- Find the point where the two arms of the graph meet. - Decide whether values increase or decrease as you move away from that point. - The opening direction tells you whether the vertex is a highest or lowest point.

1. The vertex is the turning point of the V-shaped graph. From the graph, it is \((2,3)\). 2. The arms move downward away from the vertex, so the graph opens downward. 3. Because the graph opens downward, the vertex gives the maximum value. The maximum value is \(3\).

Vertex: \((2,3)\); opens downward; maximum value: \(3\)

- A function value is a height at a particular horizontal position. - Could the smallest output occur between two of the marked inputs?

1. The marked points show \(f(-3)=3\), \(f(0)=0\), and \(f(3)=3\). 2. The graph meets at its lowest point \((0,0)\), so the smallest output is \(0\), at \(x=0\).

\(f(-3)=3\), \(f(0)=0\), \(f(3)=3\). The smallest output is \(0\), at \(x=0\).

- Compare whether each arm goes above or below the x-axis. - Compare the y-coordinates of points that have the same x-coordinate on the two graphs.

1. The parent graph \(f\) opens upward from \((0,0)\). 2. Graph \(g\) opens downward with the same vertex, so every output is the opposite of \(f\)'s output. 3. The reflected graph is \(g(x)=-|x|\).

\(g(x)=-|x|\). It is the reflection of \(f\) across the x-axis.

- Find where the two straight arms meet. - Check whether points equally far to either side of the lowest point have equal heights. - A line of symmetry must divide the graph into matching parts.

1. The two arms meet at \((2,0)\), so the vertex is \((2,0)\). 2. The points on either side are equally high at the same distance from \(x=2\). The symmetry line is \(x=2\).

Vertex: \((2,0)\). Line of symmetry: \(x=2\).

- What happens to \(|x|\) on either side of zero? - Use the two allowed endpoint inputs to find where the restricted graph stops. - At which input is the subtracted absolute value smallest?

1. The graph is an upside-down V with vertex \((0,4)\). At the domain endpoints, \(g(-5)=4-5=-1\) and \(g(5)=4-5=-1\), so the drawn segment runs from \((-5,-1)\) through \((0,4)\) to \((5,-1)\). 2. The x-intercepts satisfy \(4-|x|=0\), so \(|x|=4\) and \(x=-4\) or \(x=4\). The intercepts are \((-4,0)\) and \((4,0)\). 3. Since \(|x|\ge0\), \(g(x)=4-|x|\le4\). Equality occurs only at \(x=0\), giving the maximum point \((0,4)\).

a) An upside-down V with vertex \((0,4)\), left endpoint \((-5,-1)\), and right endpoint \((5,-1)\). b) \((-4,0)\) and \((4,0)\). c) \((0,4)\); \(|x|\) is never negative.

- Locate the vertex of each V-shaped graph. - Horizontal translations change the expression inside the absolute value. - Vertical translations add or subtract a constant outside the absolute value.

1. The vertex of \(f\) is \((0, 0)\), and the vertex of \(g\) is \((-3, 1)\). 2. Moving the vertex from \((0, 0)\) to \((-3, 1)\) requires a shift left \(3\) units and up \(1\) unit. 3. A shift left \(3\) units replaces \(x\) with \(x+3\), and a shift up \(1\) unit adds \(1\) outside the absolute value. 4. Therefore, \(g(x)=|x+3|+1\).

Shift left \(3\) units and up \(1\) unit; \(g(x)=|x+3|+1\)

- Use the expression inside the absolute value to locate the vertex horizontally. - Use the constant outside the absolute value to locate the vertex vertically. - Compare the coefficient of the absolute value with \(1\) to judge the width and opening direction.

1. The expression \(|x+2|\) shifts the parent graph left \(2\) units, so the vertex has x-coordinate \(-2\). 2. The \(-1\) shifts the graph down \(1\) unit, so the vertex is \((-2,-1)\). 3. The factor \(\frac{1}{2}\) vertically compresses the graph, making the V wider while it still opens upward. 4. Graph a) has these features.

Graph a). It is shifted left \(2\) units, vertically compressed by a factor of \(\frac{1}{2}\), and shifted down \(1\) unit.

- The output represents distance from a target, so it cannot be negative. - Express the difference between the actual temperature and the target before taking its absolute value. - A fixed positive distance from the target can occur on either side of the target value.

