Advanced Math · SAT Math

Quadratic Functions on the Digital SAT

Quadratic functions are the most-tested concept in the advanced math domain — and one of the highest-leverage topics for a score boost. This lesson covers the three forms of a quadratic, the three solution methods (factoring, the quadratic formula, and completing the square), vertex problems, and the discriminant rule — with 8 practice problems and Desmos shortcuts.
By the Prepiii Editorial TeamUpdated 2026-05-23~12 min read

The three forms of a quadratic

A quadratic function has the variable raised to the second power. The same function can be written three different ways, and the SAT tests all three — sometimes asking you to convert between them.

Standard form: f(x) = ax² + bx + c. The c is the y-intercept. Use when you need the y-intercept fast.

Factored form: f(x) = a(x - p)(x - q). The p and q are the x-intercepts (zeros / roots). Use when the question asks for x-intercepts.

Vertex form: f(x) = a(x - h)² + k. The vertex is at (h, k). Use when the question asks for the vertex, maximum, or minimum.

The number a appears in all three forms and it's the same number. If a > 0 the parabola opens up (has a minimum). If a < 0 it opens down (has a maximum).

Three ways to solve a quadratic

"Solving" a quadratic means finding the values of x that make f(x) = 0 — i.e., the x-intercepts. The SAT expects you to know three methods:

  1. Factoring. Fastest when the quadratic factors cleanly. x² - 5x + 6 = 0 → (x - 2)(x - 3) = 0 → x = 2 or x = 3.
  2. The quadratic formula. Always works. For ax² + bx + c = 0: x = (-b ± √(b² - 4ac)) / (2a). Memorize it.
  3. Completing the square. Useful when the SAT gives you a quadratic and asks for the vertex form. Take half the middle coefficient, square it, add and subtract.

When to use what. If a quadratic factors with integer roots in < 30 seconds, factor. Otherwise, use Desmos (see below) or the quadratic formula. Completing the square is mostly for conversion questions, not solving.

Finding the vertex (max or min)

The vertex is the highest or lowest point of the parabola. Two fast methods:

  • From vertex form: f(x) = a(x - h)² + k has its vertex at (h, k). Read it off directly.
  • From standard form: the x-coordinate of the vertex is x = -b / (2a). Plug that back into the function to get the y-coordinate.

SAT context: when a word problem asks for "the maximum revenue," "the maximum height," or "the minimum cost," you're looking for the y-coordinate of the vertex.

The discriminant: how many solutions?

The expression b² - 4ac (the part under the square root in the quadratic formula) is called the discriminant. It tells you how many real solutions a quadratic has — without actually solving it.

b² - 4ac > 0 → two real solutions (parabola crosses x-axis twice)

b² - 4ac = 0 → one real solution (vertex sits on x-axis)

b² - 4ac < 0 → no real solutions (parabola never touches x-axis)

The SAT loves to give you a quadratic with a variable inside — like kx² + 6x + 9 = 0 — and ask for the value of k that makes it have exactly one solution. That's a discriminant question.

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Solving with Desmos

Quadratics are where Desmos shines hardest. Most quadratic questions can be solved in < 20 seconds with the right Desmos move. Four techniques you must know:

1. Graph it — Desmos auto-shows intercepts and vertex

Type y = x² - 5x + 6, click the curve. Desmos marks x-intercepts with grey dots and the vertex automatically. Zoom-fit if the parabola is off-screen.

2. Solve for x — graph the equation set to zero

For 2x² - 7x + 3 = 0, type y = 2x² - 7x + 3 and read the x-intercepts directly. No factoring or formula needed.

3. Convert standard ↔ vertex form with a slider trick

If the question asks "which is equivalent to x² + 6x + 5?," graph it and graph each answer choice. The one that produces an identical curve is the answer.

4. Discriminant questions — graph with a slider

For "find k such that kx² + 6x + 9 = 0 has one solution": type y = kx² + 6x + 9, add a slider for k, drag until the parabola just touches the x-axis. Read k.

For the full set of Desmos techniques across the entire test, see our Desmos & Test Tools guides.

Common mistakes

Forgetting the ± when taking square roots

If x² = 16, then x = 4 OR x = -4. The SAT loves to set traps where you give only the positive root and lose the question.

Reading the wrong form's intercept

In f(x) = (x - 3)(x + 5), the x-intercepts are 3 and -5 (sign flip). In f(x) = (x - h)² + k, the vertex is (h, k) — not (-h, k).

Forgetting to set the equation to zero before factoring

If you see x² + 5x = 6, you can't factor (x)(x + 5). First move everything to one side: x² + 5x - 6 = 0, then factor.

Confusing 'a' with the y-intercept

In standard form ax² + bx + c, the y-intercept is c, not a. The a controls the parabola's width and direction (up or down).

