1. The problem asks to find equations of lines that intersect the parabola $y = x^2 + 3x + 2$ a specific number of times: 0, 1, and 2 times. 2. To find the number of intersections between a line $y = mx + b$ and the parabola $y = x^2 + 3x + 2$, set them equal: $$x^2 + 3x + 2 = mx + b$$ Rearranged: $$x^2 + (3 - m)x + (2 - b) = 0$$ This is a quadratic equation in $x$. 3. The number of solutions (intersections) depends on the discriminant $\Delta$: $$\Delta = (3 - m)^2 - 4 \cdot 1 \cdot (2 - b)$$ - If $\Delta > 0$, two distinct intersections. - If $\Delta = 0$, one intersection (tangent line). - If $\Delta < 0$, no real intersections. 4. For each case: **a. 0 intersections:** Choose $m$ and $b$ so that $$\Delta < 0 \implies (3 - m)^2 < 4(2 - b)$$ Example: Let $m=0$, then $$(3 - 0)^2 = 9 < 4(2 - b) \implies 9 < 8 - 4b \implies -4b > 1 \implies b < -\frac{1}{4}$$ Choose $b = -1$: $$y = -1$$ This line does not intersect the parabola. **b. 1 intersection:** Set $\Delta = 0$: $$(3 - m)^2 = 4(2 - b)$$ Example: Let $m=1$, then $$(3 - 1)^2 = 4 = 4(2 - b) \implies 4 = 8 - 4b \implies -4b = -4 \implies b = 1$$ Line: $$y = x + 1$$ This line is tangent to the parabola. **c. 2 intersections:** Choose $m$ and $b$ so that $$\Delta > 0$$ Example: Let $m=0$, $b=0$: $$(3 - 0)^2 - 4(2 - 0) = 9 - 8 = 1 > 0$$ Line: $$y = 0$$ This line intersects the parabola twice. Final answers: - 0 intersections: $y = -1$ - 1 intersection: $y = x + 1$ - 2 intersections: $y = 0$