Square Root Function

Taking the Principal Square Root

A square root function applies the principal square root to an expression. It has the form $f(x) = \sqrt{g(x)}$ , where $g(x)$ is the radicand, the expression under the radical sign.

The parent function is $f(x) = \sqrt{x}$ . It assigns each non-negative input its non-negative square root.

Characteristics of Square Root Functions

Square root functions share three traits. Check each one against $f(x) = \sqrt{x}$ as you read.

Limited domain: Since the square root of negative numbers is not defined in real numbers, the domain of square root functions is limited to values that make the expression inside the square root non-negative.
Recognizable parent graph: The graph of $y=\sqrt{x}$ begins at $(0,0)$ and rises more slowly as $x$ increases. Transformations can shift, stretch, or reflect this shape.
Non-negative principal root: The value of $\sqrt{g(x)}$ itself is always non-negative. An outside multiplier or translation can still make a transformed function negative.

Domain and Range of Square Root Functions

The expression under the root may not become negative, and that restriction produces the domain. The range contains only non negative values, because a square root never comes out negative.

The domain and range determine which inputs and outputs a square root function can have.

Determining Domain

The domain of a square root function $f(x) = \sqrt{g(x)}$ is all values of $x$ that make $g(x) \geq 0$ .

Steps to determine domain:

Step	Explanation	Example: $f(x) = \sqrt{x-2}$
$1$	Identify the expression inside the square root	$g(x) = x-2$
$2$	Create the inequality $g(x) \geq 0$	$x-2 \geq 0$
$3$	Solve the inequality	$x \geq 2$
$4$	Write the domain	$D_f = \{x \mid x \geq 2, x \in \mathbb{R}\}$

Determining Range

The range of a square root function is all possible output values that the function can produce.

For $f(x) = \sqrt{x-2}$ , the radicand can produce every value from zero upward. Its principal square root therefore has range

R_f = \{y \mid y \geq 0, y \in \mathbb{R}\}

Graph of Basic Square Root Function

The parent graph exists only for $x \geq 0$ and passes through $(0,0)$ , $(1,1)$ , and $(4,2)$ . To raise the output by one more unit, the input must grow by a larger amount than before.

Basic Square Root Function:

f(x) = \sqrt{x}

The graph of the basic square root function starts from point

(0,0)

and rises at an increasingly slower rate.

Transformations of Square Root Functions

Shifting, stretching, and reflecting move the starting point of the graph and the position of the curve.

Square root functions can undergo various transformations that change the shape and position of their graphs.

Horizontal Translation

The function $f(x) = \sqrt{x-h}$ shifts the graph of the basic square root function by $h \text{ units}$ to the right (if $h > 0$ ) or to the left (if $h < 0$ ).

Horizontal Translation of Square Root Functions

Comparison of

f(x) = \sqrt{x}

g(x) = \sqrt{x-2}

, and

h(x) = \sqrt{x+2}

Vertical Translation

The function $f(x) = \sqrt{x} + k$ shifts the graph of the basic square root function by $k \text{ units}$ upward (if $k > 0$ ) or downward (if $k < 0$ ).

Function	Transformation	Domain	Range
$f(x) = \sqrt{x}$	Basic function	$x \geq 0$	$y \geq 0$
$f(x) = \sqrt{x} + 2$	Shift $2 \text{ units}$ upward	$x \geq 0$	$y \geq 2$
$f(x) = \sqrt{x} - 3$	Shift $3 \text{ units}$ downward	$x \geq 0$	$y \geq -3$

Dilation

For $f(x) = a\sqrt{x}$ , a value $|a|>1$ stretches the graph vertically, while $0<|a|<1$ compresses it. A negative $a$ also reflects the graph across the $x$ -axis.

Vertical Dilation of Square Root Functions

Comparison of

f(x) = \sqrt{x}

g(x) = 2\sqrt{x}

, and

h(x) = \frac{1}{2}\sqrt{x}

General Form of Square Root Functions

For the nondegenerate case $a\neq 0$ and $b\neq 0$ , the transformed square root function has the form:

f(x) = a\sqrt{b(x-h)} + k

Here:

$a$ determines vertical scaling and, when $a<0$ , reflection across the $x$ -axis
$b$ determines horizontal scaling and, when $b<0$ , horizontal reflection
$h$ determines horizontal translation
$k$ determines vertical translation

The real domain is determined by $b(x-h)\geq 0$ . The sign of $a$ then determines whether the range extends upward or downward from $y=k$ .

