Equations of lines through a point and tangent to a function?

Homework Statement

Find the equations of all the lines through the origin that are tangent to the curve y = (some complicated cubic function)

I looked up the question in google and found a much simpler example, y = x^2 passing through (1,-1). However, I don't even get what's going on with this example even though its much more basic.

The Attempt at a Solution

I know the equations of a straight line are derived simply by slope = m = rise/run

And i'm given a point (1, -1). All I need is a slope.

The function i'm given is f(x) = x^2

I can take the derivative of f(x) to get f'(x) = 2x = Mtangent

So now I have a slope for the point (1, -1)

So I can create the equation 2x = y+1/x-1

So assuming i'm on the correct path up until this point, this is where I begin to get confused. If i'm given a point (1, -1) and a slope of say, m = 2, I can get slope intercept form of y = 8x - 9.

However, the actual Mtangent slope I have is 2x and that gives me the equation
[2x^2 - 2x -1 = y] which seems to be an equation of a curve.

But I think my goal is to find the equation of a line, not a curve. Where is my confusion? What does [2x^2 -2x - 1 = y] represent? And how would I actually get the equation of a line?

What ##f'(x)## represents is the slope of the curve represented by the function ##f## at the point ##(x, f(x))##. That is, for ##f(x)=x^3,## we have ##f'(x)=3x^2## and given a point ##(2, 8)## we can say that the slope of the graph of ##f## at this point is the value of the function's derivative ##f'(2)=12##.

But you chose a point, (2,8), that happens to exist on the graph x^3?

In my question, the point (1, -1), does not exist on the graph x^2

I'm not asking what a derivative represents, I think what i'm asking about is the equation created from taking the derivative and setting the derivative equal to a point that does not exist on the graph.

I am trying to figure out how to create an equation that connects a point on a coordinate system to a point that is tangent to the graph f(x) = x^2

EDIT: Or maybe i'm misunderstanding your response, i'm not sure.

Last edited:
Oh, my bad. Just write (-1) - f(x) = f'(x) (1 - x), so -1 - x2 = 2x(1 - x), and solving gives you x2 - 2x - 1 = 0. Both roots should lead you to valid solutions in this case.

Edit: This equation is from the point-slope formula and from knowing that m = f'(x), so you're basically solving for an x that produces a line with slope 2x which intercepts (x, x2) and (1, -1).

Last edited:
Oh, my bad. Just write (-1) - f(x) = f'(x) (1 - x), so -1 - x2 = 2x(1 - x), and solving gives you x2 - 2x - 1 = 0. Both roots should lead you to valid solutions in this case.

Edit: This equation is from the point-slope formula and from knowing that m = f'(x), so you're basically solving for an x that produces a line with slope 2x which intercepts (x, x2) and (1, -1).

Okay, that seems exactly as I already did, the difference being that I chose points (1,-1) and a random point (x,y), just as the slope is written rise/run = y-y/x-x. And you chose points (1,-1) and (x,x^2).

Does that mean that my [2x^2 -2x - 1 = y] and your [x^2 - 2x - 1 = 0] are equivelent somehow? When I graph both equations, they are different, but I see that if I substitute x^2 for y and bring it to the other side, i'll get the same equation that you produced.

Also, as per my original confusion, I thought point slope formula was a formula that represents a line. Are you saying the point slope formula represents parabolas also?

EDIT: okay, I just solved for both roots and got 1+√2 and 1-√2. Those are both points on the x axis, but what do they mean? Are they x intercepts to the lines I am trying to create an equation for?

EDIT: okay, looking at the website where I got this sample problem is starting help me figure out the last parts. However it ends up showing me the coordinates of the points on the equation fx = x^2 where the line is tangent. But my goal is not to find two points on the graph, by goal is to find an equation of the line. Would I just use those coordinates to come up with the equation?

Last edited:
Okay, so I have this equation: y-y=m(x-x), I have always seen it when a textbook is talking about straight lines. However, now it is being used to get y-(-1)=(2x)(x-1). Where this clearly seems not to be representing a straight line. First of all, a slope is a constant while the m = slope variable here is not a constant because it contains an x.

Okay so i think i get it a little bit, am I supposed to go from a general derivative to an exact slope? And the only way to do that is by finding some x value that will make that 2x turn into some constant slope?

Yeah, you're almost there.

Letting x = 1+√2 gives you an equation for a line

y - (3 + 2√2) = (2 + 2√2)(x - (1+√2)).

Plugging in (1, -1) for (x, y) shows that this line contains (1, -1). This line crosses through (1+√2, f(1+√2)) and has slope f'(1+√2), so it is tangent to the graph f at that point.

y - y1 = m(x - x1) is indeed the equation for a line, provided m, x1, y1 are constants. To find the correct constants, you can solve for them as I've done above.

HallsofIvy
Any line through (1, 1) is of the form y= m(x- 1)- 1. If that is tangent to y= f(x) at some point $(x_0, y_0)$ (it is a very bad idea to use just "x" both for the variable and a specific value), then you must have $m(x_0- 1)+ 1= f(x_0)$ (so that it touches the graph) and $m= f'(x_0)$ (so that it is tangent there). That gives you two equations to solve for m and $x_0$.
In your $f(x)= x^2$ example that would be $m(x_0- 1)+ 1= x_0^2$ and $m= 2x_0$. Replace m in the first equation by $2x_0$ to get $2x_0^2- 2x_0+ 1= x_0^2$, a quadratic equation for $x_0$.
(In fact, that equation is just $(x_0- 1)^2= 0$ which has the single root $x_0= 1$. That's because the line is tangent and does not cross through the graph.)