Help needed in finding Tangent to a graph

In summary, to find the values of m for which the line y=mx+1 is a tangent to the graph of f(x)= 3x^(2) - x + 4, we can use the Point-Slope Form and the derivative of the function to solve for m. In this case, the values of m are 5 and -7.
  • #1
xtrater
3
0

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?
 
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  • #2
xtrater said:

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...
 
  • #3
berkeman said:
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!
 
  • #4
Say you solve simultaneously and get following after simplifying ax2+bx+c=0.

We can find the roots by

[tex]x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]

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?
 
  • #5
Well said, rock.
 
  • #6
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:

[tex](y-f(a))=f'(a)(x-a)[/tex]

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

This would provide the equation.

--Elucidus
 
  • #7
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.
 

1. How do I find the tangent to a graph?

To find the tangent to a graph, you need to first identify a point on the graph where you want the tangent to be. Then, find the slope of the graph at that point by taking the derivative. The tangent will have the same slope as the derivative at that point.

2. What is the equation for the tangent to a graph?

The equation for the tangent to a graph is y = mx + b, where m is the slope of the tangent and b is the y-intercept.

3. Can I find the tangent to a graph without knowing the derivative?

No, the derivative is necessary to find the slope of the tangent at a given point on the graph.

4. Is the tangent always a straight line?

Yes, the tangent to a graph is always a straight line. However, the graph itself may not be a straight line and may have curves or bends.

5. How can finding the tangent to a graph be useful?

Finding the tangent to a graph can be useful in many applications, such as physics, engineering, and economics. It allows us to determine the rate of change at a specific point on the graph, which can help us make predictions and solve problems.

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