Square Root
Functions
Every square root function follows a completely predictable pattern. This guide shows you how to read the shape, direction, domain, and range directly from the equation — and how to work backwards from a graph to find the rule.
You already know how to square a number: 4² = 16. The square root undoes that: √16 = 4. A square root function builds a whole graph out of this — for every valid x, compute the square root and plot the result.
Square roots grow quickly at first, then slow down. √1 = 1, √4 = 2, √9 = 3, √16 = 4. To go from output 1 to 2 took a jump of 3 in x (from 1 to 4). To go from 3 to 4 took a jump of 7 (from 9 to 16). The bigger x gets, the less the function grows — that flattening curve is the signature shape of a square root.
The four parameters a, b, h, k transform this basic shape just like they transform any function — stretching, flipping, shifting. Understanding what each one does is the entire skill for this topic.
x = 1 is the starting point (vertex). The output is just k = 2.
You cannot take the square root of a negative number. x = 3 is NOT in the domain.
The domain starts exactly at x = 5. This is the vertex point (5, 0).
Domain: expression under radical is x + 3 ≥ 0, so x ≥ −3. Domain: [−3, +∞).
| Parameter | What it does | Effect on vertex / domain |
|---|---|---|
| a | Vertical stretch + direction | |a| = steepness. a < 0 → flips upside-down. |
| b | Horizontal direction | b > 0 → graph goes right. b < 0 → graph goes left. |
| h | Horizontal shift | Vertex x-coordinate. Read carefully: (x − h) means h is positive even with a minus sign. |
| k | Vertical shift | Vertex y-coordinate. Starting height of the graph. |
| vertex | Starting point of the graph | Always (h, k) |
So: y = √(4(x + 3)) − 3
a = 2, b = 1, h = −3, k = −3. Vertex = (−3, −3)
So: y = √(−9(x − 2)) + 4
a = 3, b = −1, h = 2, k = 4. Vertex = (2, 4). Goes UP and LEFT.
So: y = √3 · √(x − 4) + 1
a = 4, b = 1, h = 4, k = −3. Vertex (4, −3), opens up-right.
Before you plot a single point, you can determine the overall shape from just the signs of a and b. There are exactly four possibilities.
| Sign of a | Sign of b | Direction from vertex | Memory aid |
|---|---|---|---|
| a > 0 | b > 0 | Up and Right ↗ | The basic √x shape |
| a < 0 | b > 0 | Down and Right ↘ | Flipped vertically |
| a > 0 | b < 0 | Up and Left ↖ | Reflected horizontally |
| a < 0 | b < 0 | Down and Left ↙ | Both flipped |
a < 0 → goes DOWN. b > 0 → goes RIGHT. The graph opens down-right from the vertex.
a = −3, b = −1, h = −2, k = 6
a < 0 → goes DOWN. b < 0 → goes LEFT. Down-left from the vertex.
(A) y = 2√(x − 1) + 3 (B) y = −√(x + 2) − 1 (C) y = 4√(−x + 3) + 2 (D) y = −2√(−x) − 5
A: a = 2 > 0, b = 1 > 0 → Up-right
B: a = −1 < 0, b = 1 > 0 → Down-right
C: Factor: √(−(x − 3)), so b = −1 < 0, a = 4 > 0 → Up-left
D: √(−x) = √(−1·x), b = −1 < 0, a = −2 < 0 → Down-left
Domain and range are determined entirely by the vertex (h, k) and the directions (signs of a and b). Once you know the parameters, you can write them down immediately.
Domain and range for four different functions
| Function | a | b | h | k | Domain | Range |
|---|---|---|---|---|---|---|
| y = 3√(x − 2) + 1 | 3 | 1 | 2 | 1 | [2, +∞) | [1, +∞) |
| y = −2√(x + 1) − 3 | −2 | 1 | −1 | −3 | [−1, +∞) | (−∞, −3] |
| y = √(−x + 5) + 2 | 1 | −1 | 5 | 2 | (−∞, 5] | [2, +∞) |
| y = −√(−(x − 3)) − 4 | −1 | −1 | 3 | −4 | (−∞, 3] | (−∞, −4] |
Already factored: b = −2, h = −1, k = −5, a = 3.
Hint: factor the expression under the radical first.
a = −5, b = −1, h = 2, k = 8.
- Step 1: Rewrite in standard form — factor the coefficient of x out from under the radical and pull any perfect square factor outside.
- Step 2: Identify a, b, h, k.
- Step 3: Plot the vertex (h, k).
- Step 4: Determine direction from signs of a and b. Lightly sketch the curve going that way.
