Finding tangent lines to a parabola

[Jacobpm64]

Using a graph to help you, find the equations of all lines through the origin tangent to the parabola
y=x^2-2x+4.

Sketch the lines on the graph.

I'm not sure how to start this off, I know the the first derivative would give me the slope of the tangent line at any point. If I find that:
y=x^2-2x+4
f'(x)=2x-2

Now, I know that my line has the slope 2x-2 and it goes through the origin. So, I thought of plugging that into the slope-intercept form for a line, but it doesn't really work, I don't think:
0=(2x-2)(0)+b
b=0

The x's go out, and I'm left with an equation of:
y=(2x-2)x
y=2x^2-2x

When I graphed this parabola and the original one, I noticed that the graphs intersected at the points that the tangent lines would go through. So, to find those points, I set the equations equal to each other:
x^2-2x+4=2x^2-2x
x^2-4=0
(x-2)(x+2)=0
x=2,x=-2

Since I just got the x-values, I plugged them into the original function to get the corresponding y-values:
f(2)=(2)^2-2(2)+4=4
f(-2)=(-2)^2-2(-2)+4=12

Now, I can find the equations of the two tangent lines by using the points I found:
First tangent line:
(0, 0), (2, 4)
\frac{4-0}{2-0}=\frac{4}{2}=2
y=2x

Second tangent line:
(0, 0), (-2, 12)
\frac{12-0}{-2-0}=\frac{12}{-2}=-6
y=-6x

Is this correct? It seems like there would be an easier way to go about doing this... Does anyone know?... It seems like I made things work in some very weird, rough way.

 

[mathwonk]

Well, that's the answer I got. In fact I like your solution better than mine, but here is mine. I used the fact that the line through the origin that is tangent to the parabola is the one that only meets it once instead of twice or not at all.

Since lines through the origin have equation y = cx for some c, I set the two equations equal, to solve for their intersections, and then required there to be only one such intersection, i.e. only one common solution. So I get cx = x^2 - 2x + 4, or x^2 - (c+2)x + 4 = 0, which has only one solution exactly when the "discriminant" is zero, (AX^2 + BX + C = 0 has only one solution when B^2 = 4AC), i.e. in our case when (c+2)^2 = 4.4 = 16, or when c^2 + 4c -12 = 0, i.e. c = 2 or -6.

But you used the additional fact that the slope c = 2x-2, which seems very efficient and intelligent. Still mine looks shorter, maybe because I write more briefly.

So you used the fact that the line throiugh the origin which is also tangent to the parabola has the right slope to be the tangent line, and I used that it meets the parabola only once. Both methods work, but yours is more general since it would apply to curves more general than parabolas.

 

[fresh_42]

Here is the graph: