Finding Slope of a Tangent Line to a Parabola

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SUMMARY

The discussion centers on finding the slope of a tangent line to the parabola defined by the equation y=2x²-4x+1 at the point (-1, 7). The correct approach involves using the point-slope form of a line, y=m(x+1)+7, and equating it to the parabola's equation. The derivative of the parabola, calculated as d/dx of 2x²-4x+1, yields a slope of -8 at x=-1. This confirms that the tangent line at the specified point has a slope of -8.

PREREQUISITES
  • Understanding of parabolic equations, specifically y=2x²-4x+1
  • Knowledge of point-slope form of a line, y=m(x-x₀)+y₀
  • Familiarity with differentiation and its application to find slopes
  • Ability to manipulate polynomial equations into standard form ax²+bx+c
NEXT STEPS
  • Study the process of finding derivatives of polynomial functions
  • Learn how to apply the point-slope form of a line in various contexts
  • Explore methods for solving polynomial equations and comparing coefficients
  • Investigate the geometric interpretation of tangents to curves
USEFUL FOR

Students studying calculus, particularly those focusing on derivatives and tangent lines, as well as educators seeking to clarify concepts related to parabolas and their properties.

Cascadian
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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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