Welcome back to class! Yesterday, we started exploring the standard form of a quadratic equation, which looks like this:

$$f(x) = ax^2 + bx + c$$

We established what the general shape of the graph looks like (a parabola), how to tell if it opens up or down based on the value of $a$, and how to easily spot the y-intercept (the value of $c$). Today, we are going to expand on that foundation to find the two most critical features of the graph: the Vertex and the Axis of Symmetry.

1. The Axis of Symmetry

Parabolas are symmetric curves. The axis of symmetry is the invisible vertical line that slices the parabola perfectly in half. To find this line, we use a specific formula derived from the coefficients $a$ and $b$:

$$x = -\frac{b}{2a}$$

Helpful Hint: Before starting your calculation, make sure your equation is actually in standard form! As we saw in our class notes, sometimes equations are written out of order (e.g., $f(x) = -4x + x^2 - 3$). You must rearrange them first ($f(x) = x^2 - 4x - 3$) so you can correctly identify $a$, $b$, and $c$.

2. Finding the Vertex

The vertex is the turning point of the parabola. Since the vertex always sits exactly on the axis of symmetry, we already know the $x$-coordinate from the step above! To find the $y$-coordinate, simply substitute your $x$ value back into the original function.

So, your vertex is the point: $$\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$$

3. Minimums, Maximums, Domain, and Range

Once you have the vertex, you can determine the range of the function.

  • Happy Parabola ($U$): If $a$ is positive, the graph opens up. The vertex is a minimum value.
    The range is $y \geq k$ (where $k$ is the y-value of the vertex).
  • Sad Parabola ($\cap$): If $a$ is negative, the graph opens down. The vertex is a maximum value.
    The range is $y \leq k$.

Note: The Domain of a quadratic function is almost always all real numbers ($\mathbb{R}$).

Class Example Walkthrough

Let's look at one of the handwritten examples we did in class:

Function: $f(x) = x^2 - 4x - 3$

  1. Identify Coefficients: $a=1$, $b=-4$, $c=-3$
  2. Axis of Symmetry: $$x = \frac{-(-4)}{2(1)} = \frac{4}{2} = 2$$
  3. Vertex: Plug $2$ back into the equation:
    $$f(2) = (2)^2 - 4(2) - 3 = 4 - 8 - 3 = -7$$
    Vertex: $(2, -7)$
  4. Analysis: Since $a$ is positive ($1$), it opens up. We have a minimum at $-7$.
    Domain: $\mathbb{R}$
    Range: $y \geq -7$

Resources & Homework

Make sure to use the template provided below to keep your work organized. It helps prevent simple arithmetic errors!

  • Class Notes: Review the slides and handwritten examples attached below.
  • Homework: Textbook Pg. 72 # 5-7, 15-23.

Discussion Question of the Day: When doing the calculations to graph a parabola, what are some patterns or shortcuts you notice that might reduce the amount of work required? (Extra credit for the first distinct answers!)