Finding Slope of a Tangent Line to a Parabola

  • Thread starter Cascadian
  • Start date
  • #1
6
0

Homework Statement


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.


Homework Equations


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

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]
 

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
4,302
47
It looks like you are on the right track, but have some trouble with your bookkeeping.

If you have m(x + 1) = mx + m = 2x² - 4x + 1, start by bringing everything to one side of the equals sign: 2x² - 4x - mx + 1 - m = 0.
Now carefully compare this to the given form, ax² + bx + c. Try rewriting the equation to get this: 2x² + (....)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).
 
  • #3
symbolipoint
Homework Helper
Education Advisor
Gold Member
5,999
1,094
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.
 
  • #4
Mentallic
Homework Helper
3,798
94
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.
 

Related Threads on Finding Slope of a Tangent Line to a Parabola

  • Last Post
Replies
5
Views
2K
Replies
3
Views
2K
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
3
Views
2K
Replies
1
Views
6K
  • Last Post
Replies
3
Views
8K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
8K
  • Last Post
Replies
1
Views
4K
Top