Steps to draw the graph:

Step	Action	Example: $f(x) = 2\sqrt{x-1} + 3$
$1$	Determine starting point	$x-1 = 0 \Rightarrow x = 1$ , $f(1) = 3$ , Point: $(1, 3)$
$2$	Determine domain	$x-1 \geq 0 \Rightarrow x \geq 1$
$3$	Create value table	Choose several values $x \geq 1$
$4$	Calculate function values	For $x = 2$ : $f(2) = 2\sqrt{1} + 3 = 5$
$5$	Plot points	Plot $(1,3)$ , $(2,5)$ , $(5,7)$ , etc.
$6$	Connect points	Create a smooth curve through the points

Square Root Function Equations

To solve an equation containing a square root, isolate the radical, square both sides, solve, and check every candidate in the original equation. Squaring is not reversible for arbitrary real expressions, so it can introduce an extraneous solution.

Step	Explanation	Example: $\sqrt{2x+3} = 5$
$1$	Isolate the square root	Already isolated
$2$	Square both sides	$(\sqrt{2x+3})^2 = 5^2$
$3$	Simplify	$2x+3 = 25$
$4$	Solve	$2x = 22 \Rightarrow x = 11$
$5$	Verify	$\sqrt{2(11)+3} = \sqrt{25} = 5$ ✓

Square Root Function Inequalities

For a square root inequality, first enforce the domain. Before squaring, also check the sign of the side without the radical so the comparison remains equivalent.

Example: Solve $\sqrt{x-1} < 3$

\text{Condition: } x-1 \geq 0 \Rightarrow x \geq 1

Because both sides are non-negative on the domain, squaring preserves the inequality. Combining $x\geq 1$ with $x<10$ gives $1 \leq x < 10$ .

Practice Problems

The problems ask for the domain, the range, or the graph of a root function. Start from the condition inside the square root.

Determine the domain and range of the function $f(x) = \sqrt{3x-6}$
Draw the graph of the function $g(x) = -\sqrt{x+4} + 2$
Solve the equation $\sqrt{x+5} + \sqrt{x} = 5$
A rocket is launched vertically. Its height after $t \text{ seconds}$ is given by $h(t) = 100\sqrt{t} \text{ meters}$ . What is the height of the rocket after $9 \text{ seconds}$ ?
Determine the value of $x$ that satisfies $2\sqrt{x-3} \geq 6$

Worked Solutions

The radicand must be non-negative:
$3x-6\geq 0$
Therefore, $D_f=\{x\in\mathbb{R}\mid x\geq 2\}$ . As $x$ runs through this domain, $3x-6$ takes every non-negative value, so $R_f=\{y\in\mathbb{R}\mid y\geq 0\}$ .

Drawing the graph $g(x) = -\sqrt{x+4} + 2$

Steps to draw:

Step	Explanation	Details for $g(x) = -\sqrt{x+4} + 2$
$1$	Identify transformations	$a = -1$ (reflection across the $x$ -axis), $h = -4$ (shift $4 \text{ units}$ left), $k = 2$ (shift $2 \text{ units}$ up)
$2$	Determine starting point	$x+4 = 0 \Rightarrow x = -4$ , $g(-4) = 0 + 2 = 2$ , Starting point: $(-4, 2)$
$3$	Determine domain	$x+4 \geq 0 \Rightarrow x \geq -4$
$4$	Determine range	Since $a = -1 < 0$ , the graph decreases from the starting point, so $y \leq 2$
$5$	Create value table	Choose values $x \geq -4$

Value table:

$x$	$x+4$	$\sqrt{x+4}$	$-\sqrt{x+4}$	$g(x) = -\sqrt{x+4} + 2$
$-4$	$0$	$0$	$0$	$2$
$-3$	$1$	$1$	$-1$	$1$
$0$	$4$	$2$	$-2$	$0$
$5$	$9$	$3$	$-3$	$-1$
$12$	$16$	$4$	$-4$	$-2$

Graph of

g(x) = -\sqrt{x+4} + 2

Graph of a square root function that undergoes reflection across the

x

-axis and translation.

The graph begins at $(-4,2)$ . The negative sign reflects the parent graph across the $x$ -axis, so the curve falls to the right and its outputs stay at or below $2$ .

The domain requires $x\geq 0$ . Isolate one radical:
$\sqrt{x+5}=5-\sqrt{x}$
Check the original equation: $\sqrt{4+5}+\sqrt{4}=3+2=5$ . Thus, $x=4$ is valid.
Substitute $t=9$ into the model:

$h(9)=100\sqrt{9}=100\cdot 3=300\text{ m}.$

The model uses $t\geq 0$ , and $9$ lies in that domain. The modeled height is $300\text{ m}$ .
Record the domain $x-3\geq 0$ , so $x\geq 3$ . Divide by the positive number $2$ , then square the non-negative sides:
$2\sqrt{x-3}\geq 6$
Intersecting this result with the domain still gives $x\geq 12$ .

