Question

Graph each function using the techniques of shifting, compressing, stretching, and/or reflecting. Start with the graph of the basic function (for example, \(y=x^{2}\) ) and show all the steps. Be sure to show at least three key points. Find the domain and the range of each function. $$ h(x)=\sqrt{x+1} $$

Short answer

The function \(h(x)=\sqrt{x+1}\) is the graph of \(y=\sqrt{x}\) shifted 1 unit to the left. Domain: \([-1,\infty)\). Range: \([0,\infty)\).

Step-by-step solution

Identify the basic function

The given function is \(h(x)=\sqrt{x+1}\). The basic function to start with is \(y=\sqrt{x}\).

Shifting the basic function

Compare \(y=\sqrt{x}\) with \(h(x)=\sqrt{x+1}\). Notice that \(h(x)\) is a horizontal shift of \(y=\sqrt{x}\). Specifically, it is shifted 1 unit to the left. To shift a graph to the left by 1 unit, replace \(x\) with \(x+1\).

Plot key points of the function

Choose key points on \(y=\sqrt{x}\) and apply the shift: \1. For \(x=0\), \(y=\sqrt{0}=0\) \2. For \(x=1\), \(y=\sqrt{1}=1\) \3. For \(x=4\), \(y=\sqrt{4}=2\) \After shifting to the left by 1 unit, the points become: \1. For \(x=0-1=-1\), \(y=0\) \2. For \(x=1-1=0\), \(y=1\) \3. For \(x=4-1=3\), \(y=2\).

Determine the domain and range

The original function \(y=\sqrt{x}\) has a domain \([0,\infty)\) and range \([0,\infty)\). For the shifted function \(h(x)=\sqrt{x+1}\), the domain is shifted accordingly: \(x+1 \geq 0 \Rightarrow x \geq -1\). Therefore, the domain is \([-1, \infty)\). The range remains the same: \([0, \infty)\).

Graph the function

Using the shifted key points (-1,0), (0,1), and (3,2), plot these on a coordinate system. Draw the curve starting from \(-1,0\) and moving through the key points.

Key concepts

horizontal shift

When we talk about a horizontal shift, we're referring to moving the entire graph of a function left or right. It's like sliding a piece of paper along a table without lifting it. For the function given, \( h(x)= \sqrt{x+1} \), this involves the basic function \( y=\sqrt{x} \). A horizontal shift of 1 unit to the left occurs because we're adding 1 to x inside the square root.
In general, for any function \( f(x) \), replacing \( x \) with \( x+c \) shifts the graph horizontally:

• If \( c \) is positive, the shift is to the left.
• If \( c \) is negative, the shift is to the right.

Hence, in our case, adding 1 means the entire graph of \( y=\sqrt{x} \) is moved one unit to the left.

domain and range

Understanding the domain and range of a function is crucial. The domain represents all possible input values (x-values), while the range shows all possible output values (y-values). For our basic function \( y=\sqrt{x} \), the domain is \([0, \infty)\) because you can't take the square root of a negative number. The range is also \([0, \infty)\) because square roots cannot produce negative results.
For the shifted function \( h(x)=\sqrt{x+1} \), the domain changes because of the horizontal shift. We solve \( x+1 \geq 0 \Rightarrow x \geq -1 \), giving us a domain of \([-1, \infty)\). The range remains unchanged at \([0, \infty)\) because the values of \( y \) are still non-negative.

key points

Key points help in sketching a function's graph easily by providing reference points. For \( y=\sqrt{x} \), some important key points are:

• \( (0, 0) \)
• \( (1, 1) \)
• \( (4, 2) \)

For the shifted function \(h(x)=\sqrt{x+1}\), we apply the shift to these points:

• \( (0-1, 0) = (-1, 0) \)
• \( (1-1, 1) = (0, 1) \)
• \( (4-1, 2) = (3, 2) \)

Plotting these points helps us draw the new graph accurately. Start at \( (-1, 0) \), move to \( (0, 1) \), and finally reach \( (3, 2) \). Connect these points smoothly to visualize the entire function.