Find a formula for the function graphed.
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A) $f ( x ) = \left\{ \begin{array} { l l } \frac { 1 } { 2 } x + 1 , & - 8 \leq x \leq - 2 \ 5 , & - 2

Question

Find a formula for the function graphed.
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A) $f ( x ) = \left\{ \begin{array} { l l } \frac { 1 } { 2 } x + 1 , & - 8 \leq x \leq - 2 \ 5 , & - 2 < x < 3 \ 6 - x , & 3 \leq x \leq 8 \end{array} \right.$ 
B) $f ( x ) = \left\{ \begin{array} { l l } \frac { 1 } { 2 } x + 1 , & - 8 \leq x \leq - 2 \ 5 , & - 2 < x \leq 3 \ 6 - x , & 3 < x \leq 8 \end{array} \right.$ 
C) $f ( x ) = \left\{ \begin{array} { l r } - \frac { 1 } { 2 } x + 1 , & - 8 \leq x \leq - 2 \ 5 , & - 2 < x \leq 3 \ x - 6 , & 3 < x \leq 8 \end{array} \right.$ 
D) $f ( x ) = \left\{ \begin{array} { l l } \frac { 1 } { 2 } x + 1 , & - 8 < x \leq - 2 \ 5 , & - 2 < x \leq 3 \ 6 - x , & 3 < x < 8 \end{array} \right.$

Quizplus · Accepted Answer

From the graph, we can see that the function is a line segment from $(-8,2)$ to $(-2,6)$, at which point it jumps to the point $(3,5)$, and then it is a line segment from $(3,5)$ to $(8,-2)$.

Therefore, the formula for this function has three cases:

Case 1: $-8 \leq x \leq -2$. In this range, the function is a line with slope $\frac{6-2}{-2-(-8)} = \frac{2}{3}$ and $y$-intercept $(0,2)$. So, for $-8 \leq x \leq -2$, we have:

$$f(x) = \frac{2}{3}x + 2$$

Case 2: $-2 < x \leq 3$. In this range, the function is constant at $y=6$. So, for $-2 < x \leq 3$, we have:

$$f(x) = 5$$

Case 3: $3 < x \leq 8$. In this range, the function is a line with slope $\frac{-2-5}{8-3} = -\frac{7}{5}$ and $y$-intercept $(0,-2)$. So, for $3 < x \leq 8$, we have:

$$f(x) = -\frac{7}{5}x - 2$$

Putting these three cases together, we get:

$$f(x) = \left\{ \begin{array}{ll} \frac{2}{3}x + 2 & \mbox{if } -8 \leq x \leq -2 \ 5 & \mbox{if } -2 < x \leq 3 \ -\frac{7}{5}x - 2 & \mbox{if } 3 < x \leq 8 \end{array} ight.$$

Comparing with the given choices, we see that (B) matches this formula.

Find a Formula for the Function Graphed B) C) D)