graph the following function by moving the green and blue dots (if necessary).
y = -|x| + 1

Question

Accepted Answer

# Explanation:
## Step1: Identify the vertex of the absolute - value function
The general form of an absolute - value function is $y = a|x - h|+k$, where $(h,k)$ is the vertex. For the function $y=-\vert x\vert + 1$, we can rewrite it as $y=-|x - 0|+1$. So, the vertex $(h,k)=(0,1)$. This means the green dot (which represents the vertex) should be moved to the point $(0,1)$ from its current position at $(0,0)$.

## Step2: Analyze the slope for $x\geq0$ and $x < 0$
- For $x\geq0$, the function $y =-\vert x\vert+1$ simplifies to $y=-x + 1$. The slope $m=- 1$ for $x\geq0$? Wait, no. Wait, when $x\geq0$, $y=-x + 1$, the slope is $- 1$? Wait, no, let's correct that. Wait, the parent function of $y = |x|$ has a slope of $1$ for $x\geq0$ and $- 1$ for $x<0$. For $y=-\vert x\vert + 1$, when $x\geq0$, $y=-x + 1$, so the slope is $-1$ for $x\geq0$? Wait, no, that's a mistake. Wait, if $y =-\vert x\vert+1$, when $x\geq0$, $y=-x + 1$, so the slope is $- 1$, and when $x<0$, $y=-(-x)+1=x + 1$, so the slope is $1$ for $x < 0$.
- Let's find some points:
    - When $x = 0$, $y=-\vert0\vert + 1=1$ (vertex).
    - When $x = 1$, $y=-\vert1\vert+1=-1 + 1=0$. So the point $(1,0)$ should be on the graph for $x\geq0$ part.
    - When $x=-1$, $y=-\vert - 1\vert+1=-1 + 1=0$. So the point $(-1,0)$ should be on the graph for $x < 0$ part.
    - When $x = 2$, $y=-\vert2\vert+1=-2 + 1=-1$.
    - When $x=-2$, $y=-\vert - 2\vert+1=-2 + 1=-1$.

So for the right - hand side ($x\geq0$) of the absolute - value graph (the part where $x\geq0$), the line should have a slope of $-1$ (since $y=-x + 1$ for $x\geq0$), and for the left - hand side ($x < 0$), the line should have a slope of $1$ (since $y=x + 1$ for $x < 0$).

The original blue dots on the right - hand side (for $x\geq0$) are following a line with slope $1$ (from the graph), but we need a slope of $-1$ for $x\geq0$ in $y =-\vert x\vert+1$. Wait, no, let's re - evaluate. Let's take $x = 1$: $y=-\vert1\vert + 1=0$, so the point $(1,0)$ should be on the graph. For $x = 2$, $y=-\vert2\vert+1=-1$, so the point $(2,-1)$ should be on the graph. For $x = 3$, $y=-\vert3\vert+1=-2$, and so on. For the left - hand side, when $x=-1$, $y = 0$, $x=-2$, $y=-\vert - 2\vert+1=-1$? Wait, no, $y=-\vert - 2\vert+1=-2 + 1=-1$? Wait, no, $y=-\vert - 2\vert+1=-2 + 1=-1$? Wait, no, $\vert - 2\vert=2$, so $y=-2 + 1=-1$. So the point $(-2,-1)$ should be on the graph.

So the correct graph of $y =-\vert x\vert+1$ has its vertex at $(0,1)$, and for $x\geq0$, it's a line with slope $-1$ (passing through $(0,1)$, $(1,0)$, $(2,-1)$, etc.), and for $x < 0$, it's a line with slope $1$ (passing through $(0,1)$, $(-1,0)$, $(-2,-1)$, etc.).

So to graph the function:
    - Move the green dot (vertex) to $(0,1)$.
    - For the right - hand side ( $x\geq0$ ), the blue dots should be adjusted. For example, when $x = 1$, $y = 0$ (so the dot at $(1,1)$ should be moved to $(1,0)$), when $x = 2$, $y=-1$ (the dot at $(2,2)$ should be moved to $(2,-1)$), etc.
    - For the left - hand side ( $x < 0$ ), when $x=-1$, $y = 0$ (the dot at $(-1,1)$ should be moved to $(-1,0)$), when $x=-2$, $y=-1$ (the dot at $(-2,2)$ should be moved to $(-2,-1)$), etc.

# Answer:
To graph $y =-\vert x\vert+1$:
1. Move the green vertex dot to $(0,1)$.
2. For $x\geq0$, the line has a slope of $- 1$ (passes through points like $(1,0)$, $(2,-1)$, etc.).
3. For $x < 0$, the line has a slope of $1$ (passes through points like $(-1,0)$, $(-2,-1)$, etc.). The final graph will be a "V - shaped" graph opening downward with vertex at $(0,1)$, passing through $(1,0)$, $(-1,0)$, $(2,-1)$, $(-2,-1)$ and so on.

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QUESTION IMAGE

Question

Explanation:

Step1: Identify the vertex of the absolute - value function

Step2: Analyze the slope for \(x\geq0\) and \(x < 0\)

Answer: