Yes! If you know x=1/2 or -1/2, then you know y. So now you know x and y and the slope. Pretty easy, right?

y = just 2 and -2.

Yes, sir. Dang my silly mistake :[. Thank you!
 

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