Question

graph the function below by dragging the points. g(x) = √x + 1 show your work here

Explanation:

Step1: Identify the parent function

The parent function is \( y = \sqrt{x} \), which has a domain \( x \geq 0 \) and starts at \( (0,0) \), with points like \( (1,1) \), \( (4,2) \), etc.

Step2: Analyze the transformation

The function \( g(x)=\sqrt{x}+1 \) is a vertical shift of the parent function \( y = \sqrt{x} \) by 1 unit up. So we take the key points of \( y = \sqrt{x} \) and shift their \( y \)-coordinates up by 1.

• For \( x = 0 \): \( g(0)=\sqrt{0}+1 = 0 + 1=1 \), so the point is \( (0,1) \).
• For \( x = 1 \): \( g(1)=\sqrt{1}+1 = 1 + 1=2 \), so the point is \( (1,2) \).
• For \( x = 4 \): \( g(4)=\sqrt{4}+1 = 2 + 1=3 \), so the point is \( (4,3) \).
• For \( x = 9 \): \( g(9)=\sqrt{9}+1 = 3 + 1=4 \), so the point is \( (9,4) \).

Step3: Plot the points

Plot the points \( (0,1) \), \( (1,2) \), \( (4,3) \), \( (9,4) \) on the coordinate plane. Since the domain of \( \sqrt{x} \) is \( x\geq0 \), the graph of \( g(x) \) will start at \( (0,1) \) and increase slowly, following the shape of the square - root curve, but shifted up by 1 unit.

Answer:

To graph \( g(x)=\sqrt{x}+1 \):

1. Start with the parent function \( y = \sqrt{x} \) (domain \( x\geq0 \), passes through \( (0,0) \), \( (1,1) \), \( (4,2) \), \( (9,3) \), etc.).
2. Apply a vertical shift up by 1 unit: new points are \( (0,1) \), \( (1,2) \), \( (4,3) \), \( (9,4) \).
3. Plot these points and draw the curve starting at \( (0,1) \) and increasing, matching the shape of the square - root graph, shifted up.