What is the equation of the tangent at (3, f(3))?

[Joza]

I have a function.

I am asked to find the equation of the tangent to the curve at (3, f(3))

I think I am just confused of notation, but what is it asking here? To find the points, don't you put in the x value into the first derivative and this gets the y value, and use the formula for an equation of a line? What are those coordinates for then?

 

[radou]

Well, you're supposed to find the equation of the tangent at that very point.

 

[Joza]

So I already have my x and y, so just use y-y1=m(x-x1)?

 

[radou]

If you meant y - f(c) = f'(c)(x - c), then yes. (c = 3)

 

[sutupidmath]

If you meant y - f(c) = f'(c)(x - c), then yes. (c = 3)

As you might know f'(c)=tg@, so the derivative of the curve at the point (c,f(c)) is actually the orientation coeficient of the tagent drawn at that point.

So this is actually what radou said, just giving u some more hints.