Free graphing resources · Original practice
Plot four one-sided radical curves from y=a√(x-h)+k. Locate the endpoint, complete perfect-square samples and compare reflected and shifted roots.
Plot four one-sided radical curves from y=a√(x-h)+k. Locate the endpoint, complete perfect-square samples and compare reflected and shifted roots.
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 |
|---|---|
| SQ01 | Basic square root |
| SQ02 | Shift right |
| SQ03 | Reflected root |
| SQ04 | Narrower rise |
All examples use original fictional data and coordinate units. They are practice material, not measured classroom results, published questions or evidence about real people.
SQ01: Basic square root
| Check | Worked result |
|---|---|
| 1 | Endpoint (0,0). |
| 2 | (0,0); (1,1); (4,2); (9,3); (16,4) |
| 3 | Opens upward. |
The first case shows the checking pattern. Individual answer files contain the completed graph and result for every case.
Begin by locating the endpoint of the curve using the horizontal shift value h. This coordinate marks the leftmost point of the graph because the domain begins at h. For example, in the shift right case, the endpoint is at x equals two. Mark this point clearly on your grid before plotting other values. This step ensures the curve starts in the correct position. The vertical shift k determines the y coordinate of this endpoint. Combining h and k gives the precise starting location. This foundational step prevents misalignment when connecting subsequent points. It establishes the anchor for the entire function shape. Ensure you distinguish this from the y intercept if it exists. The endpoint is always the first value in the domain range.
Use the provided table of x values to calculate corresponding y coordinates. These x values are chosen so that horizontal offsets from h are perfect squares. This simplifies the square root calculation significantly. For instance, if h is zero, x values like one, four, and nine yield integer results. Substitute each x into the function formula to find y. Plot these points accurately on the coordinate plane. Do not connect them with straight lines yet. Each point represents a specific solution derived from listed coordinates. The supplied values determine these answers exactly. Verify that the offset from h is indeed a perfect square. This ensures the arithmetic remains manageable and precise. Plotting these points creates the skeleton of the curve.
Connect the plotted points using a smooth one-sided curve. Start at the endpoint and extend through the sample points. The curve should not have sharp corners or breaks. If the coefficient a is negative, the curve reflects vertically. This means it opens downward instead of upward. For the narrower rise case, the curve rises more steeply. Ensure the line passes through all calculated points. Do not extend the curve to the left of the endpoint. The domain restriction means the graph exists only for x greater than or equal to h. Use a steady hand to maintain smoothness. This visual representation completes the graphical solution. Check that the shape matches the function type. The curve should look natural and continuous.
| Watch for | Check instead |
|---|---|
| Extending the curve to the left of the endpoint | The domain begins at h, so the graph exists only for x values greater than or equal to h. Stop the curve at the endpoint and do not draw any line to the left of this point. The one-sided nature is a key feature of square root functions. |
| Connecting points with straight line segments | Square root functions produce curved graphs, not linear ones. Use a smooth curve to connect the plotted points. Straight lines between points create a jagged appearance that misrepresents the function. Ensure the transition between points is gradual and continuous. |
| Ignoring the vertical reflection when a is negative | A negative coefficient reflects the curve vertically across the horizontal axis. If a is negative, the curve opens downward from the endpoint. Check the sign of a before drawing. For the reflected root case, the curve decreases as x increases. |
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At x=h in y=a√(x-h)+k, the endpoint is (h,k).
They make x-h perfect squares, which makes the sample values easy to check.
It reflects the curve vertically, so it travels downward from the endpoint.
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Consider y = -2√(x - 1) + 3. The expression under the root must be nonnegative, so x - 1 ≥ 0, giving domain x ≥ 1. The endpoint occurs when the radicand is zero: x = 1, and then y = -2√0 + 3 = 3. Plot the endpoint at (1, 3) with a solid dot because the domain includes x = 1.
Choose perfect-square offsets to keep arithmetic simple. Let x - 1 = 1, so x = 2 and y = -2(1) + 3 = 1. Let x - 1 = 4, so x = 5 and y = -2(2) + 3 = -1. Let x - 1 = 9, so x = 10 and y = -2(3) + 3 = -3. These points (2, 1), (5, -1), and (10, -3) lie on the curve. Because a = -2 is negative, the curve opens downward from the endpoint; because |a| = 2, it rises or falls twice as fast as the parent square root curve.
A frequent error is plotting the endpoint at the wrong x-value. For y = a√(x - h) + k, the endpoint is always at x = h, not at x = 0 unless h = 0. If your graph starts at the origin for a shifted function, recheck the horizontal shift. Another error is drawing a two-sided curve; the square root function has only one side because the domain is restricted to x ≥ h when a is real.
Reflection is controlled by the sign of a. If a is positive, the curve extends upward from the endpoint; if a is negative, it extends downward. A common mistake is to reflect the curve but keep the endpoint in the same place without checking that the endpoint is still on the curve. The endpoint is always on the curve because the radicand is zero there, so y = k. Verify that your endpoint has y-coordinate k.
Steepness is controlled by |a|. For |a| > 1, the curve is steeper than the parent; for 0 < |a| < 1, it is flatter. Do not confuse a vertical stretch with a horizontal shift. A quick check is to compare the y-value at one perfect-square offset with the parent value multiplied by a.
To deepen understanding, graph y = √(x - 2) and y = -√(x - 2) + 4 on the same coordinate grid. The first has endpoint (2, 0) and opens upward; the second has endpoint (2, 4) and opens downward. Use perfect-square offsets 1, 4, and 9 to generate points for each. For the first, points include (3, 1), (6, 2), and (11, 3). For the second, points include (3, 3), (6, 2), and (11, 1). Notice that the second curve is the first curve reflected across a horizontal line and shifted, but you can verify each point independently.
This comparison helps distinguish horizontal shifts from vertical shifts and reflections. Label each curve clearly and mark endpoints with solid dots. Use a ruler for axes and keep equal spacing on both axes so steepness comparisons are meaningful.
Graph y = 3√(x + 2) - 1 and verify three points.
Domain: x + 2 ≥ 0, so x ≥ -2. Endpoint: set x + 2 = 0, so x = -2 and y = 3√0 - 1 = -1; endpoint (-2, -1). Choose offsets 1, 4, and 9: x + 2 = 1 gives x = -1, y = 3(1) - 1 = 2; x + 2 = 4 gives x = 2, y = 3(2) - 1 = 5; x + 2 = 9 gives x = 7, y = 3(3) - 1 = 8. Points: (-1, 2), (2, 5), (7, 8). Check (2, 5): 3√(2 + 2) - 1 = 3√4 - 1 = 3(2) - 1 = 5, correct. The curve starts at (-2, -1) and rises to the right because a = 3 is positive.
Set the expression inside the square root equal to zero and solve for x. That x-value is the endpoint's x-coordinate. Substitute it into the function to find the y-coordinate, which will equal k in y = a√(x - h) + k.
The principal square root is defined only for nonnegative radicands. For y = a√(x - h) + k, this means x - h ≥ 0, so x ≥ h. The graph therefore exists only on one side of the endpoint.
Substitute the coordinates of at least two plotted points into the original equation. Confirm the endpoint satisfies the equation, the domain matches the direction of the curve, and the sign of a matches whether the curve rises or falls from the endpoint.
