Solving Tangent Lines for 2 Graphs: x1,x2

Michele Nunes
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Homework Statement


Sketch the graphs of y=x2 and y= -x2+6x-5, and sketch the two lines that are tangent to both graphs. Find equations of these lines.

Homework Equations

The Attempt at a Solution


So I know that a tangent line to both graphs means that the tangent line will touch the first graph at (x1,y1) and the second graph at (x2, y2) and at these separate points, the derivatives will be equal. So my first equation looks like 2x1 = -2x2+6, however I'm stuck on what other equation I'm supposed to use to solve for both x-variables.
 
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You have the sketch, I suppose ?
A line is ##y = ax + b##; its intersections with the two parabolas follow from two quadratic equations that both have to have only one solution.
 
BvU said:
You have the sketch, I suppose ?
A line is ##y = ax + b##; its intersections with the two parabolas follow from two quadratic equations that both have to have only one solution.
I'm sorry but I still don't understand how that relates to what I'm supposed to do. They're supposed to have only one solution to what? If I set ax+b equal to the quadratic equations, there's still too many variables for me to solve for which is where I'm confused.
 
Maybe this will help. Let ##P=(a,a^2)## be an unknown point on the parabola ##y=x^2##. Write the equation of the tangent line at that ##P##. Of course, it will depend on the unknown ##a##, but that's OK.
Now look at a general point on the second parabola: if ##x=b## then ##y = -b^2+6b-5##, so the general point on that parabola is ##Q=(b,-b^2+6b-5)##. Now what needs to happen for the tangent line to the first parabola at ##P## to be tangent to the second parabola at ##Q##? That should give you some equations in ##a## and ##b## to work with. Sorry, I changed your notation a bit to avoid all the subscripts.
 
LCKurtz said:
Maybe this will help. Let ##P=(a,a^2)## be an unknown point on the parabola ##y=x^2##. Write the equation of the tangent line at that ##P##. Of course, it will depend on the unknown ##a##, but that's OK.
Now look at a general point on the second parabola: if ##x=b## then ##y = -b^2+6b-5##, so the general point on that parabola is ##Q=(b,-b^2+6b-5)##. Now what needs to happen for the tangent line to the first parabola at ##P## to be tangent to the second parabola at ##Q##? That should give you some equations in ##a## and ##b## to work with. Sorry, I changed your notation a bit to avoid all the subscripts.
The derivatives have to be equal at P and Q, right? So I set the derivatives equal, but I need another equation in order to solve for both variables. Is there something I can do with the original equations? I can't set them equal though because the points don't happen at the same y-value.
 
The tangent line at ##P## must pass through ##Q##.
 
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LCKurtz said:
The tangent line at ##P## must pass through ##Q##.
I finally got it, I've been working on this problem for so long. Thank you so much!
 
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