Question

2b. what is the end behavior, and the axis of symmetry, of the absolute value graph below?
as x → __, f(x) → __ and
as x → __, f(x) → __
and
x=

Explanation:

Step1: Analyze End Behavior (Left)

For the left - hand side of the absolute - value graph (as \(x\) approaches negative infinity), we observe the trend of the graph. The left part of the V - shaped graph (absolute - value graph) is a line with a negative slope. But as \(x\) gets more and more negative ( \(x ightarrow-\infty\) ), wait, no, looking at the given graph, the domain shown starts from \(x = 0\) on the left? Wait, no, the graph is drawn from \(x=- 1\) to \(x = 5\). But for an absolute - value function of the form \(y=\vert x - h\vert + k\), the end behavior: as \(x\) approaches positive infinity and negative infinity. But in the given graph, the left - most part (as \(x\) decreases, but in the visible graph, when we look at the direction of the lines: the left line goes from the vertex (at \(x = 2\)) towards the left (as \(x\) decreases) and up? Wait, no, the left line in the graph (from \(x = 2\) towards \(x = 0\)): when \(x\) decreases (moves to the left from \(x = 2\)), the \(y\) - value increases? Wait, no, the left arrow is pointing up as \(x\) decreases? Wait, no, the left line: when \(x\) approaches negative infinity (if we extend the graph), but in the given graph, the left line (from \(x = 2\) to \(x = 0\)): when \(x\) decreases (goes to the left), \(y\) increases. The right line (from \(x = 2\) to \(x = 3\)): when \(x\) increases (goes to the right), \(y\) increases. So for the end behavior: as \(x ightarrow-\infty\), wait, no, the graph's left - most part (the line on the left of the vertex \(x = 2\)): the slope of the left line: let's take two points. The vertex is \((2,3)\), and another point on the left line: say \((0,7)\) (from the graph, at \(x = 0\), \(y = 7\)). So the slope is \(\frac{7 - 3}{0 - 2}=\frac{4}{-2}=- 2\). Wait, but the arrow is pointing up, so as \(x\) decreases ( \(x ightarrow-\infty\) ), \(y\) ( \(f(x)\)) increases? Wait, no, when \(x\) decreases (moves to the left), for the left line (with slope - 2), \(y= - 2(x - 2)+3=-2x + 4 + 3=-2x+7\). So when \(x ightarrow-\infty\), \(-2x\) becomes \(+\infty\) (since \(x\) is negative, \(-2x\) is positive and large), so \(y ightarrow+\infty\). For the right line: the right line has a slope of 2 (from vertex \((2,3)\) to \((3,7)\)? Wait, no, at \(x = 3\), \(y = 7\)? Wait, the right arrow is pointing up. The right line: from \((2,3)\) to \((3,7)\)? No, the grid: each square is 1 unit. At \(x = 2\), \(y = 3\); at \(x = 3\), \(y = 7\)? Wait, the y - axis has marks at 1,2,3,4,5,6,7. The right arrow is at \(x = 3\), \(y = 7\), and the left arrow is at \(x = 0\), \(y = 7\). So the function is symmetric about \(x = 2\). So the function is \(y=\vert x - 2\vert+3\)? Wait, no, when \(x = 2\), \(y = 3\); when \(x = 0\), \(y=\vert0 - 2\vert+3=2 + 3=5\)? But in the graph, at \(x = 0\), \(y = 7\). So maybe \(y = 2\vert x - 2\vert+3\). At \(x = 0\), \(y=2\vert- 2\vert+3=4 + 3=7\), which matches. At \(x = 3\), \(y=2\vert1\vert+3=2 + 3=5\)? Wait, no, the right arrow is at \(x = 3\), \(y = 7\). Wait, maybe I misread the graph. Let's re - examine: the vertex is at \((2,3)\). The left line goes from \((2,3)\) to \((0,7)\) (so when \(x\) decreases by 2 (from 2 to 0), \(y\) increases by 4 (from 3 to 7), so slope is \(\frac{7 - 3}{0 - 2}=\frac{4}{-2}=-2\), so the equation is \(y-3=-2(x - 2)\), so \(y=-2x + 4+3=-2x + 7\). The right line goes from \((2,3)\) to \((3,7)\)? Wait, when \(x\) increases by 1 (from 2 to 3), \(y\) increases by 4 (from 3 to 7), so slope is \(\frac{7 - 3}{3 - 2}=4\)? No, that can't be. Wait, the grid: the y - axis has marks at 1,2,3,4,5,6,7. The left arrow is at \(x = 0\), \(y = 7\), and the right arrow is at \(x = 3\), \(y = 7\). So the two endpoints of the arrows (the tips) are at \((0,7)\) and \((3,7)\)? No, the left arrow is at \(x = 0\), \(y = 7\) (the arrow is pointing up, so as \(x\) decreases from 0, \(y\) increases? Wait, no, the arrow direction: the left arrow is on the line that goes from the vertex \((2,3)\) to \((0,7)\), and the arrow is pointing up, meaning as \(x\) approaches \(-\infty\) (if we extend the line), \(y\) approaches \(+\infty\). The right arrow is on the line that goes from the vertex \((2,3)\) to \((3,7)\) (wait, \(x = 3\), \(y = 7\)), and the arrow is pointing up, so as \(x\) approaches \(+\infty\), \(y\) approaches \(+\infty\).

So for end behavior:

As \(x ightarrow-\infty\), \(f(x) ightarrow+\infty\) (because the left - hand line has a negative slope, but when \(x\) becomes very negative, \(-2x\) (from the equation \(y=-2x + 7\)) becomes very large positive, so \(y\) increases to \(+\infty\)).

As \(x ightarrow+\infty\), \(f(x) ightarrow+\infty\) (because the right - hand line has a positive slope, and as \(x\) becomes very large positive, \(y\) (from the right - hand line equation, if we extend it) will increase to \(+\infty\)).

Now, the axis of symmetry: for an absolute - value graph (V - shaped), the axis of symmetry is the vertical line passing through the vertex. The vertex is at \(x = 2\) (since the lowest point of the V is at \(x = 2\), \(y = 3\)). So the axis of symmetry is \(x = 2\).

Step2: Fill in the End - Behavior Blanks

For the end - behavior:

As \(x ightarrow-\infty\), \(f(x) ightarrow+\infty\)

As \(x ightarrow+\infty\), \(f(x) ightarrow+\infty\)

And the axis of symmetry is \(x = 2\)

Answer:

As \(x
ightarrow-\infty\), \(f(x)
ightarrow+\infty\) and As \(x
ightarrow+\infty\), \(f(x)
ightarrow+\infty\)

And \(x = 2\)