Let's take a parabolic function $f(x)=x^2$ and its derivative $f'(x)=2x$ and plot them:
In Quadrant 3, the derivative is increasing but negative until it reaches 0. What is meant by negative ? It can't be a negative slope since the slope is positive.
Also, the slope of the derivative is the same for the entirety of the function, but the parabolic function clearly denotes that the slope is constantly changing. Graphically speaking, how then would the derivative be able to find the tangent points in the parabolic function when it is itself is a linear function of fixed slope ?
Best Answer
Recall that the slope is equal to $\frac{\Delta y}{\Delta x}$. The change in $x$ and $y$ is signed, which indicates whether it is decreasing or increasing. Before $x=0$, $x$ is increasing, and $y$ is decreasing. Therefore, the slope, which is equal to the derivative, is negative. This just means it's sloping downwards.
The reason the slope graph is linear is because the slope of the derivative graph represents how fast the derivative is changing, not the original function. For a parabola, the derivative changes linearly.
The derivative doesn't find the points of tangency. It just shows the slope of the tangent lines at the points of the same $x$ coordinate.
I hope this clears up any confusion. :)