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Line Tangent ro a Parabola

by soul5
Tags: derivatives
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soul5
#1
Mar23-08, 05:16 PM
P: 64
1. The problem statement, all variables and given/known data
Find the point on the parabola y= 4x^2 + 2x - 5 where the tangent line is perpendicular to the line 3x + 2y = 7.


2. Relevant equations



3. The attempt at a solution
I don't know what to do since I was away the last 3 classes since I was away. Help me please.
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TMM
#2
Mar23-08, 05:25 PM
P: 96
You want to find the slope of the line you're given and use the definition of a derivative as well as perpendicularity to solve for the x value you want.
Snazzy
#3
Mar23-08, 05:27 PM
P: 459
Two perpendicular lines have slopes that are negative reciprocals of each other, eg: a line with a slope 2 is perpendicular to a line with a slope -1/2.

Find the slope of the line, find the negative reciprocal of that slope. The derivative of a function is the slope of that graph at any point on the graph, so find the derivative of the parabola and see at what value of x it will equal the negative reciprocal of the slope you found earlier.

soul5
#4
Mar23-08, 05:39 PM
P: 64
Line Tangent ro a Parabola

Quote Quote by TMM View Post
You want to find the slope of the line you're given and use the definition of a derivative as well as perpendicularity to solve for the x value you want.
So I take the slope of this? 3x + 2y = 7

so.....

2y = -3x + 7
y= -3/2x + 7/2

slope = -3/2 so if it is perpendicular the slope is 2/3 is that my right slope?

I now I have to do more but it that right so far?
Snazzy
#5
Mar23-08, 05:41 PM
P: 459
Yes.
BrendanH
#6
Mar24-08, 01:51 PM
P: 63
Correct.

dy/dx = 8x+2
You want the value of x when dy/dx is (2/3), as you said from above.
Solving for x gets (-1/6).
Plug this value into your original equation y=4x^2 etc.


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