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 3 of 3: identify the correct graph of this piecewise - defined function.

Explanation:

Step1: Analyze the first - part of the function

For $u(x)=-2x - 6$ when $x\leq - 3$, this is a linear function. The slope $m=-2$ and the $y$-intercept (when $x = 0$) is $-6$. When $x=-3$, $u(-3)=-2\times(-3)-6=0$.

Step2: Analyze the second - part of the function

For $u(x)=\frac{1}{3}x^{2}$ when $x > - 3$, this is a quadratic function. The vertex of the parabola $y = ax^{2}+bx + c$ (here $a=\frac{1}{3},b = 0,c = 0$) is at the origin $(0,0)$. When $x=-3$, $u(-3)=\frac{1}{3}\times(-3)^{2}=3$. But this part of the function is for $x > - 3$, so the point $(-3,3)$ is not included in this part of the graph, and the graph of $y=\frac{1}{3}x^{2}$ starts just to the right of $x=-3$.

Step3: Combine the two parts

The graph of $y=-2x - 6$ is a line with a negative slope for $x\leq - 3$ and includes the point $(-3,0)$. The graph of $y=\frac{1}{3}x^{2}$ is a parabola opening upwards for $x > - 3$ and starts just after $x=-3$.

Since no graphs are provided to choose from, we can't give a specific final answer in terms of a choice. But the general description of the graph is as above. If there were options, we would look for a graph with a line $y=-2x - 6$ for $x\leq - 3$ (including the point $(-3,0)$) and a parabola $y=\frac{1}{3}x^{2}$ for $x > - 3$ (not including the point $(-3,3)$).

Answer:

Step1: Analyze the first - part of the function

For $u(x)=-2x - 6$ when $x\leq - 3$, this is a linear function. The slope $m=-2$ and the $y$-intercept (when $x = 0$) is $-6$. When $x=-3$, $u(-3)=-2\times(-3)-6=0$.

Step2: Analyze the second - part of the function

For $u(x)=\frac{1}{3}x^{2}$ when $x > - 3$, this is a quadratic function. The vertex of the parabola $y = ax^{2}+bx + c$ (here $a=\frac{1}{3},b = 0,c = 0$) is at the origin $(0,0)$. When $x=-3$, $u(-3)=\frac{1}{3}\times(-3)^{2}=3$. But this part of the function is for $x > - 3$, so the point $(-3,3)$ is not included in this part of the graph, and the graph of $y=\frac{1}{3}x^{2}$ starts just to the right of $x=-3$.

Step3: Combine the two parts

The graph of $y=-2x - 6$ is a line with a negative slope for $x\leq - 3$ and includes the point $(-3,0)$. The graph of $y=\frac{1}{3}x^{2}$ is a parabola opening upwards for $x > - 3$ and starts just after $x=-3$.

Since no graphs are provided to choose from, we can't give a specific final answer in terms of a choice. But the general description of the graph is as above. If there were options, we would look for a graph with a line $y=-2x - 6$ for $x\leq - 3$ (including the point $(-3,0)$) and a parabola $y=\frac{1}{3}x^{2}$ for $x > - 3$ (not including the point $(-3,3)$).