Finding Two Tangents for a Given Point on a Quadratic Function

  • Thread starter KayVee
  • Start date
Therefore, there is no solution for the two b values for the tangents to intersect the y-axis at (4, -6).
  • #1
KayVee
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Homework Statement



The teacher proposed a problem today during a math session. He gave the function : f(x) = x2-6x+5 - and a point P(4,-6).

Our mission was to find two tangents for the given point. I thought it was easy, just differentiate the function, and insert the differentiate function for a in : y=a*x + b.

Homework Equations



function : f(x) = x2-6x+5
differentiate : f'(x) = 2x-6
tangent: y= (2x-6)*x + b
tangent eq: -6= (2x-6)*4 + b

The Attempt at a Solution



Given the last tangent equation, my plan was to find the two b's, where the tangents met the y-axis. But I failed.

Any thoughts?
 
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  • #2
Your system does not work, in my opinion. in my opinion the prof wants two lines that go thru your 4, -6 point and are TANGENT to the parabola.

So I'd use your differential for the slope, AND use another formula for slope as follows:

m = (Y-(-6))/(X-4) which ALSO equals your slope m = 2X-6... but we have two unknowns, so for the Y, put in your original equation (X^2-6X+5) in it's place... wee bit cumberson, but cross multiply, simplify and you end up with a quadritic equation that will give you both X points that ARE tangent to the parabola AND go through your given point.

I think my X values turned out to be 4 + root of 3, and the other was 4 - root of 3.

Good luck
LarryR : )

KayVee said:

Homework Statement



The teacher proposed a problem today during a math session. He gave the function : f(x) = x2-6x+5 - and a point P(4,-6).

Our mission was to find two tangents for the given point. I thought it was easy, just differentiate the function, and insert the differentiate function for a in : y=a*x + b.

Homework Equations



function : f(x) = x2-6x+5
differentiate : f'(x) = 2x-6
tangent: y= (2x-6)*x + b
tangent eq: -6= (2x-6)*4 + b

The Attempt at a Solution



Given the last tangent equation, my plan was to find the two b's, where the tangents met the y-axis. But I failed.

Any thoughts?
 
  • #3
Yes, that is the same solution our teacher brought. But I just could not figure out, why mine did not work :) I guess it just doesn't work...
 
  • #4
KayVee said:
Yes, that is the same solution our teacher brought. But I just could not figure out, why mine did not work :) I guess it just doesn't work...
It doesn't work because your function has only one tangent line at any given point on the curve. The point (4, -6) is not on the graph of this function.
 

1. What does it mean when two tangents have the same (X,Y) coordinates?

When two tangents have the same (X,Y) coordinates, it means that they intersect at the same point on a curve or circle. This point is called the point of tangency.

2. Can two tangents have different slopes but the same (X,Y) coordinates?

Yes, two tangents can have different slopes but the same (X,Y) coordinates. This occurs when the tangents are parallel to each other.

3. How can you determine the equation of a tangent line with the same (X,Y) coordinates as another tangent?

To determine the equation of a tangent line with the same (X,Y) coordinates as another tangent, you can use the point-slope form of a line. First, find the slope of the given tangent line and then plug in the (X,Y) coordinates of the point of tangency into the equation.

4. Is it possible for a curve to have more than two tangents with the same (X,Y) coordinates?

Yes, it is possible for a curve to have more than two tangents with the same (X,Y) coordinates. This is known as a cusp and occurs when the curve changes direction abruptly.

5. How do tangents with the same (X,Y) coordinates relate to the derivative of a function?

Tangents with the same (X,Y) coordinates are related to the derivative of a function because the derivative of a function at a point is equal to the slope of the tangent line at that point. Therefore, when two tangents have the same (X,Y) coordinates, it indicates that the derivative of the function is the same at that point.

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