Question

graph the function.
$y = \frac{3}{2}\sqrt3{x}$

Explanation:

Step1: Recall the parent function

The parent function is \( y = \sqrt[3]{x} \), which has a domain of all real numbers and passes through points like \((-8, -2)\), \((-1, -1)\), \((0, 0)\), \((1, 1)\), \((8, 2)\).

Step2: Analyze the transformation

The given function is \( y=\frac{3}{2}\sqrt[3]{x} \). This is a vertical stretch of the parent function \( y = \sqrt[3]{x} \) by a factor of \( \frac{3}{2} \).

Step3: Find key points

• For \( x = -8 \): \( y=\frac{3}{2}\sqrt[3]{-8}=\frac{3}{2}(-2)= - 3 \)
• For \( x = -1 \): \( y=\frac{3}{2}\sqrt[3]{-1}=\frac{3}{2}(-1)=-\frac{3}{2}=-1.5 \)
• For \( x = 0 \): \( y=\frac{3}{2}\sqrt[3]{0}=0 \)
• For \( x = 1 \): \( y=\frac{3}{2}\sqrt[3]{1}=\frac{3}{2}(1) = 1.5 \)
• For \( x = 8 \): \( y=\frac{3}{2}\sqrt[3]{8}=\frac{3}{2}(2)=3 \)

Step4: Plot the points and draw the curve

Plot the points \((-8, - 3)\), \((-1, - 1.5)\), \((0,0)\), \((1,1.5)\), \((8,3)\) and draw a smooth curve through them, which should match the general shape of a cube - root function (symmetric about the origin, increasing throughout its domain) with a vertical stretch.

Answer:

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

1. Recognize it's a vertical stretch (\( \frac{3}{2} \) factor) of \( y=\sqrt[3]{x} \).
2. Find key points using the function:

• When \( x=-8 \), \( y = - 3 \); \( x = - 1 \), \( y=-1.5 \); \( x = 0 \), \( y = 0 \); \( x = 1 \), \( y = 1.5 \); \( x = 8 \), \( y = 3 \).

1. Plot these points and draw a smooth, increasing curve (since the cube - root function is increasing) passing through them, which should match the shape of the given graph (with the curve passing through appropriate stretched points).