How Does the Derivative of a Cubic Function Result in a Tangent Line?

[tenman]

I'm have trouble understanding a fundamental question of a derivative. So a derivate gives me a tangent line at any given point on a function.

this makes sense for me for a function y=x^2 because the derivative is y'=2x which is a straight line function.

But what about y=x^3 where the derivative is y'=3x^2 this function is a parabola how can a parabola give me a tangent line?

I'm having trouble understand this very basic concept

 

[MarkFL]

The first derivative of a function gives you the slope of the tangent line, not the tangent line itself. Thus,for some function $f(x)$, the tangent line (using the point-slope formula) at the point $(x_0,f(x_0))$ is given by:

\(\displaystyle y=f'(x_0)(x-x_0)+f(x_0)\)

 

[tenman]

Ahh ok thanks I think I get it now.

So the general formula for a straight line is y=mx+c

are you saying that the derivative of a function gives you the m value needed to plug into the straight line formula?

 

[MarkFL]

Ahh ok thanks I think I get it now.

So the general formula for a straight line is y=mx+c

are you saying that the derivative of a function gives you the m value needed to plug into the straight line formula?

Yes, the derivative of the function evaluated at the $x$-coordinate of the tangent point will give you the slope $m$ of the tangent line. :)

 

[tenman]

Thanks heaps for that!