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Finding Slope of a Tangent Line to a Parabola 
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#1
Jan1413, 05:01 PM

P: 6

1. The problem statement, all variables and given/known data
I've got the equation of a parabola [itex]y=2x^24x+1[/itex] with point (1,7) and a tangent line running through it the point. I'm supposed to find the equation of the line. Simultaneously solve this equation with that of the parabola, place the results in form [itex]ax^2+bx+c[/itex], and find the slope of the tangent line. 2. Relevant equations [itex]y=2x^24x+1[/itex] [itex]y=m(x1)+7[/itex] [itex]ax^2+bx+c[/itex] 3. The attempt at a solution I was supposed to find the equation of the line using the point slope equation and I did, I placed it above. The problem lies when I try to set the equations equal to each other [itex]m(x+1)+7=2x^24x+1[/itex]and place the results in [itex]ax^2+bx+c[/itex] form. I guessed that [itex]a=2[/itex] and it was correct. However b is not [itex]4xmx[/itex] and c is not [itex]m6[/itex] 


#2
Jan1413, 05:35 PM

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It looks like you are on the right track, but have some trouble with your bookkeeping.
If you have m(x + 1) = mx + m = 2x²  4x + 1, start by bringing everything to one side of the equals sign: 2x²  4x  mx + 1  m = 0. Now carefully compare this to the given form, ax² + bx + c. Try rewriting the equation to get this: 2x² + (....)x + (...) = 0. You will be able to read off b and c, but this time with the correct signs :) (Also, don't forget, as I initially did, that it stays an equation  after the rewrite there will be "= 0" on the right hand side). 


#3
Jan1413, 09:44 PM

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I wish using a little differentiation were justified here. for the y=parabola,
d/dx of 2x^24x+1 is 4x4. Value of derivative when x=1 becomes 8, so slope is 8 for the line. Now we have both the (given) point, and the slope of the line. I just do not see the less advanced algebra trick to solve the question. 


#4
Jan1413, 11:52 PM

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P: 3,515

Finding Slope of a Tangent Line to a Parabola



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