First, find the equation for the function.  [latex]f\left(x+3\right)+1=\sqrt[\leftroot{1}\uproot{2} ]{x+3}+1.[/latex]

Method 1:  The domain of an even root function must have non-negative values under the root symbol, so we solve the inequality

[latex]x+3\geq0.[/latex]

[latex]x\geq-3[/latex]

Therefore, the domain is all real numbers greater that or equal to negative 3 or in interval notation [latex]\left[-3,\infty\right).[/latex]

Method 2:  Alternatively, the point [latex]\left(0,0\right)[/latex] is shifted to the point [latex]\left(-3,1\right).[/latex]  The starting value for the domain is -3 and since there is no horizontal reflection the graph opens to the right like [latex]\sqrt[\leftroot{1}\uproot{2} ]{x}.[/latex]  Again the domain is [latex]\left[-3,\infty\right).[/latex]

The range of the square root will be shifted up one unit, so the range is all real numbers greater than or equal to one or in interval notation [latex]\left[1,\infty\right).[/latex]  Notice that the starting value of 1 is reflected in the shift of the point [latex]\left(0,0\right)[/latex] above and since there is no vertical reflection the interval goes in the direction of infinity.

a. Method 1: Even root functions have non-negative input, so the domain of [latex]g\left(x\right)=\sqrt[\leftroot{1}\uproot{2}4]{3-2x}[/latex] is found by solving the inequality

[latex]3-2x\geq0.[/latex]

The solution is

[latex]x\leq\frac{3}{2}[/latex].

Therefore the domain is [latex]\left(-\infty,1.5\right].[/latex]

Method 2:  The function is a horizontal shift of [latex]\sqrt[\leftroot{1}\uproot{2}4]{x}[/latex] left by 3 units followed by a horizontal compression by a factor of 1/2 and then a horizontal reflection.  The point (0,0) is mapped as follows:

[latex]\left(0,0\right)\to\left(-3, 0\right)\to\left(-1.5,0\right)\to\left(1.5,0\right)[/latex]

It is the horizontal reflection that makes the graph open to the left rather than the right.

The domain is [latex]\left(-\infty,1.5\right].[/latex]

The range remains [latex]\left(0,\infty\right)[/latex] since there are no vertical transformations.

The domain and range can be seen in the graph below.

b.  Notice that [latex]h\left(x\right)=-3\sqrt[\leftroot{1}\uproot{2}4]{x}[/latex] is a vertical reflection and stretch by a factor of 3 of the function [latex]\sqrt[\leftroot{1}\uproot{2}4]{x}.[/latex]  This tells us that the properties associated with the output value will change so we need to consider the range carefully.

The domain will be [latex]\left [0,\infty\right)[/latex] since there are no horizontal transformations.

The vertical reflection does effect the range.  Since the fourth root function outputs positive numbers or zero, when we multiply by a negative number will will have negative values or zero.  Therefore, the range is [latex]\left(-\infty,0\right][/latex].

The vertical reflection and range is shown in the graph below.

a.  The x-intercept has [latex]y=0[/latex].  We solve the equation [latex]\sqrt[\leftroot{1}\uproot{2}4]{3-2x}=0[/latex] by raising both sides to the 4^th power to get [latex]3-2x=0.[/latex] Finally, [latex]x=1.5.[/latex]  The x-intercept is [latex]\left(1.5,0\right).[/latex]

For the y-intercept, [latex]f\left(0\right)[/latex] is evaluated.  We get [latex]f\left(0\right)=\sqrt[\leftroot{1}\uproot{2}4]{3-2\left(0\right)}=\sqrt[\leftroot{1}\uproot{2}4]{3}\approx1.316.[/latex] Therefore, the y-intercept is [latex]\left(0, 1.316\right).[/latex]

b. Since there is no horizontal or vertical shift both the [latex]x[/latex] and [latex]y[/latex] intercepts are [latex]\left(0,0\right).[/latex]

c.  Evaluate [latex]f\left(0\right)[/latex] to get [latex]f\left(0\right)=\sqrt[\leftroot{1}\uproot{2} ]{0+3}+1\approx2.73[/latex].  The y-intercept is approximately [latex]\left(0,2.73\right).[/latex]

Since [latex]y=0[/latex] is not in the range because the graph was shifted up one unit, there is no x-intercept.

[latex]k\left(x\right)[/latex] is a shift left by 2 units of [latex]\sqrt[\leftroot{1}\uproot{2}6]{x}[/latex] followed by a horizontal reflection.  The shift by 2 units will not effect the end behavior but the reflection will make it so the graph opens to the left.  Since (0,0) transforms as follows

[latex]\left(0,0\right)\to\left(-2,0\right)\to\left(2,0\right)[/latex]

the domain is [latex]\left(-\infty,2\right)[/latex].  Right end behavior does not make sense for this function.

As [latex]x[/latex] goes further and further to the left the output will become larger and larger.  We write this as [latex]x\to-\infty, f\left(x\right)\to\infty.[/latex]

We can confirm this in the table and graph below.  For the table, even when we chose extremely negative values for x, the output is not really large but we see that it continues to get bigger and does not level off.

This equation is a shift of [latex]\sqrt[\leftroot{1}\uproot{2}3]{x}[/latex] left by 3 units and then a horizontal reflection.  Finally the graph is shifted up 1 unit.

Since the domain and range of [latex]\sqrt[\leftroot{1}\uproot{2}3]{x}[/latex] is all real numbers, these transformations will not effect these properties.  Therefore the domain and range of [latex]k\left(x\right)[/latex] are [latex]\left(-\infty,\infty\right).[/latex]

To find the x-intercept we solve the equation [latex]\sqrt[\leftroot{1}\uproot{2}3]{3-x}+1=0.[/latex] Begin by subtracting one from each side to get [latex]\sqrt[\leftroot{1}\uproot{2}3]{3-x}=-1.[/latex]  Next, cube both sides to get [latex]3-x=-1.[/latex]  Finally [latex]x=4.[/latex]  The x-intercept is [latex]\left(4,0\right).[/latex]

To find the y-intercept we evaluate [latex]f\left(0\right).[/latex]  We get [latex]f\left(0\right)=\sqrt[\leftroot{1}\uproot{2}3]{3-0}+1=\sqrt[\leftroot{1}\uproot{2}3]{3}+1\approx2.442.[/latex] The y-intercept is [latex]\left(0,2.442\right).[/latex]

The end behavior will be similar to [latex]\sqrt[\leftroot{1}\uproot{2}3]{-x}\textrm{ or }-\sqrt[\leftroot{1}\uproot{2}3]{x}[/latex] since odd root functions are odd.  Therefore, as [latex]x[/latex] becomes more and more negative, [latex]f\left(x\right)[/latex] increases without bound and as [latex]x[/latex] becomes larger and larger [latex]f\left(x\right)[/latex] decreases without bound.  We write, as [latex]x\to-\infty, f\left(x\right)\to\infty[/latex] and [latex]x\to\infty, f\left(x\right)\to-\infty.[/latex]

These properties can be confirmed in the graph below.