Tangent Lines to f(x)=x^2-4x+5 Through P(0,1)

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SUMMARY

The discussion focuses on finding the equations of two tangent lines to the function f(x) = x² - 4x + 5 that pass through the point P(0, 1). The derivative, f'(x) = 2x - 4, is used to determine the slope of the tangent lines at specific points on the curve. Participants emphasize the use of the point-slope form of the line equation, f(x) = f'(a)(x - a) + f(a), to derive the tangent lines. The solution involves calculating the slope at x = 0 and applying the general line equation.

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  • Practice finding tangent lines for different quadratic functions
  • Learn how to apply the derivative to find slopes at various points
  • Explore the concept of implicit differentiation for more complex functions
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Homework Statement


Given: [tex]f(x)=x^2-4x+5[/tex] find equation for two lines that are tangent to the graph and pass through the point P(0,1)

Homework Equations


[tex]f(x)=x^2-4x+5[/tex]
[tex]\frac{dy}{dx}=2x-4[/tex]
Equation of the tangent line/s
[tex]f(x)=f'(a)(x-a)+f(a)[/tex]

The Attempt at a Solution


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You seem to have everything you need. You have the derivative, which gives you the slope of the tangent line at any point you plug in. You're supposed to find the tangent line at the point (0, 1), which means you need the slope at x=0. You can use point-slope form to find the line equation if you want, or you can literally just plug the numbers into the general line equation you were given.

- Warren
 
lol... revelation I got it.
Don't you love it when you make things more complicated then they need to be?:smile:
 

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