Certainly! Let's analyze the given function:
### Given Function
[tex]\[ h(x) = \frac{1}{6} x^2 - 9 \][/tex]
### Step-by-Step Analysis
1. Identify the Form of the Function:
The given function is a quadratic function of the form [tex]\( h(x) = ax^2 + bx + c \)[/tex], where in our case [tex]\( a = \frac{1}{6} \)[/tex], [tex]\( b = 0 \)[/tex], and [tex]\( c = -9 \)[/tex].
2. Vertex Form of a Quadratic Function:
The vertex of the quadratic function [tex]\( ax^2 + bx + c \)[/tex] is given by [tex]\( x = -\frac{b}{2a} \)[/tex]. For our function:
[tex]\[ x = -\frac{0}{2 \cdot \frac{1}{6}} = 0 \][/tex]
Therefore, the vertex occurs at [tex]\( x = 0 \)[/tex].
3. Determine the Value of [tex]\( h(x) \)[/tex] at the Vertex:
To find the value of the function at the vertex, substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ h(0) = \frac{1}{6} \cdot 0^2 - 9 = -9 \][/tex]
So, the vertex of the function is at the point (0, -9).
4. Effect and Shape of the Function:
- Parabola Opening Direction: Since the coefficient of [tex]\( x^2 \)[/tex] (which is [tex]\( \frac{1}{6} \)[/tex]) is positive, the parabola opens upwards.
- Stretching of the Parabola: The coefficient [tex]\( \frac{1}{6} \)[/tex] indicates that the parabola is wider than the standard [tex]\( x^2 \)[/tex] parabola. The smaller the coefficient of [tex]\( x^2 \)[/tex], the wider the parabola.
5. Y-intercept:
The y-intercept occurs where [tex]\( x = 0 \)[/tex]:
[tex]\[ h(0) = -9 \][/tex]
Thus, the parabola crosses the y-axis at [tex]\( y = -9 \)[/tex].
### Summary of the Effects:
- The function [tex]\( h(x) = \frac{1}{6} x^2 - 9 \)[/tex] represents a parabola that opens upwards.
- The vertex of the parabola is at (0, -9).
- The y-intercept of the function is -9.
- The parabola is wider than the standard parabola of the form [tex]\( x^2 \)[/tex].
To conclude, the effect on the function [tex]\( h(x) \)[/tex] described by [tex]\( \frac{1}{6} x^2 - 9 \)[/tex] can be visualized as an upward-opening, wide parabola with its vertex at (0, -9).
