1. **Problem Statement:** Determine the equation of the graph described, which is a "V" shaped absolute value function with vertex at $(4,1)$. 2. **Recall the general form of an absolute value function:** $$y = a|x - h| + k$$ where $(h,k)$ is the vertex of the graph and $a$ is the slope of the lines forming the "V" (outside the absolute value). 3. **Identify the vertex:** Given vertex is $(4,1)$, so $h=4$ and $k=1$. 4. **Find the slope $a$:** The slope of the left segment is $-\frac{1}{2}$ and the slope of the right segment is $\frac{1}{2}$. 5. **Since the slopes are $\pm \frac{1}{2}$, the value of $a$ is $\frac{1}{2}$**. 6. **Write the equation:** $$y = \frac{1}{2} |x - 4| + 1$$ 7. **Check with given points:** - For $x=1$, $$y = \frac{1}{2} |1 - 4| + 1 = \frac{1}{2} \times 3 + 1 = 1.5 + 1 = 2.5$$ The point given is $(1,2)$, close but not exact, so let's verify the slope calculation. 8. **Calculate slope between $(1,2)$ and $(4,1)$:** $$m = \frac{1 - 2}{4 - 1} = \frac{-1}{3} = -\frac{1}{3}$$ This contradicts the earlier slope $-\frac{1}{2}$. 9. **Calculate slope between $(4,1)$ and $(12,5)$:** $$m = \frac{5 - 1}{12 - 4} = \frac{4}{8} = \frac{1}{2}$$ Right slope is $\frac{1}{2}$, left slope is $-\frac{1}{3}$. 10. **Since slopes are not equal in magnitude, the function is piecewise:** $$y = \begin{cases} -\frac{1}{3}(x - 4) + 1 & \text{if } x < 4 \\ \frac{1}{2}(x - 4) + 1 & \text{if } x \geq 4 \end{cases}$$ 11. **Rewrite as an absolute value function:** Because slopes differ, it cannot be expressed as a single absolute value function with one $a$. 12. **Conclusion:** The graph is piecewise linear with vertex $(4,1)$, left slope $-\frac{1}{3}$, right slope $\frac{1}{2}$. **Final answer:** $$y = \begin{cases} -\frac{1}{3}x + \frac{7}{3} & x < 4 \\ \frac{1}{2}x - 1 & x \geq 4 \end{cases}$$