What is the parametric form for the tangent line to y = 2x^(2)+2x-1 at x = -1?

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SUMMARY

The parametric form for the tangent line to the curve defined by the equation y = 2x² + 2x - 1 at the point x = -1 is derived through a series of steps. First, the slope of the tangent line is calculated using the derivative, yielding y' = 4x + 2, which results in a slope of -2 at x = -1. The point on the tangent line is identified as (-1, -1), leading to the tangent line equation y = -2x - 3. Finally, the parametric form is expressed as x = t and y = -2t - 3.

PREREQUISITES
  • Understanding of calculus, specifically derivatives and tangent lines
  • Familiarity with the equation of a line in slope-intercept form (y = mx + b)
  • Knowledge of parametric equations and their representation
  • Basic algebra skills for manipulating equations
NEXT STEPS
  • Learn how to compute derivatives for various functions to find slopes of tangent lines
  • Study the conversion of linear equations from slope-intercept form to parametric form
  • Explore the concept of parametric equations in greater depth, including their applications
  • Practice solving similar problems involving tangent lines and parametric forms
USEFUL FOR

Students studying calculus, particularly those focusing on derivatives and tangent lines, as well as educators seeking to enhance their teaching materials on parametric equations.

Loppyfoot
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Homework Statement



The parametric form for the tangent line to the graph of y = 2x^(2)+2x-1 at x = -1 is

Homework Equations





The Attempt at a Solution



I am confused about where to begin this problem. Any thoughts?

Thanks!
 
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Loppyfoot said:

Homework Statement



The parametric form for the tangent line to the graph of y = 2x^(2)+2x-1 at x = -1 is

Homework Equations





The Attempt at a Solution



I am confused about where to begin this problem. Any thoughts?

Thanks!
The first step would be to find the slope of the tangent line at the point (-1, f(-1)). Once you have the slope of the tangent line, and a point on the tangent line - (-1, f(-1)), you can find the equation of the tangent line.

The final step would be to write the equation of the tangent line in parametric form.
 
So the slope of the tangent line would be:

y'=4x+2...plug in x=-1.

slope of tangent line at x=-1 is y'=-2.

A point on the line would be (-1,-1).

How would I translate this data into parametric form
 
You skipped a step - you need to find the equation of the tangent line first.
 
so the equation of the tangent line is y=-2x-3.


How would I translate the y=mx+b into the parametric form?
 
Let x = t. Then you have y = -2t - 3, x = t.
 

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