Tangent To A Function - Limits

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The discussion revolves around finding the area of a triangle formed by the tangent line to the function y=3x(x-3) at the point P(2,-6) and the coordinate axes. The slope of the tangent at this point is determined to be 3, leading to the equation of the tangent line y=3x-12. The y-intercept is calculated as 12, and the x-intercept is found to be 4. Consequently, the base and height of the triangle are established, allowing for the area to be calculated as 24 square units. The solution confirms the correctness of the calculations and the area derived.
Plutonium88
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Homework Statement



The tangent to the function y=3x(x-3) at point P(2,-6) is the hypotenuse of a right triangle that forms with the coordinate axes. Find Area


The Attempt at a Solution


First of all, i know that i A=BxH/2 so i need the opposite and adjacent sides of this triangle.

Okay so personally i graphed the function at first, and then drew the tangent at the point 2,-6. I then connected the line to the x and y axes. I did this because it says it forms a triangle with the coordinate axes.I read that the coordinate axis are respectively the x and y axis, so i thought that this made sense. Can some one reassure me about this?

Next i thought that i would need the slope of the tangent at that point because it would relate to the slope of hypotenuse of the triangle.

So i took the limit.

Lim (f(x+h)-f(x))/H
H->0

where P(2,-6), so x=2

solving this limit i got

Lim 6x+h-9 = 6x-9
H->0
and when x=2
the limit is = 3

therefore the slope of the tangent is 3. mt=3

so I'm curious, does this mean that the hypotenuse also = 3 because it says the the tangent at the point is the hypotenuse. Or would the hypotenuse side be the tangent line (6x-9)?

So yea this is my dilemma..
 
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The line through (2,-6) with slope 3 is the hypotenuse. Figure out its equation (it isn't y = 6x-9) and find where its x and y intercepts are to get the legs of the triangle.
 
LCKurtz said:
The line through (2,-6) with slope 3 is the hypotenuse. Figure out its equation (it isn't y = 6x-9) and find where its x and y intercepts are to get the legs of the triangle.

okay say i have the point (2,-6) and mt=3

here is my diagram.
http://s18.postimage.org/3r9iyue7d/New_Bitmap_Image.png

y= mx+b
-6=3(2)+b
b=-12

So therefore the height or the y value of the triangle is 12. cause it is the y intercept

y=3x-12

to find x intercept, set y=0

3x = 12
x= 4

So there for the base of the triangle is 4, and the height of the triangle is twelve.

so area=bxh/2=12x4/2=24units^2
 
Last edited by a moderator:
Plutonium88 said:
okay say i have the point (2,-6) and mt=3

here is my diagram.
http://s18.postimage.org/3r9iyue7d/New_Bitmap_Image.png

y= mx+b
-6=3(2)+b
b=-12

So therefore the height or the y value of the triangle is 12. cause it is the y intercept

y=3x-12

to find x intercept, set y=0

3x = 12
x= 4

So there for the base of the triangle is 4, and the height of the triangle is twelve.

so area=bxh/2=12x4/2=24units^2

Yes. You have it correct.
 
Last edited by a moderator:
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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