1. The problem asks to create a graph of a non-proportional linear relationship with a slope of $\frac{1}{4}$. This means the line should have the form $y = \frac{1}{4}x + b$ where $b \neq 0$ to ensure it is non-proportional (does not pass through the origin except at $b=0$). 2. Since the graph starts at the origin and continues horizontally near the x-axis, the line segment shown is likely the start of the graph but does not represent the full line with slope $\frac{1}{4}$. 3. To satisfy the problem, we choose a y-intercept $b$ different from zero, for example $b=1$, so the equation is: $$y = \frac{1}{4}x + 1$$ 4. This line has slope $\frac{1}{4}$ and y-intercept 1, so it is non-proportional and will start at $(0,1)$, not the origin. 5. The graph will show a line rising slowly from $y=1$ at $x=0$ to $y=\frac{1}{4} \times 15 + 1 = 4.75$ at $x=15$. Final answer: $$y = \frac{1}{4}x + 1$$