- Step 5: Find at least one additional point (try the y-intercept if it exists, or pick a convenient x value), plot it, and draw the curve through both points.
Vertex: (4, 1). a > 0 and b > 0 → up-right.
Domain starts at x = 4. Since 0 < 4, x = 0 is NOT in the domain — no y-intercept.
No y-intercept.
a = −2, b = −1, h = 6, k = 4.
a < 0 → DOWN. b < 0 → LEFT. Opens down-left from vertex (6, 4).
a = 3 > 0 (UP), b = −2 < 0 (LEFT). Opens up-left from vertex (1, 2).
At x = −3:
Y-Intercept
Let x = 0 and evaluate. But first check: is x = 0 in the domain? If the domain starts at h > 0 (and b > 0), then x = 0 is outside the domain and there is no y-intercept.
X-Intercept (zero of the function)
Let y = 0, isolate the square root, then square both sides. Always substitute your answer back into the original equation to verify — squaring can introduce extraneous solutions.
(1) Finding the y-intercept without checking the domain — if x = 0 is outside the domain, stop immediately.
(2) Forgetting to verify the x-intercept after squaring — check it in the original equation.
Y-intercept: (0, 2√3 − 4)
No y-intercept — x = 0 is outside the domain.
Y-intercept: (0, 3)
X-intercept:
Y-intercept: Domain is [−3, +∞). Is x = 0 in domain? Yes (0 ≥ −3).
Given the vertex and one other point, you can always find the complete equation. The vertex gives you h and k. The direction (from the graph or context) gives you the sign of b. The second point lets you solve for a.
- Step 1: Read vertex → gives h and k directly.
- Step 2: Determine direction from the graph → gives sign of a and sign of b. Use b = 1 (right) or b = −1 (left) as the simplest choice unless told otherwise.
- Step 3: Write the template: y = a√(±(x − h)) + k.
- Step 4: Substitute the second known point (x, y) and solve for a.
- Step 5: Write and verify the complete rule.
Check: vertex at x = −4: y = 3√0 + 4 = 4 ✓. At x = 0: y = 3(2) + 4 = 10 ✓.
a = −1 < 0 confirms downward direction ✓
Vertex (x=3): y = 3√0 − 2 = −2 ✓. At x = −1: y = 3√4 − 2 = 6 − 2 = 4 ✓.
Template: y = a√(x + 1) + 3 (b = 1, h = −1, k = 3).
Rule: y = 3√(x + 1) + 3. Check: vertex → y = 3(0)+3 = 3 ✓. At x=3: y=3(2)+3=9 ✓.
Template: y = a√(x − 4) + 6 (b = 1, a < 0).
Rule: y = −2√(x − 4) + 6. Check: vertex → y = 0+6 = 6 ✓. At x=8: y=−4+6=2 ✓.
Questions like "for which x values is f(x) ≥ 3?" combine graphing with inequality reasoning.
- Step 1: Find the intersection point — set f(x) = c and solve for x.
- Step 2: From the graph (or by testing a point), determine which side of the intersection satisfies the inequality.
- Step 3: Remember the domain — the answer must lie within the domain.
- Step 4: Use [ ] for ≥ or ≤ (intersection included), ( ) for > or < (strict).
The function opens up-right. As x increases beyond 5, the function grows above 5. So f(x) ≥ 5 for x ≥ 5.
The function −2√(x−3)+8 starts at (3, 8) and goes down-right. The horizontal line y = 4 is hit at x = 7. To the right of x = 7, the function is below 4. This matches x > 7. ✓
Set equal: −2√(x+3)+6 = 2 → −2√(x+3) = −4 → √(x+3) = 2 → x+3 = 4 → x = 1.
Function opens downward (a < 0, b > 0). At x = 1 it equals 2. To the LEFT of x = 1 (closer to vertex) it is above 2. Domain starts at x = −3.
Domain: 5 − x ≥ 0 → x ≤ 5. The solution x ≤ −4 is within (−∞, 5].
Verify: at x = −4: √(5−(−4))+1 = √9+1 = 4 ✓. At x = −5: √10+1 ≈ 4.16 > 4 ✓. At x = 0: √5+1 ≈ 3.24 < 4 ✓.
Domain: [−5, +∞). Intersect with x ≤ −1: x ∈ [−5, −1].
Verify: at x = −5 (vertex): y = 4√0 − 2 = −2 ≤ 6 ✓. At x = −1: y = 4√4 − 2 = 6 ≤ 6 ✓. At x = 0: y = 4√5 − 2 ≈ 6.94 > 6 ✓.