1. Consider the function $y = (2x + 3)(x - 1)(x - 4)$. (a) Find the x-intercepts by setting $y=0$: $$ (2x + 3)(x - 1)(x - 4) = 0 $$ This gives $2x + 3 = 0 \Rightarrow x = -\frac{3}{2}$, $x - 1 = 0 \Rightarrow x = 1$, and $x - 4 = 0 \Rightarrow x = 4$. (b) The x-intercepts partition the number line into intervals: $$ (-\infty, -\frac{3}{2}), (-\frac{3}{2}, 1), (1, 4), (4, \infty) $$ (c) To determine the sign of $y$ on each interval, test a point in each: - For $x = -2$ in $(-\infty, -\frac{3}{2})$: $(2(-2)+3)(-2-1)(-2-4) = (-1)(-3)(-6) = -18 < 0$ - For $x = 0$ in $(-\frac{3}{2}, 1)$: $(3)(-1)(-4) = 12 > 0$ - For $x = 2$ in $(1,4)$: $(7)(1)(-2) = -14 < 0$ - For $x = 5$ in $(4, \infty)$: $(13)(4)(1) = 52 > 0$ (d) Sketch: The graph crosses the x-axis at $x = -\frac{3}{2}, 1, 4$ and alternates sign accordingly. (e) Behavior as $x \to -\infty$: Leading term is $2x^3$, so $y \to -\infty$. (f) Behavior as $x \to \infty$: $y \to \infty$. (g) Leading term: $2x^3$, leading coefficient: 2, degree: 3. --- 2. For $y = -x^3 + 2x^2 + 11x - 2$: (a) Find x-intercepts by solving $-x^3 + 2x^2 + 11x - 2 = 0$. Try rational roots: $x=1$ gives $-1 + 2 + 11 - 2 = 10 \neq 0$; $x=-1$ gives $1 + 2 - 11 - 2 = -10 \neq 0$; $x=2$ gives $-8 + 8 + 22 - 2 = 20 \neq 0$; $x=-2$ gives $8 + 8 - 22 - 2 = -8 \neq 0$. Use synthetic division or numerical methods to find roots approximately. (b) Approximate roots are $x \approx -2.5$, $x \approx 0.2$, and $x \approx 4.3$. (c) Intervals: $$ (-\infty, -2.5), (-2.5, 0.2), (0.2, 4.3), (4.3, \infty) $$ (d) Test signs on each interval. (e) Leading term: $-x^3$, leading coefficient: $-1$, degree: 3. (f) As $x \to -\infty$, $y \to \infty$; as $x \to \infty$, $y \to -\infty$. --- 3. For $y = x^4 - 26x^2 + 25$: (a) Set $y=0$: $$ x^4 - 26x^2 + 25 = 0 $$ Let $u = x^2$, then: $$ u^2 - 26u + 25 = 0 $$ Solve quadratic: $$ u = \frac{26 \pm \sqrt{26^2 - 4 \cdot 25}}{2} = \frac{26 \pm \sqrt{676 - 100}}{2} = \frac{26 \pm \sqrt{576}}{2} = \frac{26 \pm 24}{2} $$ So $u = 25$ or $u = 1$. (b) Thus $x^2 = 25 \Rightarrow x = \pm 5$, and $x^2 = 1 \Rightarrow x = \pm 1$. (c) Intervals: $$ (-\infty, -5), (-5, -1), (-1, 1), (1, 5), (5, \infty) $$ (d) Test signs on each interval. (e) Leading term: $x^4$, leading coefficient: 1, degree: 4. (f) As $x \to \pm \infty$, $y \to \infty$. --- 4. For $y = -x^4 - 5x^3 + 3x^2 + 13x - 10$: (a) Leading term: $-x^4$, leading coefficient: $-1$, degree: 4. (b) As $x \to \pm \infty$, $y \to -\infty$. (c) Find x-intercepts numerically or by factoring if possible. (d) Intervals and sign table depend on roots. --- 5. For $y = x^2 (x + 3)(x + 1)^4 (x - 1)^3$: (a) X-intercepts are at $x=0$ (multiplicity 2), $x=-3$ (multiplicity 1), $x=-1$ (multiplicity 4), and $x=1$ (multiplicity 3). (b) Intervals: $$ (-\infty, -3), (-3, -1), (-1, 0), (0, 1), (1, \infty) $$ (c) Test signs on each interval considering multiplicities. (d) Leading term: Multiply leading terms: $$ x^2 \cdot x \cdot x^4 \cdot x^3 = x^{10} $$ Leading coefficient: 1, degree: 10. (e) As $x \to \pm \infty$, $y \to \infty$. --- Summary: - For each polynomial, find x-intercepts by solving $y=0$. - Use x-intercepts to partition the number line. - Test signs of $y$ in each interval. - Sketch graph crossing or touching x-axis at intercepts depending on multiplicity. - Leading term determines end behavior. - Leading coefficient and degree characterize growth and direction.