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Finding Slope of a Tangent Line to a Parabola

by Cascadian
Tags: line, parabola, slope, tangent
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Cascadian
#1
Jan14-13, 05:01 PM
P: 6
1. The problem statement, all variables and given/known data
I've got the equation of a parabola [itex]y=2x^2-4x+1[/itex] with point (-1,7) and a tangent line running through it the point. I'm supposed to find the equation of the line. Simultaneously solve this equation with that of the parabola, place the results in form [itex]ax^2+bx+c[/itex], and find the slope of the tangent line.


2. Relevant equations
[itex]y=2x^2-4x+1[/itex]
[itex]y=m(x--1)+7[/itex]
[itex]ax^2+bx+c[/itex]

3. The attempt at a solution
I was supposed to find the equation of the line using the point slope equation and I did, I placed it above. The problem lies when I try to set the equations equal to each other [itex]m(x+1)+7=2x^2-4x+1[/itex]and place the results in [itex]ax^2+bx+c[/itex] form. I guessed that [itex]a=2[/itex] and it was correct. However b is not [itex]-4x-mx[/itex] and c is not [itex]m-6[/itex]
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CompuChip
#2
Jan14-13, 05:35 PM
Sci Advisor
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P: 4,300
It looks like you are on the right track, but have some trouble with your bookkeeping.

If you have m(x + 1) = mx + m = 2x - 4x + 1, start by bringing everything to one side of the equals sign: 2x - 4x - mx + 1 - m = 0.
Now carefully compare this to the given form, ax + bx + c. Try rewriting the equation to get this: 2x + (....)x + (...) = 0.
You will be able to read off b and c, but this time with the correct signs :)

(Also, don't forget, as I initially did, that it stays an equation -- after the rewrite there will be "= 0" on the right hand side).
symbolipoint
#3
Jan14-13, 09:44 PM
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I wish using a little differentiation were justified here. for the y=parabola,
d/dx of 2x^2-4x+1 is 4x-4.

Value of derivative when x=-1 becomes -8, so slope is -8 for the line.

Now we have both the (given) point, and the slope of the line.

I just do not see the less advanced algebra trick to solve the question.

Mentallic
#4
Jan14-13, 11:52 PM
HW Helper
P: 3,562
Finding Slope of a Tangent Line to a Parabola

Quote Quote by symbolipoint View Post
I wish using a little differentiation were justified here. for the y=parabola,
d/dx of 2x^2-4x+1 is 4x-4.

Value of derivative when x=-1 becomes -8, so slope is -8 for the line.

Now we have both the (given) point, and the slope of the line.

I just do not see the less advanced algebra trick to solve the question.
Since the line [itex]y=m(x+1)+7[/itex] passes through the point (-1,7) which we know is on the parabola, if we choose any real gradient other than the tangential gradient, it'll cut the parabola twice, while the tangent will cut the parabola once.


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