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November 6th, 2016, 12:21 AM   #1
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Is there a 3rd non-calculus approach to this problem?

Here is the problem.

(ex) Determine the y-intercept of a line with a slope of -2 that is tangent to the curve y = x^2/3 - 4x - 3.

I solved it in two ways. I am going to write them down.

(1) with calculus

dy/dx = 2x/3 - 4

Since m=-2 and dy/dx=m, solving for x gives x=3. Then, y=-12 when x=3.

Next, I set up the slope equation between the y-intercept and the point of tangency.

(y + 12)/(0 - 3) = -2.

Solving for y gives y-intercept = -6.

(2) without calculus

Equate y=x^2/3 - 4x - 3 with y=-2x + b.

I get -2x + b = x^2/3 - 4x - 3. Arranging the terms gives x^2/3 - 2x - 3 - b.

Next, I apply the discriminant and set it equal to zero.

4 + 4(3 + b)/3 = 0.

Solving for b gives b = -6.

I got the right answer using both methods. My friend doesn't know calculus and he doesn't understand my 2nd method.

Does anyone have a 3rd way to solve it without using calculus? Thanks a lot.
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November 6th, 2016, 01:59 AM   #2
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Your method w/o calculus only works because $y(x)$ has a constant second derivative and thus you are guaranteed that the tangent line will only intersect the curve at one point.

Without that guarantee your trick setting the discriminant to $0$ won't work.

Nevertheless this curve satisfies this constraint and the geometry is pretty simple to understand.

I'd just keep working at making your friend understand your method. It's not that difficult.

Just make sure they understand the connection between intersection of the tangent line and the curve at one point and the discriminant being equal to $0$.
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November 6th, 2016, 06:59 AM   #3
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I am not supposed to use the discriminant in this problem when explaining it to my friend since it will be covered next month. He is expected to solve the problem without jumping ahead of the course. Without the use of either calculus or the discriminant, I have no idea of how to solve it in another way.

Is it impossible to find another way to solve it?
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November 6th, 2016, 07:19 AM   #4
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I am not sure. I have an idea I am working on, but it will take me a while to see if it is viable.
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November 6th, 2016, 07:56 AM   #5
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$\dfrac{x^2}{3} - 4x - 3 = -2x+b$

$\dfrac{x^2}{3} - 2x = 3+b$

$x^2 - 6x = 3(3+b)$

complete the square ...

$x^2 - 6x + 9 = 3(3+b) + 9$

$(x - 3)^2 = 3(b+6)$

$x = 3 \implies b = -6$
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November 6th, 2016, 08:09 AM   #6
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Quote:
Originally Posted by skeeter View Post
$\dfrac{x^2}{3} - 4x - 3 = -2x+b$

$\dfrac{x^2}{3} - 2x = 3+b$

$x^2 - 6x = 3(3+b)$

complete the square ...

$x^2 - 6x + 9 = 3(3+b) + 9$

$(x - 3)^2 = 3(b+6)$

$x = 3 \implies b = -6$
This shows where they intersect. Why does this imply the line is tangent to the curve?
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November 6th, 2016, 08:23 AM   #7
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Quote:
Originally Posted by romsek View Post
This shows where they intersect. Why does this imply the line is tangent to the curve?
I understood it was given in the problem statement that the line with the given slope is tangent to the curve ... maybe I'm wrong?

Quote:
Here is the problem.

(ex) Determine the y-intercept of a line with a slope of -2 that is tangent to the curve y = x^2/3 - 4x - 3.
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November 6th, 2016, 11:14 AM   #8
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My problem begins where skeeter's ends. What does a student who apparently knows nothing of quadratic functions even mean by a tangent to a curve at a point?

I keep speculating that a simple (though probably not rigorous) definition of the tangent to a curve at a point will provide an answer dependent on neither calculus nor knowledge of quadratics. I have not been able to validate or invalidate that speculation as yet.
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November 6th, 2016, 04:24 PM   #9
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interesting paper on the topic ...

notes on tangents to parabolas
Thanks from davedave and Joppy
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