1. **Solve the inequalities:** a) $8x \leq x^2 + 12$ b) $-x^2 > 8x + 16$ c) $y < -(x+2)(x-4)$ --- 2. **Check if points $A(3,7)$ and $B(-1,12)$ satisfy $y > 2x^2 - 5x + 4$** --- 3. **Find $m$ such that for $(m+1)x^2 - mx + m - 1 = 0$:** - sum of roots = 4 - product of roots = -5 - no real solution --- 4. **Solve systems:** a) $\begin{cases} y < x + 2 \\ y \geq x^2 - 4 \end{cases}$ b) $\begin{cases} -x^2 + 3x + 10 \geq 0 \\ x^2 - 11x + 28 < 0 \end{cases}$ --- 5. **Find $m$ so that vertex of $f(x) = mx^2 - 2(m-2)x + m$ lies on the $y$-axis** --- 6. **Determine inequalities from the graph description:** - Parabola opens upward with vertex near $(-1,1)$ - Line passes through origin with positive slope - Shaded region is to the left of parabola and below the line - Conditions: $a > 0$, $\Delta < 0$ --- ### Step 1: Solve $8x \leq x^2 + 12$ Rewrite as: $$x^2 - 8x + 12 \geq 0$$ Factor: $$x^2 - 8x + 12 = (x - 6)(x - 2)$$ Inequality: $$(x - 6)(x - 2) \geq 0$$ This holds when $x \leq 2$ or $x \geq 6$. --- ### Step 2: Solve $-x^2 > 8x + 16$ Rewrite: $$-x^2 - 8x - 16 > 0$$ Multiply both sides by $-1$ (flip inequality): $$x^2 + 8x + 16 < 0$$ Factor: $$x^2 + 8x + 16 = (x + 4)^2$$ Since $(x + 4)^2 \geq 0$ always, and equals zero at $x = -4$, the inequality $< 0$ has no solution. --- ### Step 3: Solve $y < -(x+2)(x-4)$ Expand: $$y < -(x^2 - 2x - 8) = -x^2 + 2x + 8$$ This is the region below the parabola $y = -x^2 + 2x + 8$. --- ### Step 4: Check points for $y > 2x^2 - 5x + 4$ For $A(3,7)$: Calculate right side: $$2(3)^2 - 5(3) + 4 = 18 - 15 + 4 = 7$$ Check $7 > 7$? No, equality only, so $A$ does not satisfy strict inequality. For $B(-1,12)$: $$2(-1)^2 - 5(-1) + 4 = 2 + 5 + 4 = 11$$ Check $12 > 11$? Yes, so $B$ satisfies. --- ### Step 5: Find $m$ for sum and product of roots Equation: $$(m+1)x^2 - mx + m - 1 = 0$$ Sum of roots $S = \frac{m}{m+1}$ Product of roots $P = \frac{m-1}{m+1}$ Set sum = 4: $$\frac{m}{m+1} = 4$$ Multiply both sides: $$m = 4(m+1)$$ $$m = 4m + 4$$ $$\cancel{m} - 4m = 4$$ $$\cancel{\cancel{m}} - 4m = 4$$ $$-3m = 4$$ $$m = -\frac{4}{3}$$ Set product = -5: $$\frac{m-1}{m+1} = -5$$ Multiply both sides: $$m - 1 = -5(m + 1)$$ $$m - 1 = -5m - 5$$ $$m + 5m = -5 + 1$$ $$6m = -4$$ $$m = -\frac{2}{3}$$ For no real solution, discriminant $\Delta < 0$: $$\Delta = (-m)^2 - 4(m+1)(m-1) = m^2 - 4(m^2 - 1) = m^2 - 4m^2 + 4 = -3m^2 + 4 < 0$$ Solve: $$-3m^2 + 4 < 0$$ $$-3m^2 < -4$$ $$\cancel{-3}m^2 > \cancel{\frac{4}{3}}$$ (inequality flips because dividing by negative) $$m^2 > \frac{4}{3}$$ So: $$m < -\frac{2}{\sqrt{3}} \quad \text{or} \quad m > \frac{2}{\sqrt{3}}$$ --- ### Step 6: Solve system $\begin{cases} y < x + 2 \\ y \geq x^2 - 4 \end{cases}$ The solution is the intersection of the region below the line $y = x + 2$ and above the parabola $y = x^2 - 4$. --- ### Step 7: Solve system $\begin{cases} -x^2 + 3x + 10 \geq 0 \\ x^2 - 11x + 28 < 0 \end{cases}$ Rewrite first inequality: $$-x^2 + 3x + 10 \geq 0 \Rightarrow x^2 - 3x - 10 \leq 0$$ Factor: $$(x - 5)(x + 2) \leq 0$$ Solution: $$-2 \leq x \leq 5$$ Second inequality: $$x^2 - 11x + 28 < 0$$ Factor: $$(x - 4)(x - 7) < 0$$ Solution: $$4 < x < 7$$ Intersection: $$4 < x \leq 5$$ --- ### Step 8: Find $m$ so vertex of $f(x) = mx^2 - 2(m-2)x + m$ lies on $y$-axis Vertex $x$-coordinate: $$x_v = -\frac{b}{2a} = -\frac{-2(m-2)}{2m} = \frac{2(m-2)}{2m} = \frac{m-2}{m}$$ For vertex on $y$-axis, $x_v = 0$: $$\frac{m-2}{m} = 0 \Rightarrow m - 2 = 0 \Rightarrow m = 2$$ --- ### Step 9: Inequalities from graph description Parabola: $y = a(x+1)^2 + 1$ with $a > 0$ (opens upward) Line: $y = kx$ with $k > 0$ Shaded region: left of parabola and below line Inequalities: $$y \leq a(x+1)^2 + 1$$ $$y \leq kx$$ Since shaded region is left of parabola, $x \leq$ parabola boundary, so rewrite parabola inequality as: $$x \leq -1 - \sqrt{\frac{y - 1}{a}}$$ But typically, the region is: $$y \leq kx$$ $$y \leq a(x+1)^2 + 1$$ with $a > 0$ and discriminant $\Delta < 0$ for the parabola's quadratic form. --- **Final answers:** 1a) $x \leq 2$ or $x \geq 6$ 1b) No solution 1c) $y < -x^2 + 2x + 8$ 2) $A$ does not satisfy, $B$ satisfies 3) Sum roots = 4: $m = -\frac{4}{3}$ Product roots = -5: $m = -\frac{2}{3}$ No real roots: $m < -\frac{2}{\sqrt{3}}$ or $m > \frac{2}{\sqrt{3}}$ 4a) $y < x + 2$ and $y \geq x^2 - 4$ 4b) $4 < x \leq 5$ 5) $m = 2$ 6) $y \leq a(x+1)^2 + 1$, $y \leq kx$, with $a > 0$, $k > 0$, and $\Delta < 0$ for parabola