1. The problem is to understand the vertex form of a quadratic function, which is given by the formula: $$y = a(x - h)^2 + k$$ 2. In this formula, $a$ controls the width and direction of the parabola (if $a > 0$, it opens upwards; if $a < 0$, it opens downwards). 3. The point $(h, k)$ is the vertex of the parabola, which is the highest or lowest point depending on the sign of $a$. 4. This form is useful because it directly shows the vertex and makes graphing easier. 5. For example, if $a = 2$, $h = 3$, and $k = -1$, the function is: $$y = 2(x - 3)^2 - 1$$ 6. This parabola opens upwards (since $a=2 > 0$), has its vertex at $(3, -1)$, and is narrower than the standard parabola $y = x^2$ because $|a| > 1$. 7. To find the y-intercept, set $x=0$: $$y = 2(0 - 3)^2 - 1 = 2(9) - 1 = 18 - 1 = 17$$ So the y-intercept is at $(0, 17)$. 8. To find the x-intercepts, set $y=0$ and solve: $$0 = 2(x - 3)^2 - 1$$ $$2(x - 3)^2 = 1$$ $$\cancel{2}(x - 3)^2 = \cancel{1} \frac{1}{2}$$ $$ (x - 3)^2 = \frac{1}{2}$$ $$x - 3 = \pm \sqrt{\frac{1}{2}} = \pm \frac{\sqrt{2}}{2}$$ $$x = 3 \pm \frac{\sqrt{2}}{2}$$ So the x-intercepts are at $\left(3 + \frac{\sqrt{2}}{2}, 0\right)$ and $\left(3 - \frac{\sqrt{2}}{2}, 0\right)$. This completes the explanation of the vertex form and how to analyze the quadratic function.