Unlocking Graphing Radical Functions: A Comprehensive Guide

Example function	Type of expression	Name of the function
$y = 7$	Constant	Constant function
$y = 3 x - 2$	Linear	Linear function
$y = - x^{2} - 2 x + 1$	Quadratic	Quadratic function

Variable in a Radicand	Variable with a Rational Exponent
$y = x$	$y = x^{\frac{1}{2}}$
$y = 3 x + 1$	$y = (x + 1)^{\frac{1}{3}}$
$y = 2 4 3 x + 1 - 4$	$y = 2 (3 x + 1)^{\frac{1}{4}} - 4$

$x$	$2 4 6 - 3 x + 5$	$y$
$- 2$	$2 4 6 - 3 (- 2) + 5$	$\approx 8.72$
$- 1$	$2 4 6 - 3 (- 1) + 5$	$\approx 8.46$
$0$	$2 4 6 - 3 (0) + 5$	$\approx 8.13$
$1$	$2 4 6 - 3 (1) + 5$	$\approx 7.63$
$2$	$2 4 6 - 3 (2) + 5$	$5$

$y = a n b x + c + d,$ when $n$ is even
Sign of $a$	Range
Positive $(a > 0)$	$y \geq d$
Negative $(a < 0)$	$y \leq d$

$x$	$\frac{1}{2} 2 x$	$y$
$0$	$\frac{1}{2} 2 (0)$	$0$
$1$	$\frac{1}{2} 2 (1)$	$\approx 0.71$
$2$	$\frac{1}{2} 2 (2)$	$1$
$3$	$\frac{1}{2} 2 (3)$	$\approx 1.22$
$4$	$\frac{1}{2} 2 (4)$	$\approx 1.41$

Domain	$x \geq 0$
Range	$y \geq 0$
$x -$ intercept	$x = 0$
$y -$ intercept	$y = 0$
End Behavior	$y x \to 0 ⟶ 0$ and $y x \to + \infty ⟶ + \infty$

$x$	$5 3 x + 1 - 2$	$y$
$- 3$	$5 3 (- 3) + 1 - 2$	$\approx - 3.52$
$- 2$	$5 3 (- 2) + 1 - 2$	$\approx - 3.38$
$- 1$	$5 3 (- 1) + 1 - 2$	$\approx - 3.15$
$0$	$5 3 (0) + 1 - 2$	$- 1$
$1$	$5 3 (1) + 1 - 2$	$\approx - 0.68$
$2$	$5 3 (2) + 1 - 2$	$\approx - 0.52$
$3$	$5 3 (3) + 1 - 2$	$\approx - 0.42$

$x$	$3 x + 1$	$y$
$- 4$	$3 - 4 + 1$	$\approx - 1.44$
$- 3$	$3 - 3 + 1$	$\approx - 1.26$
$- 2$	$3 - 2 + 1$	$- 1$
$- 1$	$3 - 1 + 1$	$0$
$0$	$3 0 + 1$	$1$
$1$	$3 1 + 1$	$\approx 1.26$
$2$	$3 2 + 1$	$\approx 1.44$
$3$	$3 3 + 1$	$\approx 1.59$
$4$	$3 4 + 1$	$\approx 1.71$

Domain	All real numbers
Range	All real numbers
$x -$ intercept	$x = - 1$
$y -$ intercept	$y = 1$
End Behavior	$y x \to - \infty ⟶ - \infty$ and $y x \to + \infty ⟶ + \infty$

$x$	$3 2 x - 4 + 3$	$y$
$- 4$	$3 2 (- 4) - 4 + 3$	$\approx 0.71$
$- 3$	$3 2 (- 3) - 4 + 3$	$\approx 0.85$
$- 2$	$3 2 (- 2) - 4 + 3$	$1$
$- 1$	$3 2 (- 1) - 4 + 3$	$\approx 1.18$
$0$	$3 2 (0) - 4 + 3$	$\approx 1.41$
$1$	$3 2 (1) - 4 + 3$	$\approx 1.74$
$2$	$3 2 (2) - 4 + 3$	$3$
$3$	$3 2 (3) - 4 + 3$	$\approx 4.26$
$4$	$3 2 (4) - 4 + 3$	$\approx 4.59$

