Find the intervals on which $f(x)$ is increasing, the intervals on which $f(x)$ is decreasing, and the local extrema.
\[
f(x)=-4 x^{2}-16 x-26
\]

Step 1 ：The function given is a quadratic function. The graph of a quadratic function is a parabola. If the coefficient of the \(x^2\) term is positive, the parabola opens upwards and if it is negative, the parabola opens downwards. In this case, the coefficient of the \(x^2\) term is -4, which is negative. Hence, the parabola opens downwards.

Step 2 ：This means that the function is increasing before the vertex and decreasing after the vertex. The vertex of the parabola is the point of the graph where it reaches its maximum or minimum value. For a parabola that opens downwards, the vertex is the maximum point.

Step 3 ：The x-coordinate of the vertex can be found using the formula \(-\frac{b}{2a}\), where \(a\) and \(b\) are the coefficients of the \(x^2\) and \(x\) terms respectively. In this case, \(a=-4\) and \(b=-16\). So, the x-coordinate of the vertex is \(-\frac{-16}{2*-4}\).

Step 4 ：The function is increasing for \(x<-\frac{-16}{2*-4}\) and decreasing for \(x>-\frac{-16}{2*-4}\). The maximum value of the function is the y-coordinate of the vertex, which can be found by substituting the x-coordinate of the vertex into the function.

Step 5 ：The x-coordinate of the vertex is -2 and the y-coordinate of the vertex is -10. Therefore, the function is increasing for \(x<-2\) and decreasing for \(x>-2\). The maximum value of the function is -10, which occurs at \(x=-2\).

Step 6 ：\(\boxed{\text{The function } f(x)=-4 x^{2}-16 x-26 \text{ is increasing on the interval } (-\infty, -2) \text{ and decreasing on the interval } (-2, \infty). \text{ The maximum value of the function is -10, which occurs at } x=-2. \text{ Therefore, the local maximum is at the point } (-2, -10).}\)