A function is linear for x $\le$ 2 and also linear for x $\ge$ 2.This function has the following values: f(-4)= 8; f(2)= 5; f(4)= 11.Find formula(s)(or equation(s))which describe this function.
A) $f(x)=\left\{\begin{array}{ll}
\frac{1}{2} x+4, & x \leq 2 \
-3 x+11, & x \geq 2
\end{array}\right.$
B) $f(x)=\left\{\begin{array}{cc}
2 x+6, & x \leq 2 \
\frac{1}{3} x-11, & x \geq 2
\end{array}\right.$
C) $f(x)=\left\{\begin{array}{cc}
-2 x+9, & x \leq 2 \
3 x-1, & x \geq 2
\end{array}\right.$
D) $f(x)=\left\{\begin{array}{cl}
-\frac{1}{2} x+6, & x \leq 2 \
3 x-1, & x \geq 2
\end{array}\right.$

Question

Quizplus · Accepted Answer

The function must be linear in two segments, one for $x \leq 2$ and another for $x \geq 2$. Using the given points, for $x \leq 2$, we can use the points (-4, 8) and (2, 5) to find the slope $m = \frac{5-8}{2-(-4)} = \frac{-3}{6} = -\frac{1}{2}$. The equation for this segment is $f(x) = -\frac{1}{2}x + b$. Substituting $x = 2$ and $f(x) = 5$ gives $5 = -\frac{1}{2}(2) + b$, solving for $b$ gives $b = 6$, so the equation is $f(x) = -\frac{1}{2}x + 6$ for $x \leq 2$.For $x \geq 2$, using the points (2, 5) and (4, 11), the slope $m = \frac{11-5}{4-2} = \frac{6}{2} = 3$. The equation for this segment is $f(x) = 3x + b$. Substituting $x = 2$ and $f(x) = 5$ gives $5 = 3(2) + b$, solving for $b$ gives $b = -1$, so the equation is $f(x) = 3x - 1$ for $x \geq 2$.Thus, the correct choice is $f(x)=\left\{\begin{array}{cl}-\frac{1}{2} x+6, & x \leq 2 \3 x-1, & x \geq 2\end{array}\right.$, which is option D.

A Function Is Linear for X ≤\le≤ 2 and Also Linear for X

A Function Is Linear for X $\le$ 2 and Also Linear for X