$x$	$3 3 x - 6 - 1$	$y$
$- 3$	$3 3 (- 3) - 6 - 1$	$\approx - 3.47$
$- 2$	$3 3 (- 2) - 6 - 1$	$\approx - 3.29$
$- 1$	$3 3 (- 1) - 6 - 1$	$\approx - 3.08$
$0$	$3 3 (0) - 6 - 1$	$\approx 2.82$
$1$	$3 3 (1) - 6 - 1$	$\approx - 2.44$
$2$	$3 3 (2) - 6 - 1$	$- 1$
$3$	$3 3 (3) - 6 - 1$	$\approx 0.44$
$4$	$3 3 (4) - 6 - 1$	$\approx 0.82$
$5$	$3 3 (5) - 6 - 1$	$\approx 1.08$

$x$	$2 \frac{1}{2} x - 1 + 1$	$y$
$2$	$2 \frac{1}{2} (2) - 1 + 1$	$1$
$3$	$2 \frac{1}{2} (3) - 1 + 1$	$\approx 2.41$
$5$	$2 \frac{1}{2} (5) - 1 + 1$	$\approx 3.45$
$10$	$2 \frac{1}{2} (10) - 1 + 1$	$5$

Consider the radical function.

y = 3 2 x + 1

Find the domain of the function. Write the answer as an inequality.

Find the range of the function. Write the answer as an inequality.

Recall that the radicand of an even-indexed radical function must be non-negative. 2x+1≥ 0 We can find the domain of the function by solving this inequality. Let's do it!

The domain of the radical function is the set of all real numbers that are greater than or equal to - 12. Domain: x≥ - 1/2

Note that the greater the value of x, the greater the value of y. Therefore, this function is an increasing function. This means that the minimum value of y is reached at the domain's least value. Let's find this value by substituting x=- 12 in the given function rule.

We found that the minimum value of y is 0. Therefore, the range of the function is the set of all real numbers greater than or equal to 0. Range: y≥ 0

Consider the radical function.

y = 3 10 - 5 x

What is the domain of the function?

What is the range of the function?

Recall that there are no restrictions on the radicand of an odd-indexed radical function. This means that the radicand can be positive, negative, or zero. Therefore, the domain of the function is the set of all real numbers. Domain: All real numbers

Recall that the value of an odd-indexed radical function can be positive, negative, or zero. Therefore, the range of the function is the set of all real numbers. Range: All real numbers

Consider the radical function.

y = 2 x + 1

Which of the following graphs is the graph of this function?

We want to determine which of the given graphs is the graph of the radical function. To do this, we will graph the function and compare the graph with the choices. To graph the function, we will start by finding its domain. Recall that the radicand of a square root must be non-negative.

Now we can make a table for values of x greater than or equal to - 0.5.

x	sqrt(2x+1)	y
- 0.5	sqrt(2( - 0.5)+1)	0
0	sqrt(2( 0)+1)	1
1.5	sqrt(2( 1.5)+1)	2
4	sqrt(2( 4)+1)	3
7.5	sqrt(2( 7.5)+1)	4

Let's plot the points found in the table on a coordinate plane and connect them with a smooth curve.

This graph corresponds to choice C.

Consider the radical function.

y = 3 4 - 2 x

Which of the following graphs is the graph of this function?

