Question

g(x) = -\frac{1}{3}(x + 2)(x + 8)

Explanation:

Assuming the problem is to find the vertex of the parabola \( g(x) = -\frac{1}{3}(x + 2)(x + 8) \):

Step 1: Find the x - coordinate of the vertex

For a quadratic function in factored form \( y=a(x - r_1)(x - r_2) \), the x - coordinate of the vertex (the axis of symmetry) is given by the midpoint of the roots \( r_1 \) and \( r_2 \). The formula for the x - coordinate \( h \) is \( h=\frac{r_1 + r_2}{2} \).

From the function \( g(x)=-\frac{1}{3}(x + 2)(x + 8) \), the roots are found by setting \( g(x) = 0 \):
\( x+2 = 0\) gives \( x=-2 \) and \( x + 8=0\) gives \( x=-8 \).

So, \( r_1=-2\) and \( r_2=-8 \). Then \( h=\frac{-2+( - 8)}{2}=\frac{-10}{2}=-5 \).

Step 2: Find the y - coordinate of the vertex

Substitute \( x = h=-5 \) into the function \( g(x) \) to find the y - coordinate \( k \) of the vertex.

\( g(-5)=-\frac{1}{3}(-5 + 2)(-5 + 8) \)
First, calculate the values inside the parentheses: \( -5 + 2=-3 \) and \( -5 + 8 = 3 \).

Then \( g(-5)=-\frac{1}{3}(-3)(3) \)
Multiply \( -3\times3=-9 \), then \( -\frac{1}{3}\times(-9) = 3 \).

So the vertex of the parabola is at \( (-5,3) \).

Answer:

The vertex of the parabola \( g(x)=-\frac{1}{3}(x + 2)(x + 8) \) is \( (-5,3) \)