Taking the Principal Square Root

A square root function applies the principal square root to an expression. It has the form $f(x) = \sqrt{g(x)}$ , where $g(x)$ is the radicand, the expression under the radical sign.

The parent function is $f(x) = \sqrt{x}$ . It assigns each non-negative input its non-negative square root.

Characteristics of Square Root Functions

Square root functions share three traits. Check each one against $f(x) = \sqrt{x}$ as you read.

Limited domain: Since the square root of negative numbers is not defined in real numbers, the domain of square root functions is limited to values that make the expression inside the square root non-negative.
Recognizable parent graph: The graph of $y=\sqrt{x}$ begins at $(0,0)$ and rises more slowly as $x$ increases. Transformations can shift, stretch, or reflect this shape.
Non-negative principal root: The value of $\sqrt{g(x)}$ itself is always non-negative. An outside multiplier or translation can still make a transformed function negative.

Domain and Range of Square Root Functions

The expression under the root may not become negative, and that restriction produces the domain. The range contains only non negative values, because a square root never comes out negative.

The domain and range determine which inputs and outputs a square root function can have.

Determining Domain

The domain of a square root function $f(x) = \sqrt{g(x)}$ is all values of $x$ that make $g(x) \geq 0$ .

Steps to determine domain:

Step	Explanation	Example: $f(x) = \sqrt{x-2}$
$1$	Identify the expression inside the square root	$g(x) = x-2$
$2$	Create the inequality $g(x) \geq 0$	$x-2 \geq 0$
$3$	Solve the inequality	$x \geq 2$
$4$	Write the domain	$D_f = \{x \mid x \geq 2, x \in \mathbb{R}\}$

Determining Range

The range of a square root function is all possible output values that the function can produce.

For $f(x) = \sqrt{x-2}$ , the radicand can produce every value from zero upward. Its principal square root therefore has range

R_f = \{y \mid y \geq 0, y \in \mathbb{R}\}

Graph of Basic Square Root Function

The parent graph exists only for $x \geq 0$ and passes through $(0,0)$ , $(1,1)$ , and $(4,2)$ . To raise the output by one more unit, the input must grow by a larger amount than before.

Basic Square Root Function:

f(x) = \sqrt{x}

The graph of the basic square root function starts from point

(0,0)

and rises at an increasingly slower rate.

Transformations of Square Root Functions

Shifting, stretching, and reflecting move the starting point of the graph and the position of the curve.

Square root functions can undergo various transformations that change the shape and position of their graphs.

Horizontal Translation

The function $f(x) = \sqrt{x-h}$ shifts the graph of the basic square root function by $h \text{ units}$ to the right (if $h > 0$ ) or to the left (if $h < 0$ ).

Horizontal Translation of Square Root Functions

Comparison of

f(x) = \sqrt{x}

g(x) = \sqrt{x-2}

, and

h(x) = \sqrt{x+2}

Vertical Translation

The function $f(x) = \sqrt{x} + k$ shifts the graph of the basic square root function by $k \text{ units}$ upward (if $k > 0$ ) or downward (if $k < 0$ ).

Function	Transformation	Domain	Range
$f(x) = \sqrt{x}$	Basic function	$x \geq 0$	$y \geq 0$
$f(x) = \sqrt{x} + 2$	Shift $2 \text{ units}$ upward	$x \geq 0$	$y \geq 2$
$f(x) = \sqrt{x} - 3$	Shift $3 \text{ units}$ downward	$x \geq 0$	$y \geq -3$

Dilation

For $f(x) = a\sqrt{x}$ , a value $|a|>1$ stretches the graph vertically, while $0<|a|<1$ compresses it. A negative $a$ also reflects the graph across the $x$ -axis.

Vertical Dilation of Square Root Functions

Comparison of

f(x) = \sqrt{x}

g(x) = 2\sqrt{x}

, and

h(x) = \frac{1}{2}\sqrt{x}

General Form of Square Root Functions

For the nondegenerate case $a\neq 0$ and $b\neq 0$ , the transformed square root function has the form:

f(x) = a\sqrt{b(x-h)} + k

Here:

$a$ determines vertical scaling and, when $a<0$ , reflection across the $x$ -axis
$b$ determines horizontal scaling and, when $b<0$ , horizontal reflection
$h$ determines horizontal translation
$k$ determines vertical translation

The real domain is determined by $b(x-h)\geq 0$ . The sign of $a$ then determines whether the range extends upward or downward from $y=k$ .