We want to determine which of the given graphs is the graph of the radical function. To do so, we will graph the function and compare the graph to the options. Let's start by make a table of values. Recall that the domain of odd-indexed radical functions is the set of all real numbers. Therefore, we can assign any value to the variable x.

x	sqrt(4-2x)	y
- 2	sqrt(4-2( - 2))	2
- 1	sqrt(4-2( - 1))	≈ 1.8
0	sqrt(4-2( 0))	≈ 1.6
1	sqrt(4-2( 1))	≈ 1.3
2	sqrt(4-2( 2))	0
3	sqrt(4-2( 3))	≈ - 1.3
4	sqrt(4-2( 4))	≈ - 1.6

Let's plot the points found in the table on a coordinate plane and connect them with a smooth curve.

This graph corresponds to choice B.

Consider the radical function.

y = 3 2 x + 1

Find the

x -

intercept of the function's graph.

Find the

y -

intercept of the function's graph.

The value of the variable y at the point where the graph of a function intercepts the x-axis is 0. Therefore, to find the x-intercept of the radical function, we will substitute y=0 in the equation and solve for x.

The x-intercept of the function is x=- 12.

The value of the x-variable at the point where the graph of a function intercepts the y-axis is 0. Therefore, to find the y-intercept of the radical function, we will substitute x=0 in the equation and evaluate the right-hand side.

The y-intercept of the function is y=3.

Consider the radical function.

y = 3 10 - 5 x

Find the

x -

intercept of the graph of the function.

Find the

y -

intercept of the graph of the function. Round the answer to two decimal places.

At the point where the graph of the function intercepts the x-axis, the value of the variable y is 0. Therefore, to find the x-intercept of the radical function, we will substitute y=0 into the equation and solve for x.

The x-intercept of the graph is x=2.

At the point where the graph of a function intercepts the y-axis, the value of the x-variable is 0. Therefore, to find the y-intercept of the radical function, we will substitute x=0 into the equation and evaluate the right-hand side.

The y-intercept of the graph of the function is y≈2.15.

Describe the end behavior of the radical function.

y = - 2 - x

To find the end behavior of the radical function, we will draw its graph. To do so, we will start by finding its domain. Recall that the radicand of an even-indexed root must be non-negative. - x≥ 0 ⇔ x≤ 0 Therefore, the domain of the function is the set of all real numbers less than or equal to 0. With this in mind, we can make a table of values to find points on the curve.

x	- 2sqrt(- x)	y
- 16	- 2sqrt(- ( - 16))	- 8
- 9	- 2sqrt(- ( - 9))	- 6
- 4	- 2sqrt(- ( - 4))	- 4
- 1	- 2sqrt(- ( - 1))	- 2
0	- 2sqrt(- 0)	0

Next, we will plot the points in the table on a coordinate plane and connect them with a smooth curve.

We see in the graph that when x tends to 0, the variable y also tends to 0. When x tends to negative infiniye, y also tends to negative infinity. y x→ 0 ⟶ 0 and y x→ - ∞ ⟶ - ∞

Describe the end behavior of the radical function.

y = - 5 \frac{1}{2} x + 1

To find the end behavior of the function, we will draw its graph. To do so, we will make a table of values to find points on the curve. Since the domain of odd-indexed radical functions is the set of all real numbers, the variable x can take any value.

x	- sqrt(1/2x+1)	y
- 10	- sqrt(1/2( - 10)+1)	≈ 1.3
- 5	- sqrt(1/2( - 5)+1)	≈ 1.1
- 2	- sqrt(1/2( - 2)+1)	0
- 1	- sqrt(1/2( - 1)+1)	≈ - 0.9
0	- sqrt(1/2( 0)+1)	- 1
1	- sqrt(1/2( 1)+1)	≈ - 1.1
5	- sqrt(1/2( 5)+1)	≈ - 1.3
10	- sqrt(1/2( 10)+1)	≈ - 1.4

Next we will plot the points found in the table on a coordinate plane and connect them with a smooth curve.

