Question

graph the function.
$y = \sqrt3{x} + 2$

Explanation:

Step1: Identify the parent function

The parent function is \( y = \sqrt[3]{x} \). Its graph passes through the origin \((0,0)\), and has points like \((-1, -1)\), \((1, 1)\), \((-8, -2)\), \((8, 2)\) since \( \sqrt[3]{-1}=-1 \), \( \sqrt[3]{1}=1 \), \( \sqrt[3]{-8}=-2 \), \( \sqrt[3]{8}=2 \).

Step2: Analyze the transformation

The given function is \( y=\sqrt[3]{x}+2 \), which is a vertical shift of the parent function \( y = \sqrt[3]{x} \) upward by 2 units. So we take each point of the parent function and add 2 to the y - coordinate.

• For the point \((0,0)\) on \( y = \sqrt[3]{x} \), the new point is \((0,0 + 2)=(0,2)\).
• For the point \((-1,-1)\) on \( y=\sqrt[3]{x} \), the new point is \((-1,-1 + 2)=(-1,1)\).
• For the point \((1,1)\) on \( y=\sqrt[3]{x} \), the new point is \((1,1 + 2)=(1,3)\).
• For the point \((-8,-2)\) on \( y=\sqrt[3]{x} \), the new point is \((-8,-2 + 2)=(-8,0)\).
• For the point \((8,2)\) on \( y=\sqrt[3]{x} \), the new point is \((8,2 + 2)=(8,4)\).

Step3: Plot the points

Plot the points \((-8,0)\), \((-1,1)\), \((0,2)\), \((1,3)\), \((8,4)\) on the coordinate plane. Then draw a smooth curve through these points, following the shape of the cube - root function (which has a horizontal and vertical asymptote? No, the cube - root function is defined for all real numbers and its graph is a smooth curve that passes through these points).

Answer:

To graph \( y=\sqrt[3]{x}+2 \):

1. Start with the parent function \( y = \sqrt[3]{x} \) and find key points: \((-8,-2)\), \((-1,-1)\), \((0,0)\), \((1,1)\), \((8,2)\).
2. Shift each point up by 2 units: \((-8,0)\), \((-1,1)\), \((0,2)\), \((1,3)\), \((8,4)\).
3. Plot these shifted points and draw a smooth curve through them. The graph will have a shape similar to the cube - root function but shifted up 2 units, passing through the points \((-8,0)\), \((-1,1)\), \((0,2)\), \((1,3)\), \((8,4)\) with the characteristic "S - shaped" curve of the cube - root function, shifted vertically.