Graph f(x) = -|x + 5| - 4. Identify the vertex, the direction it opens, the intercepts, and the range.

Step-by-step solution

1. 1

   Match the equation to vertex form. Comparing with $f(x)=a\left|x-h\right|+k$:

   $a=-1,\qquad h=-5\ \ (\text{since }x+5=x-(-5)),\qquad k=-4$

   so the vertex is $(-5,-4)$. Rewriting $x+5$ as $x-(-5)$ before reading $h$ prevents the usual sign mistake.

2. 2

   Determine the direction and steepness from $a$. Because $a=-1<0$ the graph opens downward — an upside-down V — and because $|a|=1$ the arms have slopes $+1$ (left of the vertex) and $-1$ (right of it), the same steepness as the parent $|x|$.

3. 3

   Build the graph as a sequence of transformations. Starting from $y=|x|$: shift left $5$ (because $x+5$ is zero at $x=-5$), reflect across the $x$-axis, then shift down $4$:

   $|x|\;\to\;|x+5|\;\to\;-|x+5|\;\to\;-|x+5|-4$

4. 4

   Plot a few points either side of the vertex. Using the slope $\pm 1$ from $(-5,-4)$:

   $f(-7)=-2-4=-6,\quad f(-6)=-1-4=-5,\quad f(-5)=-4,\quad f(-4)=-5,\quad f(-3)=-6$

   The symmetry about the vertical line $x=-5$ is visible in these values, which is a good check that the horizontal shift was applied correctly.

5. 5

   Read off the intercepts and range. The $y$-intercept is $f(0)=-|5|-4=-9$, giving the point $(0,-9)$. For $x$-intercepts you would need $-|x+5|-4=0$, i.e. $|x+5|=-4$, which is impossible since an absolute value is never negative — so there are none. The maximum output is $-4$ at the vertex, so the range is

   $(-\infty,-4]$