To understand the effect of the coefficient [tex]\( \frac{1}{3} \)[/tex] in the function [tex]\( v = \frac{1}{3} x^2 \)[/tex], let's compare it to the standard quadratic function [tex]\( y = x^2 \)[/tex].
1. Function Analysis:
- The standard quadratic function [tex]\( y = x^2 \)[/tex] represents a parabola with its vertex at the origin (0,0), opening upwards. The general form of a quadratic function is [tex]\( y = ax^2 \)[/tex], where [tex]\( a = 1 \)[/tex] in this case.
2. Effect of Coefficient [tex]\( \frac{1}{3} \)[/tex]:
- In the given function [tex]\( v = \frac{1}{3} x^2 \)[/tex], the coefficient [tex]\( \frac{1}{3} \)[/tex] affects the term [tex]\( ax^2 \)[/tex]. Here, [tex]\( a = \frac{1}{3} \)[/tex].
- When [tex]\( a \)[/tex] is a number between 0 and 1 (i.e., [tex]\( 0 < a < 1 \)[/tex]), it causes the graph to shrink vertically.
3. Vertical Compression:
- Specifically, [tex]\( v = \frac{1}{3} x^2 \)[/tex] means that for any given value of [tex]\( x \)[/tex], the corresponding value of [tex]\( v \)[/tex] will be [tex]\( \frac{1}{3} \)[/tex] of what it would be in the function [tex]\( y = x^2 \)[/tex].
- This implies that the height of the graph at any point is reduced to [tex]\( \frac{1}{3} \)[/tex] of its original height in the graph of [tex]\( y = x^2 \)[/tex].
4. Summary:
- The coefficient [tex]\( \frac{1}{3} \)[/tex] causes a vertical shrinking or compression of the graph.
Therefore, the correct answer is:
D. It shrinks the graph vertically to [tex]\( \frac{1}{3} \)[/tex] its original height.
