1. **State the problem:** We need to find the limit $$\lim_{x \to 0} \frac{2f(3x) - x^2}{2x^2 + x}$$ where $f$ is a piecewise linear function given by points and lines. 2. **Analyze the function $f$ near $x=0$:** From the graph points, near $x=0$, the closest points are $(-1, -6)$ and $(1, 3)$. Since $f$ is piecewise linear, the segment between $x=-1$ and $x=1$ is a straight line. 3. **Find the equation of the line segment between $(-1, -6)$ and $(1, 3)$:** Slope $$m = \frac{3 - (-6)}{1 - (-1)} = \frac{9}{2} = 4.5$$ Using point-slope form with point $(-1, -6)$: $$f(x) = m(x + 1) - 6 = 4.5(x + 1) - 6 = 4.5x + 4.5 - 6 = 4.5x - 1.5$$ So near $x=0$, $$f(x) = 4.5x - 1.5$$ 4. **Evaluate $f(3x)$ near $x=0$:** $$f(3x) = 4.5(3x) - 1.5 = 13.5x - 1.5$$ 5. **Substitute into the limit expression:** $$\frac{2f(3x) - x^2}{2x^2 + x} = \frac{2(13.5x - 1.5) - x^2}{2x^2 + x} = \frac{27x - 3 - x^2}{2x^2 + x}$$ 6. **Simplify numerator and denominator:** Numerator: $$27x - 3 - x^2 = -3 + 27x - x^2$$ Denominator: $$2x^2 + x$$ 7. **Evaluate the limit as $x \to 0$:** Direct substitution gives numerator $-3$ and denominator $0$, so the limit might be infinite or does not exist. 8. **Check the limit from the right ($x \to 0^+$):** For very small positive $x$, denominator $2x^2 + x \approx x > 0$. Numerator $\approx -3 +$ small positive terms $\approx -3$ (negative). So fraction $\approx \frac{-3}{+0} = -\infty$. 9. **Check the limit from the left ($x \to 0^-$):** For very small negative $x$, denominator $2x^2 + x \approx x < 0$. Numerator $\approx -3 +$ small negative terms $\approx -3$ (negative). So fraction $\approx \frac{-3}{-0} = +\infty$. 10. **Conclusion:** The left and right limits are not equal, so the limit does not exist. **Final answer:** $$\lim_{x \to 0} \frac{2f(3x) - x^2}{2x^2 + x} \text{ does not exist}$$