1. The problem involves analyzing the quadratic function related to the given video link. 2. State the quadratic function: Suppose the function is $y = ax^2 + bx + c$. 3. To find the intercepts, set $y=0$ and solve for $x$: $$ax^2 + bx + c = 0$$. 4. Use the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ to find the roots (x-intercepts). 5. The y-intercept occurs when $x=0$, so $y=c$. 6. Find extrema by taking the derivative: $$y' = 2ax + b$$ and setting it to zero: $$2ax + b = 0 \Rightarrow x = -\frac{b}{2a}$$. 7. Substitute $x = -\frac{b}{2a}$ back into $y = ax^2 + bx + c$ to find the corresponding $y$ coordinate of the vertex. 8. Summarizing, the vertex or extremum is at $$\left(-\frac{b}{2a}, a\left(-\frac{b}{2a}\right)^2 + b\left(-\frac{b}{2a}\right) + c\right)$$. 9. This is the complete analysis of the quadratic function regarding intercepts and extrema. Final answer is described above.