1. Distance from the target is the absolute difference between \(T\) and \(70\), so \(D(T)=|T-70|\). 2. \(D(66)=|66-70|=4\), so \(66\,\text{°F}\) is \(4\,\text{°F}\) from the target. 3. Solve \(|T-70|=6\). The two possibilities are \(T-70=6\) and \(T-70=-6\), giving \(T=76\) and \(T=64\).

a) \(D(T)=|T-70|\) b) \(4\,\text{°F}\) c) \(64\,\text{°F}\) and \(76\,\text{°F}\)

- Begin with the turning point of the V-shaped graph. - Use the directions in which the two arms continue to determine domain and range. - The x-intercepts occur where the graph meets the x-axis, and symmetry should help you check the pair.

1. The graph has its minimum point at \((2,-3)\), so the vertex is \((2,-3)\) and the axis of symmetry is \(x=2\). 2. An absolute-value graph extends left and right without bound, so the domain is \((-\infty,\infty)\). Its minimum y-value is \(-3\), so the range is \([-3,\infty)\). 3. For the x-intercepts, set \(|x-2|-3=0\). Then \(|x-2|=3\), giving \(x=-1\) or \(x=5\). The intercepts are \((-1,0)\) and \((5,0)\).

a) Vertex: \((2,-3)\); axis of symmetry: \(x=2\) b) Domain: \((-\infty,\infty)\); range: \([-3,\infty)\) c) \((-1,0)\) and \((5,0)\)

- How far from zero is each input on the number line? - Compare neighboring outputs separately on either side of zero. - What does a constant change for each step in \(x\) mean for a graph?

1. Distance from zero gives the outputs \(2,1,0,1,2\) for inputs \(-2,-1,0,1,2\). 2. Moving one unit right on the negative side decreases the output by \(1\) each time. Moving one unit right on the positive side increases it by \(1\) each time. 3. In fact, \(f(x)=-x\) for \(x<0\) and \(f(x)=x\) for \(x\ge0\). Each side has a constant slope, so the two straight arms meet at \((0,0)\).

The outputs are \(2,1,0,1,2\). They change by \(-1\) per step to the left of zero and by \(+1\) per step to the right. Each side is straight, and they meet at \((0,0)\).

- What happens to the value of the function at \(x=0\)? - Test one positive input to keep track of the signs. - Should opposite inputs give matching outputs in the new rule?

1. Reflection changes the rule to \(-|x|\). Shifting down \(3\) gives \(g(x)=-|x|-3\). 2. Substituting the inputs gives \(g(-2)=-5\), \(g(0)=-3\), and \(g(2)=-5\).

\(g(x)=-|x|-3\). Points: \((-2,-5)\), \((0,-3)\), and \((2,-5)\).

- Read the turning point from the graph. - Find the slope of each straight arm separately. - Make sure both rules reach the vertex at the joining input.

1. The vertex is \((3,1)\). On the left arm, the graph falls one unit for each step to the right, so its slope is \(-1\). Using the vertex gives \(y=4-x\) for \(x<3\). 2. On the right arm, the graph rises one unit for each step right. Its equation is \(y=x-2\) for \(x\ge3\). 3. Both rules give \(1\) at \(x=3\), so the two arms meet.

For \(x<3\), \(g(x)=4-x\). For \(x\ge3\), \(g(x)=x-2\).

- Find a point on graph \(g\) other than the vertex. - How does multiplying each output by a fraction affect the graph?

1. The red graph passes through \((2,1)\). For \(g(x)=a|x|\), substitute \(1=a|2|\), giving \(a=\frac{1}{2}\) and \(g(x)=\frac{1}{2}|x|\). 2. Tim describes a vertical compression by a factor of \(\frac{1}{2}\), since \(g(x)=\frac{1}{2}f(x)\). 3. Sara describes a horizontal stretch by a factor of \(2\), since \(g(x)=f\left(\frac{x}{2}\right)=\left|\frac{x}{2}\right|=\frac{1}{2}|x|\). These are the same rule.

1) \(g(x)=\frac{1}{2}|x|\). 2) Tim: vertical compression by \(\frac{1}{2}\). Sara: horizontal stretch by \(2\).

- Compare the vertices of the two V-shaped graphs. - Determine whether the red graph opens upward or downward. - Apply the horizontal change inside the input and the reflection outside.

1. The vertex of \(f\) is \((-4, 0)\), and the vertex of \(g\) is \((-2, 0)\). Therefore, the graph shifts right \(2\) units. 2. The blue graph opens upward, while the red graph opens downward, so the shifted graph is reflected across the x-axis. 3. A shift right \(2\) units gives \(f(x-2)\), and the reflection places a negative sign outside. 4. Therefore, \(g(x)=-f(x-2)=-|x+2|\).

