Finding tangent lines for 𝑓(𝑥) = 𝑥^3 − 𝑥 + 6

[ttpp1124]

Homework Statement

Consider the curve defined by 𝑓(𝑥) = 𝑥^3 − 𝑥 + 6. What is the slope of tangent(s) line to the graph of 𝑓(𝑥) from the point 𝑃 (−2, 8)?

Relevant Equations

n/a

not quite sure if this is right.. can someone confirm?

 

[andrewkirk]

It is not correct. The point (-2,8) does not lie on the curve. You need to find the slopes of the line(s) that go through that point and touch the curve somewhere at a tangent.

I suggest you start by sketching the curve, either by identifying any extrema, inflection point and limits as x goes to +/- infinity and sketching it yourself, or using Wolfram to sketch it for you. With a sketch you should be able to tell how many lines that touch the curve at a tangent there will be.

Then you need to write some equations. Set (a,b) as the (x,y) coordinates of a tangent point. You will get equations from the fact that (a,b) lies on the curve, and that the line through (-2,8) and (a,b) has the same slope as the curve at (a,b). That's two equations and two unknowns, which you can solve to find a and b. Then calculate the slope at that point.

Edit: On second thoughts, you don't need to sketch it. Just make the equations and solve. There will be one or two solutions according to whether there are one or two tangents from the point to the curve.

 

[ttpp1124]

It is not correct. The point (-2,8) does not lie on the curve. You need to find the slopes of the line(s) that go through that point and touch the curve somewhere at a tangent.

I suggest you start by sketching the curve, either by identifying any extrema, inflection point and limits as x goes to +/- infinity and sketching it yourself, or using Wolfram to sketch it for you. With a sketch you should be able to tell how many lines that touch the curve at a tangent there will be.

Then you need to write some equations. Set (a,b) as the (x,y) coordinates of a tangent point. You will get equations from the fact that (a,b) lies on the curve, and that the line through (-2,8) and (a,b) has the same slope as the curve at (a,b). That's two equations and two unknowns, which you can solve to find a and b. Then calculate the slope at that point.

Edit: On second thoughts, you don't need to sketch it. Just make the equations and solve. There will be one or two solutions according to whether there are one or two tangents from the point to the curve.

Is this better?

 

[Mark44]

@ttpp1124, both images in this thread are posted sideways. Many helpers won't bother looking at them if they have to crane their heads sideways to read what you've written.

 

[ttpp1124]

@ttpp1124, both images in this thread are posted sideways. Many helpers won't bother looking at them if they have to crane their heads sideways to read what you've written.

sorry, here's the upright version!

 

[Mark44]

Your answers for the two slopes look fine to me.