Using a graph to help you, find the equations of all lines through the origin tangent to the parabola(adsbygoogle = window.adsbygoogle || []).push({});

[tex]y=x^2-2x+4[/tex].

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:

[tex]y=x^2-2x+4[/tex]

[tex]f'(x)=2x-2[/tex]

Now, I know that my line has the slope [tex]2x-2[/tex] 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:

[tex]0=(2x-2)(0)+b[/tex]

[tex]b=0[/tex]

The [tex]x[/tex]'s go out, and I'm left with an equation of:

[tex]y=(2x-2)x[/tex]

[tex]y=2x^2-2x[/tex]

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:

[tex]x^2-2x+4=2x^2-2x[/tex]

[tex]x^2-4=0[/tex]

[tex](x-2)(x+2)=0[/tex]

[tex]x=2,x=-2[/tex]

Since I just got the x-values, I plugged them in to the original function to get the corresponding y-values:

[tex]f(2)=(2)^2-2(2)+4=4[/tex]

[tex]f(-2)=(-2)^2-2(-2)+4=12[/tex]

Now, I can find the equations of the two tangent lines by using the points I found:

First tangent line:

(0, 0), (2, 4)

[tex]\frac{4-0}{2-0}=\frac{4}{2}=2[/tex]

[tex]y=2x[/tex]

Second tangent line:

(0, 0), (-2, 12)

[tex]\frac{12-0}{-2-0}=\frac{12}{-2}=-6[/tex]

[tex]y=-6x[/tex]

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.

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# Homework Help: Finding tangent lines to a parabola.

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