Question

use the quadratic function $f(x)=\frac{1}{3}x^2 + 6x + 4$ to answer the following questions.
a) use the vertex formula to determine the vertex.
the vertex is $(-9, -23)$.
(type an ordered pair. simplify your answer.)
b) does the graph \open up\ or \open down\?
$circ$ down
$circ$ up

Explanation:

Part a)

Step 1: Recall Vertex Formula for Quadratic

For a quadratic function \( f(x) = ax^2 + bx + c \), the x - coordinate of the vertex is given by \( x = -\frac{b}{2a} \).
Given \( f(x)=\frac{1}{3}x^{2}+6x + 4 \), we have \( a=\frac{1}{3} \), \( b = 6 \), \( c = 4 \).
First, calculate the x - coordinate of the vertex:
\( x=-\frac{b}{2a}=-\frac{6}{2\times\frac{1}{3}}=-\frac{6}{\frac{2}{3}}=-6\times\frac{3}{2}=-9 \)

Step 2: Find the y - coordinate of the vertex

Substitute \( x=-9 \) into the function \( f(x)=\frac{1}{3}x^{2}+6x + 4 \):
\( f(-9)=\frac{1}{3}\times(-9)^{2}+6\times(-9)+4 \)
\(=\frac{1}{3}\times81-54 + 4 \)
\(=27-54 + 4=-23 \)
So the vertex is \( (-9,-23) \)

For a quadratic function \( f(x)=ax^{2}+bx + c \), if \( a>0 \), the parabola opens up; if \( a < 0 \), the parabola opens down.
In the function \( f(x)=\frac{1}{3}x^{2}+6x + 4 \), \( a=\frac{1}{3}\). Since \( \frac{1}{3}>0 \), the graph of the quadratic function opens up.

Answer:

\((-9, - 23)\)

Part b)