Question

the graph of g(x) is shown below. (a) write an equation that represents the function g(x). 8 points

Explanation:

Step1: Identify function type

The graph appears to be a piece - wise function.

Step2: Analyze left - hand part

The left - hand part from $x = - 4$ to $x=0$ seems to be a quadratic function. Let the quadratic function be $y = a(x - h)^2 + k$. The vertex of this parabola is around $( - 2,4)$. So $h=-2,k = 4$. Using the point $( - 4,2)$: $2=a(-4 + 2)^2+4$, $2 = 4a+4$, $4a=-2$, $a=-\frac{1}{2}$. So the quadratic part is $y=-\frac{1}{2}(x + 2)^2+4$ for $-4\leq x<0$.

Step3: Analyze middle part

The middle part from $x = 0$ to $x = 2$ is a horizontal line at $y=-1$. So the equation is $y=-1$ for $0\leq x<2$.

Step4: Analyze right - hand part

The right - hand part from $x = 2$ to $x=8$ is a linear function. The two - point form of a line is $y - y_1=\frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$. Using the points $(2,3)$ and $(8,-3)$: The slope $m=\frac{-3 - 3}{8 - 2}=\frac{-6}{6}=-1$. Using the point - slope form with $(2,3)$: $y - 3=-(x - 2)$, $y=-x + 5$ for $2\leq x<8$.

The piece - wise function $g(x)$ is:
\[g(x)=

$$\begin{cases}-\frac{1}{2}(x + 2)^2+4, &-4\leq x<0\\-1, &0\leq x<2\\-x + 5, &2\leq x<8\end{cases}$$

\]

Answer:

\[g(x)=

$$\begin{cases}-\frac{1}{2}(x + 2)^2+4, &-4\leq x<0\\-1, &0\leq x<2\\-x + 5, &2\leq x<8\end{cases}$$

\]