1. **Solve the inequalities:** **Problem 1:** Solve $8x \leq x^2 + 12$ **Step 1:** Rewrite the inequality as $$x^2 - 8x + 12 \geq 0$$ **Step 2:** Factor the quadratic: $$x^2 - 8x + 12 = (x - 6)(x - 2)$$ **Step 3:** The inequality becomes $$(x - 6)(x - 2) \geq 0$$ **Step 4:** The product is non-negative when both factors are positive or both are negative. - For $x - 6 \geq 0$ and $x - 2 \geq 0$, we get $x \geq 6$ - For $x - 6 \leq 0$ and $x - 2 \leq 0$, we get $x \leq 2$ **Answer:** $x \leq 2$ or $x \geq 6$ **Problem 2:** Solve $-x^2 > 8x + 16$ **Step 1:** Rewrite as $$-x^2 - 8x - 16 > 0$$ Multiply both sides by $-1$ (remember to reverse inequality): $$x^2 + 8x + 16 < 0$$ **Step 2:** Factor the quadratic: $$x^2 + 8x + 16 = (x + 4)^2$$ **Step 3:** Since $(x + 4)^2 \geq 0$ for all $x$, the inequality $(x + 4)^2 < 0$ has no solution. **Answer:** No solution. **Problem 3:** Solve $y < -(x+2)(x-4)$ **Step 1:** Expand the right side: $$-(x+2)(x-4) = -(x^2 - 4x + 2x - 8) = -x^2 + 2x + 8$$ **Step 2:** The inequality is $$y < -x^2 + 2x + 8$$ **Answer:** The solution is all points $(x,y)$ such that $y < -x^2 + 2x + 8$. 2. **Check if points $A(3,7)$ and $B(-1,12)$ satisfy $y > 2x^2 - 5x + 4$** **Step 1:** Calculate right side for $x=3$: $$2(3)^2 - 5(3) + 4 = 18 - 15 + 4 = 7$$ Check if $7 > 7$ for point A: No, equality but not greater. **Step 2:** Calculate right side for $x=-1$: $$2(-1)^2 - 5(-1) + 4 = 2 + 5 + 4 = 11$$ Check if $12 > 11$ for point B: Yes. **Answer:** Point A does not satisfy the inequality; Point B satisfies it. 3. **Find $m$ for equation $(m+1)x^2 - m x + m - 1 = 0$ with conditions:** - Sum of roots $S = 4$ - Product of roots $P = -5$ - No real solution **Step 1:** Identify coefficients: $$a = m+1, \quad b = -m, \quad c = m - 1$$ **Step 2:** Sum of roots formula: $$S = -\frac{b}{a} = -\frac{-m}{m+1} = \frac{m}{m+1}$$ Set equal to 4: $$\frac{m}{m+1} = 4$$ Multiply both sides by $m+1$: $$m = 4(m+1)$$ $$m = 4m + 4$$ Subtract $4m$: $$m - 4m = 4$$ $$\cancel{m} - 4\cancel{m} = 4$$ $$-3m = 4$$ Divide both sides by $-3$: $$m = -\frac{4}{3}$$ **Step 3:** Product of roots formula: $$P = \frac{c}{a} = \frac{m - 1}{m + 1}$$ Set equal to $-5$: $$\frac{m - 1}{m + 1} = -5$$ Multiply both sides by $m+1$: $$m - 1 = -5(m + 1)$$ $$m - 1 = -5m - 5$$ Add $5m$ and $1$ to both sides: $$m + 5m = -5 + 1$$ $$6m = -4$$ Divide both sides by 6: $$m = -\frac{2}{3}$$ **Step 4:** For no real solution, discriminant $\Delta < 0$: $$\Delta = b^2 - 4ac = (-m)^2 - 4(m+1)(m-1) = m^2 - 4(m^2 - 1) = m^2 - 4m^2 + 4 = -3m^2 + 4$$ Set $\Delta < 0$: $$-3m^2 + 4 < 0$$ $$-3m^2 < -4$$ Divide by $-3$ (reverse inequality): $$m^2 > \frac{4}{3}$$ **Answer:** - Sum of roots 4: $m = -\frac{4}{3}$ - Product of roots -5: $m = -\frac{2}{3}$ - No real solution: $m < -\frac{2}{\sqrt{3}}$ or $m > \frac{2}{\sqrt{3}}$ 4. **Solve systems:** **System 1:** $$\begin{cases} y < x + 2 \\ y \geq x^2 - 4 \end{cases}$$ **Step 1:** The solution is the set of points where $y$ is less than the line $x+2$ and greater or equal to the parabola $x^2 - 4$. **System 2:** $$\begin{cases} -x^2 + 3x + 10 \geq 0 \\ x^2 - 11x + 28 < 0 \end{cases}$$ **Step 2:** Solve first inequality: $$-x^2 + 3x + 10 \geq 0 \Rightarrow x^2 - 3x - 10 \leq 0$$ Factor: $$(x - 5)(x + 2) \leq 0$$ Solution: $$-2 \leq x \leq 5$$ **Step 3:** Solve second inequality: $$x^2 - 11x + 28 < 0$$ Factor: $$(x - 4)(x - 7) < 0$$ Solution: $$4 < x < 7$$ **Step 4:** Intersection of both: $$[-2, 5] \cap (4, 7) = (4, 5]$$ **Answer:** $x \in (4, 5]$ 5. **Find $m$ so vertex of $f(x) = m x^2 - 2(m-2)x + m$ lies on y-axis** **Step 1:** Vertex $x$-coordinate: $$x_v = -\frac{b}{2a} = -\frac{-2(m-2)}{2m} = \frac{2(m-2)}{2m} = \frac{m-2}{m}$$ **Step 2:** For vertex on y-axis, $x_v = 0$: $$\frac{m-2}{m} = 0$$ Multiply both sides by $m$: $$m - 2 = 0$$ $$m = 2$$ 6. **Determine inequalities from graph description:** - Parabola opens upward with vertex at $(-1,1)$ - Line passes through $(0,0)$ and $(1,3)$, slope $3$ - Shaded region is left/below parabola and left of line **Step 1:** Equation of parabola vertex form: $$y = a(x + 1)^2 + 1$$ Since it passes through $(0,0)$ (approximate), find $a$: $$0 = a(0 + 1)^2 + 1 \Rightarrow 0 = a + 1 \Rightarrow a = -1$$ Parabola: $$y = -(x + 1)^2 + 1$$ **Step 2:** Equation of line through $(0,0)$ and $(1,3)$: $$y = 3x$$ **Step 3:** Inequalities: - Left/below parabola: $$y \leq -(x + 1)^2 + 1$$ - Left of line: $$x \leq 0$$ **Answer:** $$\begin{cases} y \leq -(x + 1)^2 + 1 \\ x \leq 0 \end{cases}$$