Steps to draw the graph:

Step	Action	Example: $f(x) = 2\sqrt{x-1} + 3$
$1$	Determine starting point	$x-1 = 0 \Rightarrow x = 1$ , $f(1) = 3$ , Point: $(1, 3)$
$2$	Determine domain	$x-1 \geq 0 \Rightarrow x \geq 1$
$3$	Create value table	Choose several values $x \geq 1$
$4$	Calculate function values	For $x = 2$ : $f(2) = 2\sqrt{1} + 3 = 5$
$5$	Plot points	Plot $(1,3)$ , $(2,5)$ , $(5,7)$ , etc.
$6$	Connect points	Create a smooth curve through the points

Square Root Function Equations

Step	Explanation	Example: $\sqrt{2x+3} = 5$
$1$	Isolate the square root	Already isolated
$2$	Square both sides	$(\sqrt{2x+3})^2 = 5^2$
$3$	Simplify	$2x+3 = 25$
$4$	Solve	$2x = 22 \Rightarrow x = 11$
$5$	Verify	$\sqrt{2(11)+3} = \sqrt{25} = 5$ ✓

Square Root Function Inequalities

For a square root inequality, first enforce the domain. Before squaring, also check the sign of the side without the radical so the comparison remains equivalent.

Example: Solve $\sqrt{x-1} < 3$

\text{Condition: } x-1 \geq 0 \Rightarrow x \geq 1

Because both sides are non-negative on the domain, squaring preserves the inequality. Combining $x\geq 1$ with $x<10$ gives $1 \leq x < 10$ .

Practice Problems

The problems ask for the domain, the range, or the graph of a root function. Start from the condition inside the square root.

Determine the domain and range of the function $f(x) = \sqrt{3x-6}$
Draw the graph of the function $g(x) = -\sqrt{x+4} + 2$
Solve the equation $\sqrt{x+5} + \sqrt{x} = 5$
A rocket is launched vertically. Its height after $t \text{ seconds}$ is given by $h(t) = 100\sqrt{t} \text{ meters}$ . What is the height of the rocket after $9 \text{ seconds}$ ?
Determine the value of $x$ that satisfies $2\sqrt{x-3} \geq 6$

Worked Solutions

The radicand must be non-negative:
$3x-6\geq 0$
Therefore, $D_f=\{x\in\mathbb{R}\mid x\geq 2\}$ . As $x$ runs through this domain, $3x-6$ takes every non-negative value, so $R_f=\{y\in\mathbb{R}\mid y\geq 0\}$ .

Drawing the graph $g(x) = -\sqrt{x+4} + 2$

Steps to draw:

Step	Explanation	Details for $g(x) = -\sqrt{x+4} + 2$
$1$	Identify transformations	$a = -1$ (reflection across the $x$ -axis), $h = -4$ (shift $4 \text{ units}$ left), $k = 2$ (shift $2 \text{ units}$ up)
$2$	Determine starting point	$x+4 = 0 \Rightarrow x = -4$ , $g(-4) = 0 + 2 = 2$ , Starting point: $(-4, 2)$
$3$	Determine domain	$x+4 \geq 0 \Rightarrow x \geq -4$
$4$	Determine range	Since $a = -1 < 0$ , the graph decreases from the starting point, so $y \leq 2$
$5$	Create value table	Choose values $x \geq -4$

Value table:

$x$	$x+4$	$\sqrt{x+4}$	$-\sqrt{x+4}$	$g(x) = -\sqrt{x+4} + 2$
$-4$	$0$	$0$	$0$	$2$
$-3$	$1$	$1$	$-1$	$1$
$0$	$4$	$2$	$-2$	$0$
$5$	$9$	$3$	$-3$	$-1$
$12$	$16$	$4$	$-4$	$-2$

Graph of

g(x) = -\sqrt{x+4} + 2

Graph of a square root function that undergoes reflection across the

x

-axis and translation.

The graph begins at $(-4,2)$ . The negative sign reflects the parent graph across the $x$ -axis, so the curve falls to the right and its outputs stay at or below $2$ .

The domain requires $x\geq 0$ . Isolate one radical:
$\sqrt{x+5}=5-\sqrt{x}$
Check the original equation: $\sqrt{4+5}+\sqrt{4}=3+2=5$ . Thus, $x=4$ is valid.
Substitute $t=9$ into the model:

$h(9)=100\sqrt{9}=100\cdot 3=300\text{ m}.$

The model uses $t\geq 0$ , and $9$ lies in that domain. The modeled height is $300\text{ m}$ .
Record the domain $x-3\geq 0$ , so $x\geq 3$ . Divide by the positive number $2$ , then square the non-negative sides:
$2\sqrt{x-3}\geq 6$
Intersecting this result with the domain still gives $x\geq 12$ .