We see in the graph that when x tends to negative infinity, the variable y tends to positive infinity. When x tends to positive infinity, y tends to negative infinity. y x→ - ∞ ⟶ + ∞ and y x→ + ∞ ⟶ - ∞

Consider the radical inequality.

y \leq \frac{1}{2} x - 4

Which of the graphs corresponds to this inequality?

We want to determine which graph corresponds to the given radical inequality. To do so, we will draw the inequality on a coordinate plane and see which of the given options matches this graph. First, we need to draw the boundary curve. To do this, we start by replacing the inequality sign with an equals sign. cc Inequality & Boundary Curve [0.6em] y ≤ 1/2sqrt(x-4) & y = 1/2sqrt(x-4) Next, we will make a table of values for the curve. Recall that the radicand of a square root must be greater than or equal to zero. x-4≥ 0 ⇔ x≥ 4 We can now make the table for values of x greater than or equal to 4.

x	1/2sqrt(x-4)	y
4	1/2sqrt(4-4)	0
5	1/2sqrt(5-4)	0.5
6	1/2sqrt(6-4)	≈ 0.7
7	1/2sqrt(7-4)	≈ 0.9
8	1/2sqrt(8-4)	1

Let's plot the points and connect them with a smooth curve. Since we have a non-strict inequality, the boundary curve will be solid.

Now we have to shade a region on the plane. To determine which region to shade, we will test a point not on the curve by substituting its coordinates into the given inequality and simplifying. If we get a true statement, then we shade the region that contains the point. If not, we shade the opposite region. Let's test the point (5,0).

Because we ended with a true statement, we will shade the region that contains the point (5,0). Also, since the variable x can only take values that are greater than or equal to 4, we will shade to the right of this x-value.

Our graph matches option B.

Consider the radical inequality.

y > 2 3 x + 1

Which of the graphs corresponds to this inequality?

We want to determine which graph corresponds to the radical inequality. To do so, we will draw the inequality on a coordinate plane and see which of the given options matches this graph. First, we need to draw the boundary curve. To do this, we start by replacing the inequality sign with an equals sign. cc Inequality & Boundary Curve [0.3em] y > 2sqrt(x+1) & y = 2sqrt(x+1) Next, we will make a table of values for the curve. Because the radicand of a cube root can be negative, positive or zero, we can assign any value to the variable x.

x	y= 2sqrt(x+1)	y
- 9	y= 2sqrt(- 9+1)	- 4
- 2	y= 2sqrt(- 2+1)	- 2
- 1	y= 2sqrt(- 1+1)	0
0	y= 2sqrt(0+1)	2
7	y= 2sqrt(7+1)	4

Let's plot the points we found in the table and connect them with a smooth curve. Since we have a strict inequality, the boundary curve will be dashed.

Now we have to shade a region on the plane. To do this, we will test a point not on the curve by substituting its coordinates into the given inequality and simplifying. If we get a true statement, then we shade the region that contains the point. If not, we shade the opposite region. Let's test the point (0,0).

Because we got a false statement, we will shade the region that does not contain the point (0,0).

Our graph matches option D.

Graphing Radical Functions and Inequalities

Catch-Up and Review

Reflecting the Graph of a Quadratic Function

Radical Functions and Square Root Functions

Radical Function

Square Root Function

Domain and Range of Even Indexed Radical Functions

Football and Square Root Functions

Answer

Hint

Solution

Cube Root Functions

Domain and Range of Odd Indexed Radical Functions

Football and Cube Root Functions

Answer

Hint

Solution

Finding the Domain and Range of Radical Functions

Radical Inequalities

Graphing a Radical Inequality in Two Variables

Football and Cube Root Inequalities

Answer

Hint

Solution

Football and Square Root Inequalities

Answer

Hint

Solution

Comparing Radical Functions with Even and Odd Indices

Odd-Indexed Radical Functions

Even-Indexed Radical Functions

Graphing Radical Functions and Inequalities

Recommended exercises

	13 Theory slides
	16 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions