Finding tangent lines that pass through given points

In summary, the problem requires finding the coordinates of all points on the curve f(x)= x3 whose tangent lines pass through the point (a,0). Using the equation f'(x) = 3x2, two point-slope equations can be set up to solve for the coordinates. The resulting points are (-x3, x3) and (a, 0).
  • #1
methionine
3
0

Homework Statement


Find the co-ordinates of all points on the curve f(x)= x3 whose tangent lines pass through the point (a,0)


Homework Equations


f '(x) = nxn-1


The Attempt at a Solution


I am really not sure how to attack this question. My initial thoughts are to find f '(x) then, given the points (a,0), create a point-slope equation for a line. But, this would only give me one equation. On the other hand, when I look at the graph of x^3 I can't see how two tangent lines could share a point. I'm not sure how I would find the co-ordinates either.

f '(x) = 3x2

y = 3x2 (x-a)
 
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  • #2
methionine said:

Homework Statement


Find the co-ordinates of all points on the curve f(x)= x3 whose tangent lines pass through the point (a,0)


Homework Equations


f '(x) = nxn-1


The Attempt at a Solution


I am really not sure how to attack this question. My initial thoughts are to find f '(x) then, given the points (a,0), create a point-slope equation for a line. But, this would only give me one equation. On the other hand, when I look at the graph of x^3 I can't see how two tangent lines could share a point. I'm not sure how I would find the co-ordinates either.

f '(x) = 3x2

y = 3x2 (x-a)
... and y = x3 .

Hello methionine. Welcome to PF !
 
  • #3
I'm sorry, what do you mean? "...and y = x3
 
  • #4
methionine said:
I'm sorry, what do you mean? "...and y = x3
Your line passes through the point (x, x3) as well as the point (a, 0).
 
  • #5
methionine said:
On the other hand, when I look at the graph of x^3 I can't see how two tangent lines could share a point.

I believe there are indeed two lines tangent to the curve that go through the point (a,0). Look carefully.
 
  • #6
Hey guys, I spent the last day mulling this question over in my head, and decided to attack it in a slightly different way.

Basically, did what I was doing up to this point, set a point-slope form equation using x and x and x3 and x1 and y1...

y-x3 = 3x2(x-x3)

Now, I know the line has to pass through (a,0) as well, so I plugged those values into X and Y and ended up with two points.

-x3 = 3x2(a-x3)

Kind of an interesting question for me. I hope my work checks out!
 

What is the purpose of finding tangent lines that pass through given points?

The purpose of finding tangent lines that pass through given points is to determine the slope of the curve at those specific points. This can provide valuable information for understanding the behavior and characteristics of the curve.

What is the equation for finding the slope of a tangent line through a given point?

The equation for finding the slope of a tangent line through a given point is y = f'(x0)(x-x0) + f(x0), where x0 is the x-coordinate of the given point and f'(x0) is the derivative of the function at that point.

How do you find the equation of a tangent line through a given point on a graph?

To find the equation of a tangent line through a given point on a graph, you first need to determine the slope of the tangent line using the derivative of the function at that point. Then, use the point-slope form of a line (y-y0 = m(x-x0)) to plug in the slope and the given point to find the equation.

What is the difference between a tangent line and a secant line?

A tangent line is a line that touches a curve at one specific point, while a secant line is a line that intersects a curve at two points. A tangent line represents the slope of the curve at that point, while a secant line represents the average rate of change of the curve between the two points.

Can you find the equation of a tangent line through a given point if the function is not continuous at that point?

No, the equation of a tangent line can only be found if the function is continuous at the given point. If the function is not continuous, it means that the slope of the curve at that point is undefined, and a tangent line cannot be determined.

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