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?