Free graphing resources · Original practice
Plot four functions in y=a|x-h|+k. Identify each vertex and symmetry line, calculate sample values and compare upward, downward and wider V shapes.
Plot four functions in y=a|x-h|+k. Identify each vertex and symmetry line, calculate sample values and compare upward, downward and wider V shapes.
Choose Letter or A4, then download a student sheet first. Keep the separate reference or answer file closed until you have attempted the case. Every worksheet, SVG and answer is free without signup.
Original worksheet collection. Letter or A4 PDF and editable SVG.
Original worksheet collection. Letter or A4 PDF and editable SVG.
Original worksheet collection. Letter or A4 PDF and editable SVG.
Original worksheet collection. Letter or A4 PDF and editable SVG.
Complete free PDF pack.
| Case | Focus |
|---|---|
| AV01 | A centered V |
| AV02 | A shifted V |
| AV03 | A reflected V |
| AV04 | A wider V |
All examples use original fictional data and coordinate units. They are practice material, not measured classroom results, published questions or evidence about real people.
AV01: A centered V
| Check | Worked result |
|---|---|
| 1 | Vertex (0,0); axis x=0. |
| 2 | (-3,3); (-1,1); (0,0); (1,1); (3,3) |
| 3 | Opens up. |
The first case shows the checking pattern. Individual answer files contain the completed graph and result for every case.
Begin by identifying the vertex coordinates h and k from the function definition y equals a absolute value of x minus h plus k. For the centered V case, the vertex is at zero zero. For the shifted V, it is at two negative one. Mark this point clearly on your grid. Do not rely on physical ruler measurements to find this spot. Instead, use the listed coordinates provided in the reference PDF. This ensures the sharp corner aligns exactly with the mathematical definition. The vertex serves as the anchor for all subsequent plotting steps. Accurate placement here prevents distortion in the rest of the graph. Verify the coordinates against the supplied values before proceeding to the next step.
Use the table of symmetric inputs around the vertex h to calculate y values. For the reflected V case with vertex at negative one three, use inputs negative four, negative two, negative one, zero, and two. Calculate each output using the function rule. Plot these points on the coordinate plane. Connect the points with straight lines to form the V shape. Ensure the lines meet at a sharp corner at the vertex. Do not smooth the corner. The symmetry of the inputs ensures the graph balances correctly around the vertical axis of symmetry. Check your plotted points against the answer PDF to confirm accuracy. This method relies on calculation, not physical estimation.
Adjust the width of the V by applying the coefficient a. For the wider V case, use a equals one half. This value stretches the graph vertically compared to the standard V. Calculate the y values using this factor. The vertex remains at one negative two. The inputs are negative two, zero, one, two, and four. Plot these calculated points. The graph will appear broader than the standard case. Do not use a physical ruler to estimate the width. Rely on the calculated coordinates. The negative sign in other cases flips the graph downward. Verify the shape matches the reference image. The scale factor directly controls the steepness of the lines.
| Watch for | Check instead |
|---|---|
| Using a physical ruler to measure distances on the printed page to determine the vertex location. | Use the digital coordinates provided in the reference PDF to locate the vertex exactly on the grid. |
| Smoothing the corner at the vertex to make the graph look like a curve. | Plot a sharp corner at the vertex by connecting the straight lines directly at the calculated point. |
| Assuming the graph width is standard when the coefficient a is one half. | Apply the scale factor a equals one half to stretch the graph vertically, making it wider than the standard case. |
Choose the PDF that matches your paper. Print at Actual size / 100% and turn off Fit, Shrink and browser headers. SVG is editable but an editor can rescale it; the PDF is the simpler print route. Physical printer, classroom and curriculum tests have not been completed.
Match the expression with x-h. In |x-2| the vertex x-coordinate is 2; in |x+2| it is -2.
No. It changes the opening direction and makes the vertex a maximum, but h and k still locate the corner.
No. Plot the samples, then use two straight arms with the correct slopes.
Yes. The individual student, reference and answer files plus collection packs are free without signup. Credit Zumola under CC BY 4.0 when sharing or adapting.
For any absolute value graph y = a|x - h| + k, the vertex is at (h, k). The two linear pieces have slopes a and -a. When graphing an inequality such as y < a|x - h| + k, first graph the boundary y = a|x - h| + k. Use a dashed boundary if the inequality is strict (< or >) and a solid boundary if it includes equality (≤ or ≥). Then test one point not on the boundary to decide which region to shade.
A common error is shading the wrong side because the test point lies on the boundary or because the inequality symbol was misread. Choose a test point clearly inside one region, such as the vertex shifted vertically by one unit, and substitute it into the original inequality. If the statement is true, shade that region; if false, shade the other side.
Also check the domain and range. The domain of any absolute value function of this form is all real numbers. The range depends on the sign of a: if a > 0, the range is y ≥ k; if a < 0, the range is y ≤ k. For inequalities, the shaded region may extend beyond these bounds, so state the boundary's range separately from the shaded set.
Consider the inequality y ≤ 2|x - 1| - 3. The boundary is y = 2|x - 1| - 3, with vertex (1, -3) and slopes 2 and -2. Because the inequality includes equality (≤), draw a solid V. To decide shading, test the point (1, 0), which is above the vertex. Substitute: 0 ≤ 2|1 - 1| - 3 = -3, which is false. Therefore shade the region below the V, not above it.
Verify with another point: (0, -1). Left side: -1. Right side: 2|0 - 1| - 3 = 2(1) - 3 = -1. The statement -1 ≤ -1 is true, so (0, -1) lies on the boundary and should be included. A point just below the boundary, such as (0, -2), gives -2 ≤ -1, true, confirming the shaded region is below.
This example also exposes a frequent mistake: assuming the shaded region is always above the V. The test point method prevents that error. Record the test point and its substitution on the worksheet so the reasoning can be checked independently.
After graphing, verify three independent features: the vertex coordinates, one point on each branch, and the direction of opening. For y = a|x - h| + k, the vertex must satisfy the equation exactly. Substitute x = h and confirm y = k. Then choose x = h + 1 and x = h - 1; both substitutions give y = a|1| + k = k + a, so the two points are symmetric about x = h and their y-value is k + a. If a is negative, this value is below k, which confirms the downward opening.
Check intercepts only when they exist. The y-intercept is found by setting x = 0: y = a| -h| + k = a|h| + k. The x-intercepts require solving a|x - h| + k = 0, which has solutions only if -k/a ≥ 0. If no real solution exists, state that the graph has no x-intercept rather than forcing one.
Finally, confirm that the graph is a function: it must pass the vertical line test. Absolute value graphs of this form always do, but inequality regions may not represent functions. Label the boundary and shaded region clearly so the reader can distinguish the function from the solution set.
Graph the inequality y > |x + 2| - 1 and determine the shaded region.
The boundary is y = |x + 2| - 1, with vertex (-2, -1) and slopes 1 and -1. Since the inequality is strict (>), draw a dashed V. Test the point (-2, 0): 0 > |0| - 1 = -1 is true, so shade above the V. Check (-2, -2): -2 > -1 is false, confirming the region below is excluded. The boundary is not part of the solution.
Graph the boundary first, then test one point not on the boundary. If the test point makes the inequality true, shade the region containing that point; otherwise shade the other region.
Yes. For y = a|x - h| + k, x-intercepts exist only when -k/a is greater than or equal to zero. If it is negative, the graph does not cross the x-axis.
Because a = -2 is negative, the graph opens downward with vertex (-1, 4). The range is y ≤ 4.
