1. **Problem statement:** Find the equation of the parabola passing through the given points for 2b and 2c, and then solve problem 3 about shifting the coordinate system for the parabola $y=0.25x^2$. --- ### Problem 2b: Points N(-0.5|0), B(0|-2), N2|0) (assuming N2 means (2|0)) 2b.1. We want a quadratic function $y = ax^2 + bx + c$ passing through points $(-0.5,0)$, $(0,-2)$, and $(2,0)$. 2b.2. Substitute each point into the equation: - For $(0,-2)$: $-2 = a\cdot0^2 + b\cdot0 + c \Rightarrow c = -2$ - For $(-0.5,0)$: $0 = a(-0.5)^2 + b(-0.5) + c = 0.25a - 0.5b - 2$ - For $(2,0)$: $0 = a(2)^2 + b(2) + c = 4a + 2b - 2$ 2b.3. Write the system: $$\begin{cases} 0.25a - 0.5b - 2 = 0 \\ 4a + 2b - 2 = 0 \end{cases}$$ 2b.4. Simplify first equation: $$0.25a - 0.5b = 2$$ Multiply both sides by 4: $$\cancel{4} \times 0.25a - \cancel{4} \times 0.5b = 4 \times 2$$ $$a - 2b = 8$$ 2b.5. Second equation: $$4a + 2b = 2$$ 2b.6. Solve system: From first: $a = 8 + 2b$ Substitute into second: $$4(8 + 2b) + 2b = 2$$ $$32 + 8b + 2b = 2$$ $$32 + 10b = 2$$ $$10b = 2 - 32 = -30$$ $$b = -3$$ 2b.7. Find $a$: $$a = 8 + 2(-3) = 8 - 6 = 2$$ 2b.8. Final function: $$y = 2x^2 - 3x - 2$$ --- ### Problem 2c: Points A(-2|0), B(-1|3), C(1|3) 2c.1. Assume $y = ax^2 + bx + c$. 2c.2. Substitute points: - $0 = a(-2)^2 + b(-2) + c = 4a - 2b + c$ - $3 = a(-1)^2 + b(-1) + c = a - b + c$ - $3 = a(1)^2 + b(1) + c = a + b + c$ 2c.3. System: $$\begin{cases} 4a - 2b + c = 0 \\ a - b + c = 3 \\ a + b + c = 3 \end{cases}$$ 2c.4. Subtract second from third: $$(a + b + c) - (a - b + c) = 3 - 3$$ $$a + b + c - a + b - c = 0$$ $$2b = 0 \Rightarrow b = 0$$ 2c.5. Substitute $b=0$ into second: $$a - 0 + c = 3 \Rightarrow a + c = 3$$ 2c.6. Substitute $b=0$ into first: $$4a - 0 + c = 0 \Rightarrow 4a + c = 0$$ 2c.7. Subtract equations: $$(4a + c) - (a + c) = 0 - 3$$ $$3a = -3 \Rightarrow a = -1$$ 2c.8. Find $c$: $$a + c = 3 \Rightarrow -1 + c = 3 \Rightarrow c = 4$$ 2c.9. Final function: $$y = -x^2 + 4$$ --- ### Problem 3: Parabola $y = 0.25x^2$ with origin shifted to points B, C, or D 3.1. Original parabola: $$y = \frac{1}{4}x^2$$ 3.2. If the origin is shifted to point $(h,k)$, new coordinates $(X,Y)$ relate to old by: $$x = X + h, \quad y = Y + k$$ 3.3. Substitute into original: $$Y + k = \frac{1}{4}(X + h)^2$$ 3.4. Solve for $Y$: $$Y = \frac{1}{4}(X + h)^2 - k$$ 3.5. This is the new parabola equation in shifted coordinates. 3.6. For each point B, C, D (coordinates not given explicitly in user message), plug in their $(h,k)$ to get the new equation. --- **Final answers:** - 2b: $$y = 2x^2 - 3x - 2$$ - 2c: $$y = -x^2 + 4$$ - 3: $$Y = \frac{1}{4}(X + h)^2 - k$$ where $(h,k)$ is the new origin point (B, C, or D).