How does a Gradient points to the steepest ascent of a curve?

Question

What I understand about a Gradient is:

A gradient is an angle which points to the direction of the steepest ascent of a curve.

Let us take a look at the plot of the following function:

$$ \bbox[lightgray] {f(x) = -x^2+4}\qquad (1)$$

The 1st derivative of the function is:

$$ \bbox[lightgray] {\frac{dy}{dx} = -2x}\qquad (2)$$

Putting $x=-1$ in $(2)$ we obtain,

$\implies \frac{dy}{dx} = \frac{rise}{run}= 2$
$\implies tan \theta = 2$
$\implies \theta = tan^{-1}(2)$
$\implies \theta = 0.964 $ radian

So, $\theta = 55.23 ^\circ$

Similarly, putting $x=-2$ in $(2)$ we obtain,

So, $\theta = 57.25 ^\circ$

Now, as we can see, red and blue lines are representative of two angles which represent the ascent of the curve at points $(-1, 3)$ and $(-2, 0)$ respectively.

So, my question is, How are these two angles pointing towards the steepest ascent (Coz, these are just angles. They are not pointing towards anything)?