1. **Problem Statement:** We are given the function $y = (3 - 2x)(2 + x)$ and a linear equation $3y - 5x + 6 = 0$. We need to analyze these graphs and solve related questions. 2. **Expand the quadratic function:** $$y = (3 - 2x)(2 + x) = 3 \times 2 + 3 \times x - 2x \times 2 - 2x \times x = 6 + 3x - 4x - 2x^2 = 6 - x - 2x^2$$ 3. **Rewrite the linear equation in terms of $y$:** $$3y - 5x + 6 = 0 \implies 3y = 5x - 6 \implies y = \frac{5x - 6}{3}$$ 4. **Roots of the quadratic equation $y=0$:** Set $y=0$: $$0 = 6 - x - 2x^2$$ Rearranged: $$2x^2 + x - 6 = 0$$ Use quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \times 2 \times (-6)}}{2 \times 2} = \frac{-1 \pm \sqrt{1 + 48}}{4} = \frac{-1 \pm 7}{4}$$ So, $$x_1 = \frac{-1 + 7}{4} = \frac{6}{4} = 1.5$$ $$x_2 = \frac{-1 - 7}{4} = \frac{-8}{4} = -2$$ 5. **Find $x$ for which $3(3 - 2x)(2 + x) = 5x - 6$:** Rewrite: $$3y = 5x - 6$$ Substitute $y$: $$3(6 - x - 2x^2) = 5x - 6$$ Expand left side: $$18 - 3x - 6x^2 = 5x - 6$$ Bring all terms to one side: $$18 - 3x - 6x^2 - 5x + 6 = 0$$ Simplify: $$24 - 8x - 6x^2 = 0$$ Rewrite: $$-6x^2 - 8x + 24 = 0$$ Divide both sides by $-2$: $$\cancel{-2} \times 3x^2 + \cancel{-2} \times 4x - \cancel{-2} \times 12 = 0 \implies 3x^2 + 4x - 12 = 0$$ Use quadratic formula: $$x = \frac{-4 \pm \sqrt{4^2 - 4 \times 3 \times (-12)}}{2 \times 3} = \frac{-4 \pm \sqrt{16 + 144}}{6} = \frac{-4 \pm \sqrt{160}}{6} = \frac{-4 \pm 4\sqrt{10}}{6} = \frac{-2 \pm 2\sqrt{10}}{3}$$ 6. **Maximum value of $y$ and corresponding $x$:** Since $y = 6 - x - 2x^2$ is a quadratic with negative leading coefficient, it opens downward and has a maximum at vertex: $$x = -\frac{b}{2a} = -\frac{-1}{2 \times (-2)} = -\frac{-1}{-4} = -\frac{1}{4} = -0.25$$ Calculate $y$ at $x = -0.25$: $$y = 6 - (-0.25) - 2(-0.25)^2 = 6 + 0.25 - 2(0.0625) = 6.25 - 0.125 = 6.125$$ 7. **Range of $x$ for which $y > 0$:** Recall roots are $x = -2$ and $x = 1.5$. Since parabola opens downward, $y > 0$ between roots: $$-2 < x < 1.5$$ 8. **Equation of intersection of graphs:** Set $y$ equal from both equations: $$(3 - 2x)(2 + x) = \frac{5x - 6}{3}$$ Multiply both sides by 3: $$3(3 - 2x)(2 + x) = 5x - 6$$ From step 5, this leads to: $$3x^2 + 4x - 12 = 0$$ **Final answers:** - Roots of $(3 - 2x)(2 + x) = 0$ are $x = -2$ and $x = 1.5$. - Values of $x$ satisfying $3(3 - 2x)(2 + x) = 5x - 6$ are $x = \frac{-2 + 2\sqrt{10}}{3}$ and $x = \frac{-2 - 2\sqrt{10}}{3}$. - Maximum value of $y$ is $6.125$ at $x = -0.25$. - $y > 0$ for $-2 < x < 1.5$. - Intersection equation in standard form is $3x^2 + 4x - 12 = 0$.