My Math Forum Is there a 3rd non-calculus approach to this problem?

 Algebra Pre-Algebra and Basic Algebra Math Forum

 November 6th, 2016, 12:21 AM #1 Senior Member   Joined: Apr 2008 Posts: 194 Thanks: 3 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.
 November 6th, 2016, 01:59 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,505 Thanks: 1374 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 $0$.
 November 6th, 2016, 06:59 AM #3 Senior Member   Joined: Apr 2008 Posts: 194 Thanks: 3 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?
 November 6th, 2016, 07:19 AM #4 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 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.
 November 6th, 2016, 07:56 AM #5 Math Team     Joined: Jul 2011 From: Texas Posts: 2,982 Thanks: 1575 $\dfrac{x^2}{3} - 4x - 3 = -2x+b$ $\dfrac{x^2}{3} - 2x = 3+b$ $x^2 - 6x = 3(3+b)$ complete the square ... $x^2 - 6x + 9 = 3(3+b) + 9$ $(x - 3)^2 = 3(b+6)$ $x = 3 \implies b = -6$
November 6th, 2016, 08:09 AM   #6
Senior Member

Joined: Sep 2015
From: USA

Posts: 2,505
Thanks: 1374

Quote:
 Originally Posted by skeeter $\dfrac{x^2}{3} - 4x - 3 = -2x+b$ $\dfrac{x^2}{3} - 2x = 3+b$ $x^2 - 6x = 3(3+b)$ 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?

November 6th, 2016, 08:23 AM   #7
Math Team

Joined: Jul 2011
From: Texas

Posts: 2,982
Thanks: 1575

Quote:
 Originally Posted by romsek 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.

 November 6th, 2016, 11:14 AM #8 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 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.
 November 6th, 2016, 04:24 PM #9 Math Team     Joined: Jul 2011 From: Texas Posts: 2,982 Thanks: 1575 interesting paper on the topic ... notes on tangents to parabolas Thanks from davedave and Joppy

 Tags 3rd, approach, noncalculus, problem

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post alikim Elementary Math 6 June 10th, 2015 09:35 AM mahjk17 Linear Algebra 4 September 20th, 2012 06:16 PM JohnA Algebra 3 March 10th, 2012 06:31 AM JohnA Algebra 2 February 19th, 2012 09:29 AM ashachar Calculus 0 July 24th, 2011 05:44 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top