MHB Equation of the Tangent Line? (Derivatives)

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The equation of the tangent line, y = f(a) + f'(a)(x - a), incorporates (x - a) to reflect the horizontal distance from the point of tangency, a. This adjustment allows for the calculation of the line's slope, f'(a), at any point x, rather than just at x = 0. The discussion clarifies that f(a) represents the y-coordinate at the point a, making it the correct starting point for the tangent line. Understanding this formulation helps in visualizing how the tangent line behaves around the point of tangency. The equation effectively shows how the tangent line shifts based on the value of x relative to a.
Velo
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So, I can't wrap around my head of why the Equation of the Tangent Line is:
[M]y = f(a) + f'(a)(x - a)[/M]
I get it that it's the equation of a line, and so it should be something like [M]y = mx + b[/M]. I also understand why f(a) = b (since it's a point in that line) and why f'(a) = m (since it's the slope), but where did the (x-a) come from? Shouldn't it just be x?
Thanks for the help in advanced :T
 
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Velo said:
So, I can't wrap around my head of why the Equation of the Tangent Line is:
[M]y = f(a) + f'(a)(x - a)[/M]
I get it that it's the equation of a line, and so it should be something like [M]y = mx + b[/M]. I also understand why f(a) = b (since it's a point in that line) and why f'(a) = m (since it's the slope), but where did the (x-a) come from? Shouldn't it just be x?
Thanks for the help in advanced :T

Hi Velo!

The slope of a line is how much it goes up when we move 1 point to the right.
f'(a) is the slope at a.
We're not going 1 point to the right though, but as much as x is bigger than a, that is (x-a).
 
I like Serena said:
Hi Velo!

The slope of a line is how much it goes up when we move 1 point to the right.
f'(a) is the slope at a.
We're not going 1 point to the right though, but as much as x is bigger than a, that is (x-a).

Oh, I think I got it now... So in the equation [M]y = mx + b[/M], b is the y when [M]x = 0[/M] in that equation, correct? :o And then, since our starting point in the tangent line's equation doesn't actually have to be[M]x = 0[/M], we move around that point instead?
 
Velo said:
Oh, I think I got it now... So in the equation [M]y = mx + b[/M], b is the y when [M]x = 0[/M] in that equation, correct? :o And then, since our starting point in the tangent line's equation doesn't actually have to be[M]x = 0[/M], we move around that point instead?

Yep. (Nod)
 
Thanks a lot :3 Was really struggling with this for some reason, even though it was actually pretty simple >..<
 
You could have checked that y= f'(a)x+ f(a), at x= a, is y= f'(a)a+ f(a), NOT f(a). With y= f'(a)(x- a)+ f(a) when x= a. y= f'(a)(a- a)+ f(a)= f'(a)(0)+ f(a)= f(a).
 

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