Tangent intersection question

[8monday]

Find the coordinates of the points on the graph of y=f(x) such that the tangents to the graph at these points intersect at [math](-1,\frac{7}{2})[/math]. [math]f(x)=\frac{1}{2x-4}+3[/math]
If anyone knows the process I can take to solve this problem, or any ideas on where I can start, I would really appreciate it. Problem from VCAA (VCE Mathematical Methods 2012 Exam 2)
Thank you!

 

[khansaheb]

Find the coordinates of the points on the graph of y=f(x) such that the tangents to the graph at these points intersect at [math](-1,\frac{7}{2})[/math]. [math]f(x)=\frac{1}{2x-4}+3[/math]
If anyone knows the process I can take to solve this problem, or any ideas on where I can start, I would really appreciate it. Problem from VCAA (VCE Mathematical Methods 2012 Exam 2)
Thank you!

1) Assume that the points of tangencies are S (x_s , y_s) and V (x_v, y_v)
2) Given point of intersection(I) of tangents is (-1, 0.5)
3) Calculate the equation of the tangent lines (SI and VI) and the co-ordinates of their point of intersection.
4) Continue.... if still stuck share your work.

 

[Aion]

1) Assume that the points of tangencies are S (x_s , y_s) and V (x_v, y_v)
2) Given point of intersection(I) of tangents is (-1, 0.5)
3) Calculate the equation of the tangent lines (SI and VI) and the co-ordinates of their point of intersection.
4) Continue.... if still stuck share your work.

I think there’s a mistake — the intersection point should be [imath](-1, \tfrac{7}{2})[/imath], not [imath](-1, 0.5)[/imath]. Also, since that point is already given, I don’t see why we would calculate it again in step 3. Shouldn’t the idea be to use that point to relate the slope of the tangent to the point of tangency?

 

[khansaheb]

the intersection point should be (−1,72)(-1, \tfrac{7}{2})(−1,27), not (−1,0.5)(-1, 0.5)(−1,0.5)

Correct .... That was a mistake .

From step (3), equations relating S (x_s , y_s) and V (x_v, y_v) can be obtained and then use the numerical values that has been given.

 

[Aion]

Correct .... That was a mistake .

From step (3), equations relating S (x_s , y_s) and V (x_v, y_v) can be obtained and then use the numerical values that has been given.

Thanks, I think I understand now. I found it clearer to think in terms of a single point ([imath]x=a[/imath]), and then use that the slope of the tangent [imath]f'(a)[/imath] must equal the slope from [imath](a, f(a))[/imath] to [imath](-1, \tfrac{7}{2})[/imath]. That leads directly to an equation in [imath]a[/imath], and the number of solutions explains why there are two points.

 

[jonah2.0]

Beer drenched reaction follows.

Find the coordinates of the points on the graph of y=f(x) such that the tangents to the graph at these points intersect at [math](-1,\frac{7}{2})[/math]. [math]f(x)=\frac{1}{2x-4}+3[/math]
If anyone knows the process I can take to solve this problem, or any ideas on where I can start, I would really appreciate it. Problem from VCAA (VCE Mathematical Methods 2012 Exam 2)
Thank you!

The age of AI is upon us.
You could, in addition to posts # 2 to 5, also ask the Google AI god or any number of AI gods around.