1. We start with the first problem: find for which values of $x$ the graph of $f(x)$ lies below the graph of $g(x)$. 2. For (a), the functions are $f(x) = x^2$ and $g(x) = -4x + 5$. 3. We want to find where $f(x) < g(x)$, so: $$x^2 < -4x + 5$$ 4. Rearrange the inequality to one side: $$x^2 + 4x - 5 < 0$$ 5. Factor the quadratic expression: $$x^2 + 4x - 5 = (x + 5)(x - 1)$$ 6. The inequality becomes: $$(x + 5)(x - 1) < 0$$ 7. This product is less than zero when the factors have opposite signs. So, between the roots: $$-5 < x < 1$$ 8. Therefore, the graph of $f$ lies below $g$ for $x$ in the interval $(-5, 1)$. 9. For (b), the functions are $f(x) = x^2 - 4x + 2$ and $g(x) = -2x^2 + 8x - 14$. 10. Find where $f(x) < g(x)$: $$x^2 - 4x + 2 < -2x^2 + 8x - 14$$ 11. Bring all terms to one side: $$x^2 - 4x + 2 + 2x^2 - 8x + 14 < 0$$ 12. Simplify: $$3x^2 - 12x + 16 < 0$$ 13. Divide entire inequality by 3: $$\cancel{3}x^2 - \cancel{3}4x + \cancel{3} \frac{16}{3} < 0$$ which is $$x^2 - 4x + \frac{16}{3} < 0$$ 14. Calculate the discriminant $\Delta$ to check if the quadratic can be negative: $$\Delta = (-4)^2 - 4 \cdot 1 \cdot \frac{16}{3} = 16 - \frac{64}{3} = \frac{48}{3} - \frac{64}{3} = -\frac{16}{3} < 0$$ 15. Since $\Delta < 0$ and the leading coefficient is positive, the quadratic is always positive. 16. Therefore, $3x^2 - 12x + 16$ is never less than zero, so $f(x)$ is never less than $g(x)$. 17. Final answers: - (a) $f(x) < g(x)$ for $x \in (-5, 1)$. - (b) No values of $x$ satisfy $f(x) < g(x)$. 18. Graphical check: plotting both functions confirms these intervals.