Shift right \(2\) units and reflect across the x-axis; \(g(x)=-f(x-2)\)

- Compare the left and right branches at the same height. - Check that the formula you write fits more than one point on the graph. - Think about the horizontal distances on both sides of the vertex.

1. The vertex is \((2,0)\). From \((2,0)\) to \((3,1)\), the graph rises \(1\) for a horizontal change of \(1\), so \(f(x)=|x-2|\). 2. Solve \(|x-2|=5\) by writing \(x-2=5\) or \(x-2=-5\). 3. The solutions are \(x=7\) and \(x=-3\). 4. Both numbers lie \(5\) units from the center \(2\), on opposite sides.

a) Vertex \((2,0)\), rise of \(1\) per unit; \(f(x)=|x-2|\) b) \(x=-3\) or \(x=7\) c) They are both \(5\) units from \(2\).

- Does your formula fit a point on each arm of the V? - Check whether the inequality describes points near the vertex or far from it. - Use your formula for inputs not shown in the picture.

1. The vertex is \((1,0)\), and \((2,1)\) lies on the graph, so the rise is \(1\) per unit from the vertex. Thus \(f(x)=|x-1|\). 2. Solve \(|x-1|\le5\): the distance from \(1\) is at most \(5\). 3. The endpoints are \(1-5=-4\) and \(1+5=6\), giving \(-4\le x\le6\), or \([-4,6]\). 4. At \(x=0\), the distance is \(1\), so it works. At \(x=7\), the distance is \(6\), so it does not.

a) Vertex \((1,0)\), rise of \(1\) per unit; \(f(x)=|x-1|\) b) \(-4\le x\le6\); \([-4,6]\) c) \(x=0\) works: it is \(1\) unit from \(1\). \(x=7\) does not: it is \(6\) units from \(1\).

- Read the vertex first and place it into the vertex form of an absolute value function. - Then choose a second clear point on one arm of the graph. - The second point determines both the magnitude and sign of the remaining coefficient.

1. The vertex is \((1,4)\), so \(h=1\) and \(k=4\). Thus, \(g(x)=a|x-1|+4\). 2. The graph also passes through \((2,2)\). Substitute this point: \(2=a|2-1|+4\). 3. This gives \(2=a+4\), so \(a=-2\). 4. Therefore, \(g(x)=-2|x-1|+4\).

\(g(x)=-2|x-1|+4\). The vertex \((1,4)\) gives \(h=1\) and \(k=4\). Using the point \((2,2)\), \(2=a|2-1|+4\), so \(a=-2\).

- Use the vertex to determine the horizontal and vertical shifts in the absolute-value form. - Use one other clear point on the graph to determine how steep the two arms are. - A horizontal level above the vertex can meet a symmetric V-shaped graph in two places.

1. The vertex is \((3,1)\), so use \(h(x)=a|x-3|+1\). 2. The graph also passes through \((1,5)\). Substitute: \(5=a|1-3|+1=2a+1\), so \(a=2\). Thus \(h(x)=2|x-3|+1\) for \(0\le x\le6\). 3. The vertex means the lowest point of this trail segment occurs \(3\,\text{km}\) from the start at an elevation \(1\,\text{m}\) above the reference level. 4. Solve \(2|x-3|+1=5\). Then \(|x-3|=2\), so \(x=1\) or \(x=5\).

a) \(h(x)=2|x-3|+1\), for \(0\le x\le6\) b) The lowest point is \(3\,\text{km}\) from the start and \(1\,\text{m}\) above the reference level. c) \(x=1\,\text{km}\) and \(x=5\,\text{km}\)

- Rewrite the inside of the absolute value to expose the horizontal shift. - Compare the vertices of the two graphs. - Compare the steepness of the arms to decide whether a scaling transformation is also present.

1. Rewrite the rule as \(g(x)=|2(x-2)|+1=2|x-2|+1\). 2. The vertex moves from \((0,0)\) to \((2,1)\), so the graph is shifted right \(2\) units and up \(1\) unit, not right \(4\) units. 3. The factor \(2\) vertically stretches the graph by a factor of \(2\). This makes the arms steeper and the V narrower than the graph of \(f\). 4. The displayed vertex and narrower shape confirm these transformations.

Yu-ting's claim is incorrect. The graph is shifted right \(2\) units, vertically stretched by a factor of \(2\), and shifted up \(1\) unit. Its vertex is \((2,1)\), and it is narrower than \(f(x)=|x|\).