Graphing Radical Functions and Inequalities

Since the index of the radical expression on the right-hand side of the equation is 2, the given radical function is a square root function. Therefore, the radicand must be non-negative. 2x ≥ 0 ⇔ x≥ 0 This means that the domain of the function is the set of all real numbers greater than or equal to 0. To determine its range, let's compare the given function to the general form of a square root function. General Form:& asqrt(bx+ c)+ d Given Function:& 1/2sqrt(2x+ 0)+ 0 Since a= 12 is positive, the range of the function is all real numbers greater than or equal to 0. To determine the intercepts and end behavior, the function will be first drawn on a coordinate plane. To do so, a make a table of values. Be careful to use only x-values that belong to the domain in the table!

x	1/2sqrt(2x)	y
0	1/2sqrt(2( 0))	0
1	1/2sqrt(2( 1))	≈ 0.71
2	1/2sqrt(2( 2))	1
3	1/2sqrt(2( 3))	≈ 1.22
4	1/2sqrt(2( 4))	≈ 1.41

Next, the obtained points can be plotted and connected with a smooth curve.

From the graph, it is seen that the x-intercept and the y-intercept both occur at the origin. It can also be seen that this function increases over its entire domain and that y tends to infinity as x tends to infinity. With this information, the desired characteristics can be written. Recall that the domain and range are both all non-negative real numbers!

Domain	x≥ 0
Range	y≥ 0
x-intercept	x=0
y-intercept	y=0
End Behavior	y x→ 0 ⟶ 0 and y x→ + ∞ ⟶ + ∞

The radicand of a cube root can be any real number. Therefore, the domain of the function is the set of all real numbers. To determine the range, intercepts, and end behavior, the function will be first drawn on a coordinate plane. To do so, a make a table of values using both positive and negative values!

x	sqrt(x+1)	y
- 4	sqrt(- 4+1)	≈ - 1.44
- 3	sqrt(- 3+1)	≈ - 1.26
- 2	sqrt(- 2+1)	- 1
- 1	sqrt(- 1+1)	0
0	sqrt(0+1)	1
1	sqrt(1+1)	≈ 1.26
2	sqrt(2+1)	≈ 1.44
3	sqrt(3+1)	≈ 1.59
4	sqrt(4+1)	≈ 1.71

Next, the calculated points can be plotted and connected with a smooth curve.

The graph suggests that the range is the set of all real numbers. It shows that the x-intercept occurs at x=- 1 and the y-intercept at y=1. It can also be seen that y tends to negative infinity as x tends to negative infinity, and that y tends to infinity as x tends to infinity. With this information, the desired characteristics can be written.

Domain	All real numbers
Range	All real numbers
x-intercept	x=- 1
y-intercept	y=1
End Behavior	y x→ - ∞ ⟶ - ∞ and y x→ + ∞ ⟶ + ∞

The first step to graph a radical inequality is to draw the boundary curve. The equation for this curve is written by replacing the inequality sign with an equals sign. cc Inequality & Equation [0.5em] y+1 > sqrt(3x-6) & y+1 = sqrt(3x-6) To graph the equation, the radical expression must be first isolated. y+1 = sqrt(3x-6) ⇕ y= sqrt(3x-6)-1 Because the index of the root is 3, this radical function is a cube root function. Therefore, the domain for this function is the set of all real numbers. With this information in mind, a make a table of values to find points on the curve, using both positive and negative values for the variable x.

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

Next, the points obtained in the table can be plotted and connected with a smooth curve. Because the given inequality is a strict inequality, the curve will be dashed.

Now the correct region must be shaded. To determine which region to shade, a point not on the boundary curve will be tested. The point (0,0) seems like the easiest choice.

The point (0,0) satisfies the given inequality. Therefore, the region that contains this point will be shaded.

To graph the inequality, start by drawing the boundary curve. To find the boundary curve, replace the inequality sign with an equals sign. Inequality y ≥ 2sqrt(1/2x-1)+1 [1em] Equation y = 2sqrt(1/2x-1)+1 To graph the function, a table of values can be used. However, the domain of the function must be found first. To do so, recall that the radicand of a square root must be greater than or equal to 0.

The domain of the function is the set of all real numbers greater than or equal to 2. Now the make the table of values, keeping in mind that only x-values that belong to the domain can be used.

x	2sqrt(1/2x-1)+1	y
2	2sqrt(1/2( 2)-1)+1	1
3	2sqrt(1/2( 3)-1)+1	≈ 2.41
5	2sqrt(1/2( 5)-1)+1	≈ 3.45
10	2sqrt(1/2( 10)-1)+1	5

The points can be plotted and connected with a smooth curve. Remember that since the inequality is not strict, the curve will be solid.

Finally, the correct region should be shaded. To determine which region to shade, a point not on the curve must be tested. The point (10,3) looks like a good choice. This point will be evaluated in the given inequality.

The test point does not satisfy the inequality. Therefore, the region that does not contain this point should be shaded. Keep in mind that the domain of the related function is the set of real numbers greater than or equal to 2. Therefore, to shade the region, only the x-values greater than 2 must be considered.