Finding Two Tangents for a Given Point on a Quadratic Function

[KayVee]

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?

 

[LareeRudi]

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 : )

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?

 

[KayVee]

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...

 

[Mark44]

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.