Finding Slope of a Tangent Line to a Parabola

AI Thread Summary
To find the slope of the tangent line to the parabola defined by the equation y=2x^2-4x+1 at the point (-1,7), the point-slope form of the line is used: y=m(x+1)+7. Setting this equal to the parabola's equation leads to the equation 2x² - 4x - mx + 1 - m = 0, which should be rearranged to match the form ax² + bx + c. The derivative of the parabola, 4x-4, evaluated at x=-1 gives a slope of -8 for the tangent line. The discussion emphasizes that only the tangent line will intersect the parabola at one point, while other slopes will intersect at two points.
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Homework Statement


I've got the equation of a parabola y=2x^2-4x+1 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 ax^2+bx+c, and find the slope of the tangent line.


Homework Equations


y=2x^2-4x+1
y=m(x--1)+7
ax^2+bx+c

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 m(x+1)+7=2x^2-4x+1and place the results in ax^2+bx+c form. I guessed that a=2 and it was correct. However b is not -4x-mx and c is not m-6
 
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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).
 
I wish using a little differentiation were justified here. for the y=parabola,
d/dx of 2x^2-4x+1 is 4x-4.

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.
 
symbolipoint said:
I wish using a little differentiation were justified here. for the y=parabola,
d/dx of 2x^2-4x+1 is 4x-4.

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.

Since the line y=m(x+1)+7 passes through the point (-1,7) which we know is on the parabola, if we choose any real gradient other than the tangential gradient, it'll cut the parabola twice, while the tangent will cut the parabola once.
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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