Help needed in finding Tangent to a graph

[xtrater]

Homework Statement

If f(x)= 3x^(2) - x + 4, find the values of m for which line y=mx+1 is a tangent to the graph of f?

Homework Equations

None

The Attempt at a Solution

Well first you have to get the derivative of the function which is 6x-1...Then what? There are no points given...I tried solving in terms of x and plugging it back in the equation but it didn't work:

x=(m+1)/6 and y=6x^(2)-x+1...I tried substituting these but nothing happened...Any ideas?

 

[berkeman]

Homework Statement

If f(x)= 3x^(2) - x + 4, find the values of m for which line y=mx+1 is a tangent to the graph of f?

Homework Equations

None

The Attempt at a Solution

Well first you have to get the derivative of the function which is 6x-1...Then what? There are no points given...I tried solving in terms of x and plugging it back in the equation but it didn't work:

x=(m+1)/6 and y=6x^(2)-x+1...I tried substituting these but nothing happened...


Any ideas?

Welcome to the PF. Try drawing a graph of the original function -- that should help you visualize the problem and how to get to the solution...

 

[xtrater]

Welcome to the PF. Try drawing a graph of the original function -- that should help you visualize the problem and how to get to the solution...

okay, I solved the question through graphing! I got 5 and -7, however, I still can't figure out the algebraic solution!

 

[rock.freak667]

Say you solve simultaneously and get following after simplifying ax²+bx+c=0.

We can find the roots by

$$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

If b²-4ac>0, we get two real and distinct solutions.

eg. x= (-1±1)/2 → (-1+1)/2 or (-1-1)/2

if b²-4ac=0 we get one solution.

if b²-4ac<0, we get two complex solutions.

A tangent to a graph touches the graph at how many points? Thus how many solutions do you expect?

 

[berkeman]

Well said, rock.

 

[Elucidus]

A way to do this (among others) is to let (a, f(a)) be a generic point on the curve. Find the derivative of f and evaluate this when x = a. This is the tangential slope to the curve at (a, f(a)). The equation y = mx + 1 does tell us the tangent passes through (0, 1) (the y-intercept) so we can use the Point-Slope Form:

$$(y-f(a))=f'(a)(x-a)$$

and the fact that (0, 1) must be a solution of this equation to back-solve for a.

This would provide the equation.

--Elucidus

 

[xtrater]

Thank you all for your help. It turns out that I kept using a different number while solving and therefore my substitution was wrong which kept giving me wrong answers! :P

I've solved the question algebraically. You just need to substitute the derivative in for m and equate both equations which will give you: 3x^(2)-3
Then solving for X, you get x=+-1. Plugging these 2 values into the derivative will give you the 2 values of m which are 5 and -7.