1. The problem is to understand how the coefficients $a$, $b$, and $c$ in the quadratic equation $$ax^2 + bx + c$$ affect the shape and position of the parabola on the graph. 2. The quadratic formula is $$y = ax^2 + bx + c$$ 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 point where the parabola crosses the y-axis. 3. For each parabola: - Red: Opens upward with vertex near $(0,4)$, so $a > 0$, $b = 0$ (vertex on y-axis), $c \\approx 4$. - Blue: Opens upward with vertex near $(0,-3)$, so $a > 0$, $b = 0$, $c \\approx -3$. - Green: Opens downward with vertex near $(0,2)$, so $a < 0$, $b = 0$, $c \\approx 2$. - Purple: Opens upward with vertex near $(-2,-6)$, so $a > 0$, $b \\neq 0$ (vertex not on y-axis), $c$ is the y-intercept. - Black: Opens downward with vertex near $(2,2)$, so $a < 0$, $b \\neq 0$, $c$ is the y-intercept. 4. The vertex formula is $$x = -\frac{b}{2a}$$ and $$y = c - \frac{b^2}{4a}$$ which helps find $b$ and $c$ given $a$ and vertex coordinates. 5. Example for Purple parabola with vertex $(-2,-6)$ and $a > 0$: - Using vertex formula: $$-2 = -\frac{b}{2a} \Rightarrow b = 4a$$ - Substitute vertex into equation: $$-6 = a(-2)^2 + b(-2) + c = 4a - 2b + c$$ - Substitute $b=4a$: $$-6 = 4a - 2(4a) + c = 4a - 8a + c = -4a + c$$ - Solve for $c$: $$c = -6 + 4a$$ 6. Similarly for Black parabola with vertex $(2,2)$ and $a < 0$: - $$2 = -\frac{b}{2a} \Rightarrow b = -4a$$ - Substitute vertex: $$2 = a(2)^2 + b(2) + c = 4a + 2b + c$$ - Substitute $b = -4a$: $$2 = 4a + 2(-4a) + c = 4a - 8a + c = -4a + c$$ - Solve for $c$: $$c = 2 + 4a$$ 7. Summary: - $a$ controls opening direction and width. - $b$ shifts vertex horizontally. - $c$ shifts parabola vertically. This explains how $a$, $b$, and $c$ affect the parabola's graph.