1. The problem is to identify the values of $a$, $b$, and $c$ in the quadratic equation $ax^2 + bx + c$ for each colored parabola based on their graph shapes and vertex positions. 2. Recall the quadratic function formula: $$y = ax^2 + bx + c$$ where: - $a$ determines the direction and width of the parabola (if $a > 0$, it opens upward; if $a < 0$, it opens downward). - $b$ affects the horizontal position of the vertex. - $c$ is the $y$-intercept (the value of $y$ when $x=0$). 3. The vertex of a parabola given by $ax^2 + bx + c$ is at $$x = -\frac{b}{2a}$$ and the $y$-coordinate of the vertex is $$y = c - \frac{b^2}{4a}$$. 4. Using the vertex positions and opening directions from the graph: - Red parabola: vertex at approximately $(0,4)$ and opens upward. Since vertex $x=0$, $$-\frac{b}{2a} = 0 \Rightarrow b=0$$. The vertex $y$ is 4, so $$4 = c - \frac{b^2}{4a} = c$$ (since $b=0$). Since it opens upward, $a > 0$. Choose $a=1$ for simplicity. Thus, Red: $a=1$, $b=0$, $c=4$. - Blue parabola: vertex at approximately $(0,-3)$ and opens upward. Similarly, $b=0$ and $c = -3$, $a > 0$. Choose $a=1$. Blue: $a=1$, $b=0$, $c=-3$. - Green parabola: vertex at approximately $(0,2)$ and opens downward. $b=0$, $c=2$, $a < 0$. Choose $a=-1$. Green: $a=-1$, $b=0$, $c=2$. - Purple parabola: vertex at approximately $(-2,-6)$ and opens upward. Use vertex formula for $x$: $$-2 = -\frac{b}{2a} \Rightarrow b = 4a$$. Vertex $y$: $$-6 = c - \frac{b^2}{4a} = c - \frac{(4a)^2}{4a} = c - 4a$$. Choose $a=1$ (opens upward), then $b=4$, and $c = -6 + 4 = -2$. Purple: $a=1$, $b=4$, $c=-2$. - Black parabola: vertex at approximately $(2,2)$ and opens downward. $$2 = -\frac{b}{2a} \Rightarrow b = -4a$$. Vertex $y$: $$2 = c - \frac{b^2}{4a} = c - \frac{(-4a)^2}{4a} = c - 4a$$. Choose $a=-1$ (opens downward), then $b = -4(-1) = 4$, and $c = 2 + 4 = 6$. Black: $a=-1$, $b=4$, $c=6$. 5. Final values: - Red: $a=1$, $b=0$, $c=4$ - Blue: $a=1$, $b=0$, $c=-3$ - Green: $a=-1$, $b=0$, $c=2$ - Purple: $a=1$, $b=4$, $c=-2$ - Black: $a=-1$, $b=4$, $c=6$