Finding Coordinates with same gradient

[TheAkuma]

Okay, I have two points on my graph - (-2,72) and (0,64). Here is the question on my assignment - "Find another two points on your curve which have the same gradient as those in Parts 3 and 4 and find the equations of the tangents to the curve at these points."
The problem I have is I don't know any methods that will find a point with the same gradient. I can do the rest of the question, but I can't find the points as my teacher never showed me how to.

I could find it by trial and error - By picking a number on the x-axis and do some number crunching. But that would take ages and my draft is due like tomorrow!



I got those first two points by using this method -
1.) Sub value of "x" into equation y=x blah blah blah... (which would give me the y value)
2.) Sub value of "x" into the derivative to get the gradient
3.) Put all the variables into the equation y=mx+c to show the equation of the tangent at x=blah blah blah

If anyone could help, it would be much appreciated as my assignment is due tomorrow.

 

[rock.freak667]

Well if you are given two points, you can find the equation of the line passing through those two points, y=mx+c and then you can pick any x value to find the corresponding y value to get one point on the line.

But you said tangents to a curve, so I am not sure I am understanding your dilemma properly.

 

[TheAkuma]

Well if you are given two points, you can find the equation of the line passing through those two points, y=mx+c and then you can pick any x value to find the corresponding y value to get one point on the line.

But you said tangents to a curve, so I am not sure I am understanding your dilemma properly.

I honestly don't know either. I'm suppose to find two points with the same gradient as the ones I got. In the next question, I'm suppose to connect the coordinates to make a parallelogram or just a four sided figure. But it says they have to have the same gradient. The equation I'm using is y=x³-4x²-16x+64 if that helps.

 

[rock.freak667]

well in that case gradient at (-2,72)=m (you found m)

dy/dx = m you will get two x values.

Do the same for the next point (0,64)

 

[TheAkuma]

well in that case gradient at (-2,72)=m (you found m)

dy/dx = m you will get two x values.

Do the same for the next point (0,64)

Oh yeah! that dy/dx stuff. I never really understood it although I wrote like 4 pages worth of notes on the stuff. I'll go look back at them and see if I can solve it. Thanks

 

[TheAkuma]

Damn it! Hey, I don't know how to work it out. There is nothing like that in my notes. The gradient for (-2,72) is 12. Can you at least tell me the method when using that dx/dy stuff please?

 

[TheAkuma]

don't worry about it. I learned how to do it by adding the gradient to the derivative of the quadtratic formula then factorizing it. I got two "x" values, the original and the new one. Anyway, thanks for all your help. I'll try and contribute as much as I can to physics forums.

 

[CaffeineJunky]

Step 1: Determine the Gradient of your function.
Step 2: Evaluate the Gradient at the given points.
Step 3: Find the set of all points whose gradient is the same value (i.e. set the gradient equal to the values given above and solve for possible values of x. You need only produce one, but it is a good idea to try and describe each of them, if possible.)

Now, the gradient is perpendicular to the level surfaces of your original function. Hence the tangent line should be perpendicular to your gradient. It is easy to find an equation for the line if you know a vector that is perpendicular to the line.

 

[Mark44]

Step 1: Determine the Gradient of your function.
Step 2: Evaluate the Gradient at the given points.
Step 3: Find the set of all points whose gradient is the same value (i.e. set the gradient equal to the values given above and solve for possible values of x. You need only produce one, but it is a good idea to try and describe each of them, if possible.)

Now, the gradient is perpendicular to the level surfaces of your original function. Hence the tangent line should be perpendicular to your gradient. It is easy to find an equation for the line if you know a vector that is perpendicular to the line.

My guess is that by "gradient" TheAkuma really means the slope of a function of one variable. If that's the case, the advice about gradient and level curves is overkill, especially given that he posted this problem in the Precalc forum.

 

[TheAkuma]

My guess is that by "gradient" TheAkuma really means the slope of a function of one variable. If that's the case, the advice about gradient and level curves is overkill, especially given that he posted this problem in the Precalc forum.

Oops. Sorry about that, I just didn't know where to post it.

 

[CaffeineJunky]

I apologize for that, I didn't realize that this was the pre-calculus forum until after I posted that. Either way, it seems kind of awkward to use gradient in place of slope when talking about functions of a single variable. Disregard my post.