MHB Equation of the Tangent Line? (Derivatives)

[Velo]

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

 

[I like Serena]

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).

 

[Velo]

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? 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?

 

[I like Serena]

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? 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)

 

[Velo]

Thanks a lot :3 Was really struggling with this for some reason, even though it was actually pretty simple >..<

 

[HallsofIvy]

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).