Practice problems

8 problems adapted from College Board released questions and internal Prepiii sets. Click each one to reveal the solution.

1

What are the solutions to x² - 7x + 12 = 0?

  1. x = -3, -4
  2. x = 3, 4
  3. x = -3, 4
  4. x = 3, -4

Click to reveal solution →

Answer: (B) x = 3, 4

Factor: we need two numbers that multiply to 12 and add to -7. Those are -3 and -4. So (x - 3)(x - 4) = 0, giving x = 3 or x = 4.

2

The function f(x) = 2(x - 3)² + 5 has its minimum value at what point?

  1. (-3, 5)
  2. (3, 5)
  3. (3, -5)
  4. (-3, -5)

Click to reveal solution →

Answer: (B) (3, 5)

Vertex form: f(x) = a(x - h)² + k, vertex at (h, k). Here h = 3 (sign flip!) and k = 5. Since a = 2 > 0, the parabola opens up, so the vertex is a minimum.

3

What is the sum of the solutions to x² + 6x - 16 = 0?

  1. -8
  2. -6
  3. 2
  4. 8

Click to reveal solution →

Answer: (B) -6

Factor: (x + 8)(x - 2) = 0, so x = -8 or x = 2. Sum = -8 + 2 = -6.

Shortcut: for ax² + bx + c = 0, the sum of solutions is -b/a. Here -6/1 = -6.

4

For what value of k does kx² + 12x + 9 = 0 have exactly one solution?

Click to reveal solution →

Answer: k = 4

One solution means the discriminant is zero: b² - 4ac = 0. Here a = k, b = 12, c = 9.

So 144 - 4(k)(9) = 0 → 144 = 36k → k = 4.

5

The function h(t) = -16t² + 64t + 80 models the height (in feet) of a projectile t seconds after launch. What is the maximum height the projectile reaches?

  1. 80 feet
  2. 96 feet
  3. 128 feet
  4. 144 feet

Click to reveal solution →

Answer: (D) 144 feet

Find the vertex. x-coordinate of vertex: t = -b/(2a) = -64/(2 · -16) = 2.

Plug back in: h(2) = -16(4) + 64(2) + 80 = -64 + 128 + 80 = 144.

6

Which expression is equivalent to x² + 8x + 15?

  1. (x + 3)(x + 5)
  2. (x + 15)(x + 1)
  3. (x - 3)(x - 5)
  4. (x + 4)² + 1

Click to reveal solution →

Answer: (A) (x + 3)(x + 5)

Two numbers that multiply to 15 and add to 8 are 3 and 5. So x² + 8x + 15 = (x + 3)(x + 5).

7

What is the y-intercept of f(x) = 3x² - 5x + 7?

  1. -7
  2. 3
  3. 5
  4. 7

Click to reveal solution →

Answer: (D) 7

The y-intercept is f(0): f(0) = 3(0) - 5(0) + 7 = 7. In standard form ax² + bx + c, the y-intercept is always c.

8

A model rocket is launched from a 4-meter platform. Its height is modeled by h(t) = -5t² + 20t + 4. How many seconds after launch does the rocket hit the ground?

Click to reveal solution →

Answer: t ≈ 4.19 seconds

Set h(t) = 0 and solve -5t² + 20t + 4 = 0 with the quadratic formula:

t = (-20 ± √(400 - 4(-5)(4))) / (2(-5)) = (-20 ± √480) / -10.

Taking the positive root (time can't be negative): t ≈ 4.19 seconds.

Frequently asked questions

What is a quadratic function?

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A quadratic function is one where the variable is raised to the second power. The standard form is f(x) = ax² + bx + c, and its graph is always a parabola. On the SAT, quadratics appear in roughly 1 in 8 math questions.

What is the quadratic formula?

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For any quadratic ax² + bx + c = 0, the solutions are x = (-b ± √(b² - 4ac)) / (2a). Memorize this — it always works, even when factoring fails.

When should I factor vs. use the quadratic formula?

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Factor when the roots are clean integers and you can spot them in under 30 seconds. Use the quadratic formula (or Desmos) when the roots are fractions, irrationals, or the numbers are awkward.

What does the discriminant tell me?

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The discriminant b² - 4ac tells you how many real solutions a quadratic has. If it's positive, there are two real solutions. If it's zero, there's exactly one. If it's negative, there are no real solutions.

How do I find the vertex of a parabola?

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From vertex form f(x) = a(x - h)² + k, the vertex is (h, k). From standard form ax² + bx + c, the x-coordinate is -b/(2a); plug that back in to get the y-coordinate.

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All SAT Math topics

Browse every Digital SAT math topic

Linear Functions

The other most-tested function type on the SAT

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