Free math worksheets

- 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: \(3\)

- 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 \((5,0)\)

- Rewrite the absolute value function as two linear pieces if helpful. - An x-intercept occurs where \(g(x)=0\). - Use the fact that \(|x|\ge0\).

1. The graph of \(g(x)=4-|x|\) is an upside-down V with vertex \((0, 4)\). For \(x\ge0\), \(g(x)=4-x\); for \(x<0\), \(g(x)=4+x\). 2. Set \(4-|x|=0\). Then \(|x|=4\), so \(x=4\) or \(x=-4\). The x-intercepts are \((4, 0)\) and \((-4, 0)\). 3. Because \(|x|\ge0\), \(4-|x|\le4\). Equality occurs at \(x=0\), so the maximum point is \((0, 4)\).

a) An upside-down V with vertex \((0,4)\) b) \((-4,0)\) and \((4,0)\) c) The maximum point is \((0,4)\) because \(|x|\ge0\), so \(4-|x|\le4\), with equality only at \(x=0\).

- Use a point on the red graph to determine its coefficient. - Write a vertical scaling in the form \(af(x)\). - Write a horizontal scaling in the form \(f(bx)\). - Use the rule \(|ab|=|a||b|\) to compare the two expressions.

1. The red graph passes through \((2, 1)\). Using \(g(x)=a|x|\), substitute the point: \(1=a|2|\). Therefore, \(a=\frac{1}{2}\), so \(g(x)=\frac{1}{2}|x|\). 2. Tim describes a vertical compression by a factor of \(\frac{1}{2}\): \(g(x)=\frac{1}{2}f(x)\). 3. Sara describes a horizontal stretch by a factor of \(2\): \(g(x)=f\left(\frac{x}{2}\right)=\left|\frac{x}{2}\right|=\frac{1}{2}|x|\). 4. Both transformations produce the same function because absolute value is homogeneous for a positive scale factor.

1) \(g(x)=\frac{1}{2}|x|\) 2) Vertical compression factor: \(\frac{1}{2}\), since \(g(x)=\frac{1}{2}f(x)\); horizontal stretch factor: \(2\), since \(g(x)=f\left(\frac{x}{2}\right)\)

- 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; \(g(x)=-f(x-2)\)

- 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 \(x=5\,\text{km}\)

- Factor the coefficient of \(x\) inside the absolute value before interpreting 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.

The student'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 \(f(x)=|x|\).