Question

consider the following function.
u(x)=\begin{cases}-2x - 6&\text{if }xleq - 3\\frac{1}{3}x^{2}&\text{if }x > - 3end{cases}
step 2 of 3: identify the general shape and direction of the graph of this function on the interval ((-3,infty)).

Explanation:

Step1: Analyze the function for $x > - 3$

The function for $x > - 3$ is $u(x)=\frac{1}{3}x^{2}$. This is a quadratic function of the form $y = ax^{2}+bx + c$ where $a=\frac{1}{3}$, $b = 0$, and $c = 0$.

Step2: Recall properties of quadratic functions

For a quadratic function $y=ax^{2}+bx + c$, when $a>0$, the graph is a parabola. Since $a=\frac{1}{3}>0$, the parabola opens upwards.

Answer:

The graph is a parabola that opens upwards.