SAT Math · Advanced Math
Quadratic Functions
A quadratic function is any function whose graph is a parabola. On the SAT you will see the same parabola written three different ways. Knowing which form to reach for — and how to flip between them — is the entire game.
The Three Forms
Same parabola, three disguises. Each form makes one feature of the parabola obvious for free.
Standard
$$f(x) = ax^2 + bx + c$$
Reveals the y-intercept directly: $(0, c)$.
Vertex
$$f(x) = a(x-h)^2 + k$$
Reveals the vertex directly: $(h, k)$.
Factored
$$f(x) = a(x-r_1)(x-r_2)$$
Reveals the roots directly: $x = r_1, r_2$.
Try it yourself
- Find the three sliders for $a$, $h$, and $k$ in the panel on the left of the graph.
- Drag $h$ to slide the vertex left and right. The whole parabola moves with it.
- Drag $k$ to slide the vertex up and down.
- Drag $a$ to change the width and direction. Try $a < 0$ to flip the parabola upside down.
Notice: The vertex is always at $(h, k)$ — you can read it directly off the formula. That's the whole point of vertex form: each parameter does one clear visual job. Compare with standard form, where the vertex requires computation.
Strategic move
Read the question first. If it asks for the vertex, get to vertex form. If it asks for x-intercepts, get to factored form. Don't default to expanding everything into standard form — that's almost always the slowest path.
The Quadratic Formula
When factoring is ugly or impossible, this formula always works on $ax^2 + bx + c = 0$.
Quadratic Formula
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
The discriminant tells you what to expect
The expression under the square root, $b^2 - 4ac$, is called the discriminant. Before computing anything, peek at its sign.
| Discriminant | Real solutions | What the parabola does |
|---|---|---|
| $b^2 - 4ac > 0$ | Two distinct real roots | Crosses the x-axis twice |
| $b^2 - 4ac = 0$ | One repeated real root | Touches the x-axis at the vertex |
| $b^2 - 4ac < 0$ | No real roots | Sits entirely above or below the x-axis |
Finding the Vertex Quickly
Two reliable methods. Pick whichever fits the form you already have.
Vertex from standard form
$$x_{\text{vertex}} = -\frac{b}{2a}, \qquad y_{\text{vertex}} = f\!\left(-\frac{b}{2a}\right)$$
Vertex from factored form (axis of symmetry sits at the midpoint of the roots)
$$x_{\text{vertex}} = \frac{r_1 + r_2}{2}$$
Worked Example
SAT-style problem
The function $f$ is defined by $f(x) = 2x^2 - 12x + 7$. What is the minimum value of $f$?
01
Recognize that since $a = 2 > 0$, the parabola opens upward, so the minimum is at the vertex.
02
Find the x-coordinate of the vertex.
$$x = -\frac{b}{2a} = -\frac{-12}{2(2)} = 3$$
03
Plug $x = 3$ back into $f$ to get the y-coordinate.
$$f(3) = 2(3)^2 - 12(3) + 7 = 18 - 36 + 7 = -11$$
04
The minimum value of $f$ is $\boxed{-11}$.
Common Student Errors
- Dropping the ± in the quadratic formula Students compute one root and stop. Always check whether the problem wants both, the sum, or the product.
- Sign errors when reading vertex form In $f(x) = (x - 3)^2 + 5$, the vertex is $(3, 5)$, not $(-3, 5)$. The sign flips because the form is $(x - h)$.
- Forgetting to balance when completing the square Adding $\left(\tfrac{b}{2}\right)^2$ inside parentheses with a leading coefficient $\neq 1$ changes the equation. You must subtract $a \cdot \left(\tfrac{b}{2}\right)^2$ outside to compensate.
- Confusing roots with the y-intercept Roots are where $y = 0$ (x-intercepts). The y-intercept is where $x = 0$. The SAT loves to see if you reverse these.
SAT trap
When a question gives you a quadratic in factored form like $f(x) = (x+4)(x-2)$ and asks for the y-intercept, plug in $x = 0$. The y-intercept is $f(0) = (4)(-2) = -8$, not the constant term of an expanded form.