1. **State the problem:** We are given two linear functions: $$y=\frac{1}{4}x+6$$ and $$y=-2x+\frac{3}{2}$$ We want to find the values of $y$ for given $x$ values in each function and understand their behavior. 2. **Calculate $y$ values for the first function $y=\frac{1}{4}x+6$ at $x=-4,0,4$: ** - For $x=-4$: $$y=\frac{1}{4}(-4)+6=-1+6=5$$ - For $x=0$: $$y=\frac{1}{4}(0)+6=0+6=6$$ - For $x=4$: $$y=\frac{1}{4}(4)+6=1+6=7$$ 3. **Calculate $y$ values for the second function $y=-2x+\frac{3}{2}$ at $x=-1,0,1$: ** - For $x=-1$: $$y=-2(-1)+\frac{3}{2}=2+1.5=3.5$$ - For $x=0$: $$y=-2(0)+\frac{3}{2}=0+1.5=1.5$$ - For $x=1$: $$y=-2(1)+\frac{3}{2}=-2+1.5=-0.5$$ 4. **Summary of points:** - For $y=\frac{1}{4}x+6$: Points are $(-4,5)$, $(0,6)$, $(4,7)$. - For $y=-2x+\frac{3}{2}$: Points are $(-1,3.5)$, $(0,1.5)$, $(1,-0.5)$. 5. **Explanation:** - The first line has a gentle positive slope $\frac{1}{4}$, so $y$ increases slowly as $x$ increases. - The second line has a steeper negative slope $-2$, so $y$ decreases quickly as $x$ increases. These points can be plotted on the Cartesian plane with $x$ from $-4$ to $4$ and $y$ from $-1$ to $7$ as described. **Final answers:** For $y=\frac{1}{4}x+6$: $$y(-4)=5,\quad y(0)=6,\quad y(4)=7$$ For $y=-2x+\frac{3}{2}$: $$y(-1)=3.5,\quad y(0)=1.5,\quad y(1)=-0.5$$