1. Problem 16a: Find values of $m$ such that the quadratic equation $$x^2 + (3 - m)x + 2m - 1 = 0$$ has real roots. 2. For real roots, the discriminant must be non-negative: $$\Delta = b^2 - 4ac \geq 0$$ Here, $a=1$, $b=3 - m$, $c=2m - 1$. 3. Calculate the discriminant: $$\Delta = (3 - m)^2 - 4(1)(2m - 1) = (3 - m)^2 - 8m + 4$$ 4. Expand and simplify: $$(3 - m)^2 = 9 - 6m + m^2$$ So, $$\Delta = 9 - 6m + m^2 - 8m + 4 = m^2 - 14m + 13$$ 5. Set the inequality for real roots: $$m^2 - 14m + 13 \geq 0$$ 6. Solve the quadratic inequality by finding roots: $$m^2 - 14m + 13 = 0$$ Use quadratic formula: $$m = \frac{14 \pm \sqrt{196 - 52}}{2} = \frac{14 \pm \sqrt{144}}{2} = \frac{14 \pm 12}{2}$$ 7. Roots are: $$m = \frac{14 + 12}{2} = 13, \quad m = \frac{14 - 12}{2} = 1$$ 8. Since the parabola opens upward ($a=1>0$), the inequality $m^2 - 14m + 13 \geq 0$ holds for: $$m \leq 1 \quad \text{or} \quad m \geq 13$$ --- 9. Problem 16b: If one root is 1, find the other root. 10. Substitute $x=1$ into the equation: $$1^2 + (3 - m)(1) + 2m - 1 = 0$$ Simplify: $$1 + 3 - m + 2m - 1 = 0 \Rightarrow 3 + m = 0 \Rightarrow m = -3$$ 11. With $m = -3$, the quadratic becomes: $$x^2 + (3 - (-3))x + 2(-3) - 1 = x^2 + 6x - 7 = 0$$ 12. Use sum and product of roots: Sum of roots $= -b/a = -6$ One root is 1, so the other root $= -6 - 1 = -7$ --- 13. Problem 17a: Given $$f(x) = \frac{x + 4}{x + 3}$$ find the domain. 14. The function is undefined where denominator is zero: $$x + 3 = 0 \Rightarrow x = -3$$ 15. Domain is all real numbers except $x = -3$: $$\{x \in \mathbb{R} : x \neq -3\}$$ --- 16. Problem 17b: Find $x$-intercept and $y$-intercept. 17. $x$-intercept: set numerator zero: $$x + 4 = 0 \Rightarrow x = -4$$ So, $x$-intercept is $(-4, 0)$. 18. $y$-intercept: set $x=0$: $$f(0) = \frac{0 + 4}{0 + 3} = \frac{4}{3}$$ So, $y$-intercept is $(0, \frac{4}{3})$. --- 19. Problem 17c: Find relative asymptotes. 20. Vertical asymptote where denominator zero: $$x = -3$$ 21. Horizontal asymptote is limit as $x \to \pm \infty$: $$\lim_{x \to \infty} \frac{x + 4}{x + 3} = \lim_{x \to \infty} \frac{1 + \frac{4}{x}}{1 + \frac{3}{x}} = 1$$ 22. So horizontal asymptote is: $$y = 1$$ --- 23. Problem 17d: Determine if $f(x)$ has extrema. 24. Find derivative: $$f'(x) = \frac{(1)(x + 3) - (x + 4)(1)}{(x + 3)^2} = \frac{x + 3 - x - 4}{(x + 3)^2} = \frac{-1}{(x + 3)^2}$$ 25. Since denominator squared is always positive, $f'(x) = -1/(x+3)^2 < 0$ for all $x \neq -3$. 26. Derivative is always negative, so $f(x)$ is strictly decreasing and has no relative maxima or minima. --- **Final answers:** - 16a) $m \leq 1$ or $m \geq 13$ - 16b) Other root is $-7$ when $m = -3$ - 17a) Domain: $x \neq -3$ - 17b) $x$-intercept: $(-4,0)$, $y$-intercept: $(0, \frac{4}{3})$ - 17c) Vertical asymptote: $x = -3$, Horizontal asymptote: $y = 1$ - 17d) No extrema, function strictly decreasing.