Question

graph the function. g(x) = \frac{1}{3}(x - 6)^2 + 1

Explanation:

Step1: Identify Parent Function

The parent function here is \( y = \frac{1}{3}x^3 \), a cubic function. Its graph has an inflection point at the origin, increasing for all \( x \), with a horizontal tangent at \( x = 0 \).

Step2: Analyze Transformations

The given function is \( g(x)=\frac{1}{3}(x - 6)^3+1 \).

• Horizontal shift: \( (x - 6) \) shifts the parent graph 6 units to the right (since we replace \( x \) with \( x - 6 \)).
• Vertical shift: \( +1 \) shifts the graph 1 unit up.
• Vertical stretch/compression: The \( \frac{1}{3} \) is a vertical compression (since \( 0<\frac{1}{3}<1 \)), but for cubic functions, the shape remains similar, just "flatter" than \( y = x^3 \).

Step3: Plot Key Points

• For the parent \( y=\frac{1}{3}x^3 \), when \( x = 0 \), \( y = 0 \); \( x = 3 \), \( y=\frac{1}{3}(27)=9 \); \( x=-3 \), \( y=\frac{1}{3}(-27)= - 9 \).
• After shifting: The inflection point (origin for parent) moves to \( (6,1) \) (since \( x - 6=0\Rightarrow x = 6 \), \( y = 0 + 1=1 \)).
• For \( x=6 + 3=9 \), \( g(9)=\frac{1}{3}(3)^3+1=\frac{1}{3}(27)+1 = 9 + 1=10 \).
• For \( x=6 - 3=3 \), \( g(3)=\frac{1}{3}(-3)^3+1=\frac{1}{3}(-27)+1=-9 + 1=-8 \).
• Connect these points smoothly, maintaining the cubic curve shape, shifted right 6 and up 1, with vertical compression.

Answer:

To graph \( g(x)=\frac{1}{3}(x - 6)^3+1 \):

1. Start with the parent cubic \( y=\frac{1}{3}x^3 \) (inflection at \((0,0)\), increasing, smooth curve).
2. Shift the graph 6 units right (replace \( x \) with \( x - 6 \)) and 1 unit up (add 1 to \( y \)).
3. Plot the new inflection point at \((6,1)\), and use transformed points (e.g., \((3, - 8)\) and \((9,10)\)) to sketch the smooth, increasing cubic curve with vertical compression, shifted right 6 and up 1.