Every quadratic is a parabola, and almost every SAT quadratic question is asking for one of three features: the roots (where it crosses the x-axis), the vertex (its turning point), or the axis of symmetry. The trick is knowing which form of the equation hands you which.
Key takeaways
- Roots / x-intercepts: factored form a(x - r_1)(x - r_2) shows them directly.
- Vertex / min or max: vertex form a(x - h)^2 + k gives the vertex (h, k).
- Axis of symmetry is the vertical line x = h through the vertex.
- Stuck? Graph it in Desmos and read the features off.
See it once
Take f(x) = x^2 - 2x - 3. It factors to (x - 3)(x + 1), so its roots are x = 3 and x = -1; its vertex sits at (1, -4). Here’s the whole story in one picture:
Try one
The function f(x) = x^2 - 2x - 3 has how many real x-intercepts, and what are they?
Which form for which feature
| You want… | Use this form | Why |
|---|---|---|
| Roots / x-intercepts | Factored: a(x - r₁)(x - r₂) | Each factor = 0 gives a root |
| Vertex (min/max) | Vertex: a(x - h)² + k | Vertex is (h, k) directly |
| Y-intercept | Standard: ax² + bx + c | c is the y-intercept |
Word problems ask for the same three features in costume. "The maximum height of the ball" is the vertex’s y-value. "The time at which it reaches that height" is the vertex’s x-value. "When does it hit the ground" means find the positive root. Translate the phrase to the feature first, then pick the form that hands it over (or practice it until that translation is automatic).
From standard form ax^2 + bx + c, the axis of symmetry (and the vertex’s x-value) is x = -\dfrac{b}{2a}, no factoring needed.
In y = (x - 3)^2 + 2 the vertex is at x = 3, not x = -3. Vertex form is a(x - h)^2 + k with the minus already built in, so (x + 5)^2 means h = -5. Misreading that sign is the most common vertex mistake on the test, and both signs always appear in the choices.
Counting solutions: the discriminant
When a question asks how many real solutions there are (or for the constant that produces exactly one), you don’t need the solutions themselves. The discriminant b^2 - 4ac answers it: positive means two real solutions, zero means exactly one (the parabola just touches the x-axis at its vertex), negative means none (it never reaches the axis).
For what value of k does the equation x^2 + 6x + k = 0 have exactly one real solution?
Work parabolas, roots, and vertices with a tutor that ties form to feature.
Practice Quadratics →Frequently asked questions
- How do I find the roots of a quadratic on the SAT?
- Factor it into a(x - r₁)(x - r₂) and set each factor to zero, or graph it and read where it crosses the x-axis. You can also use the quadratic formula when it won’t factor cleanly.
- How do I find the vertex of a parabola?
- Vertex form a(x - h)² + k gives the vertex (h, k) directly. From standard form ax² + bx + c, the vertex’s x-value is x = -b/2a; plug it in to get the y-value.
- What is the axis of symmetry?
- The vertical line through the vertex, x = h (equivalently x = -b/2a). The parabola is a mirror image across this line.
- How do I know how many solutions a quadratic has?
- Check the discriminant, b² − 4ac. Positive means two real solutions, zero means exactly one, and negative means none. You never need to finish solving to answer a how-many question.