What does the Advanced Math domain test on the Digital SAT?

Advanced Math is about 35% of the section, tied with Algebra as the largest domain. It covers equivalent expressions (factoring, expanding, exponent rules), nonlinear equations in one variable (quadratics, radicals, exponentials), nonlinear functions (especially exponential growth versus linear), and systems that include a nonlinear equation.

When should you use the quadratic formula instead of factoring?

Try factoring first; it is faster when the quadratic factors with integer roots. If no integer factors work, or the discriminant is not a perfect square, use the quadratic formula. The discriminant also tells you the number of real solutions: positive gives two, zero gives one, negative gives none.

How do you tell linear growth from exponential growth?

Linear growth adds a constant amount each step (a constant difference in a table). Exponential growth multiplies by a constant factor each step (a constant ratio). Exponential growth starts slow but eventually overtakes any linear growth. In a model, a growth factor of 1.05 means a 5% increase per step.

Which form of a quadratic shows the vertex, and which shows the zeros?

Vertex form, a(x - h)^2 + k, shows the vertex (h, k). Factored form, a(x - r1)(x - r2), shows the x-intercepts (zeros). Standard form, ax^2 + bx + c, shows the y-intercept c. Pick the form that displays the feature the question asks for.

How do you solve a system with a line and a parabola?

Substitute the linear equation into the nonlinear one to get a single quadratic, solve it, and back-substitute for the coordinates. The number of intersection points equals the number of real solutions: two (the line cuts the parabola), one (tangent), or none. The discriminant counts them, or you can graph both in Desmos.

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Digital SAT Advanced Math: a complete guide to equivalent expressions, nonlinear equations and functions

A deep-dive guide to the Digital SAT Advanced Math domain: rewriting equivalent expressions, solving nonlinear equations (quadratic, radical, exponential), reading quadratic graphs, distinguishing linear from exponential growth, interpreting nonlinear functions, and solving line-and-parabola systems, with Desmos throughout.

Generated by Claude Opus 4.817 min readDSAT-ADVUpdated 2026-06-14

Reviewed by: AI editorial process; not yet individually human-reviewed

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What the Advanced Math domain demands
Equivalent expressions
Nonlinear equations in one variable
Quadratic graphs
Nonlinear functions and growth
Systems with a nonlinear equation
How Advanced Math is examined
Check your knowledge

What the Advanced Math domain demands

Advanced Math is, with Algebra, one of the two largest Digital SAT Math domains, about 35% of the section. It extends linear algebra into the nonlinear world: quadratics, exponentials, and the algebra needed to manipulate them. This guide ties together the matching dot-point pages, each with its own practice: equivalent expressions, nonlinear equations in one variable, quadratic functions and their graphs, nonlinear functions, and systems of nonlinear equations.

Equivalent expressions

Two expressions are equivalent if they are equal for all valid inputs. The core moves are expanding (FOIL, distribution), factoring (common factor, difference of squares $a^2 - b^2 = (a-b)(a+b)$ , quadratic factoring), simplifying rational expressions (factor and cancel), and the exponent laws ( $x^a x^b = x^{a+b}$ , $\frac{x^a}{x^b} = x^{a-b}$ , $(x^a)^b = x^{ab}$ , $x^{-a} = \frac{1}{x^a}$ , $x^{1/n} = \sqrt[n]{x}$ ). The reason it matters: different forms reveal different features, so rewrite to display whatever the question needs.

Nonlinear equations in one variable

For a quadratic $ax^2 + bx + c = 0$ , try factoring, then the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , or complete the square. The discriminant $b^2 - 4ac$ counts real solutions (positive: two; zero: one; negative: none). For a radical equation, isolate and square, then check for extraneous solutions. For an exponential equation, match the bases and equate exponents.

Quadratic graphs

A quadratic graphs as a parabola: it opens up if $a > 0$ , down if $a < 0$ . The three forms expose different features.

Standard $ax^2 + bx + c$ : $y$ -intercept $c$ .
Factored $a(x - r_1)(x - r_2)$ : zeros $r_1, r_2$ .
Vertex $a(x - h)^2 + k$ : vertex $(h, k)$ , the optimum.

The axis of symmetry is $x = -\frac{b}{2a}$ , and the discriminant gives the number of $x$ -intercepts.

Nonlinear functions and growth

The headline contrast is linear versus exponential. A linear function adds a constant amount (constant difference); an exponential function $f(x) = a \cdot b^x$ multiplies by a constant factor (constant ratio), with $a$ the initial value and $b$ the growth factor ( $b > 1$ growth, $0 < b < 1$ decay). Exponential growth always overtakes linear growth eventually. Read graph features: intercepts, asymptotes, increasing or decreasing intervals, and end behaviour.

Systems with a nonlinear equation

Substitute the line into the parabola to get a quadratic, solve it, and back-substitute for the points. The number of intersection points (two, one tangent, or none) equals the number of real solutions, which the discriminant counts. For "for what value makes the line tangent" questions, set the combined quadratic's discriminant to zero.

How Advanced Math is examined

Equivalent expressions. Expand, factor, simplify rational expressions, apply exponent rules; pick the form that shows the feature.
Nonlinear equations. Solve quadratics (factoring, formula, completing the square), radicals (square and check), exponentials (match bases).
Quadratic graphs. Read the vertex, zeros, intercept, axis of symmetry; use the discriminant.
Nonlinear functions. Distinguish linear from exponential growth; interpret parameters and graph features.
Nonlinear systems. Substitute to a quadratic; count intersections with the discriminant.

Check your knowledge

Work these under timed conditions, then read the solutions.

Expand $(x - 4)(x + 6)$ . (2 marks)
Solve $x^2 + 2x - 15 = 0$ . (2 marks)
A \ $500 investment grows 6% per year. Write its value after$ t$ years and state the growth factor. (2 marks)
Find the vertex of $f(x) = (x + 1)^2 - 9$ . (1 mark)
At how many points do $y = x^2 - 4$ and $y = 5$ intersect? (2 marks)

Solutions to the five practice questions

These solutions show the full working for each check-your-knowledge question above.

Step 1: Q1 - Expand $(x - 4)(x + 6)$

To expand two brackets, multiply each term in the first bracket by each term in the second (the FOIL pattern: First, Outer, Inner, Last). This converts the factored form into standard form.

(x - 4)(x + 6) = x \cdot x + x \cdot 6 + (-4) \cdot x + (-4) \cdot 6 = x^2 + 6x - 4x - 24

Combining the like terms $6x$ and $-4x$ :

x^2 + 2x - 24

Final answer: $x^2 + 2x - 24$ .

Step 2: Q2 - Solve $x^2 + 2x - 15 = 0$

Factoring is the fastest method here. We need two numbers that multiply to $-15$ (the constant term) and add to $2$ (the coefficient of $x$ ). Testing pairs: $5 \times (-3) = -15$ and $5 + (-3) = 2$ . Those are the numbers, so we write:

(x + 5)(x - 3) = 0

Setting each factor to zero gives the solutions:

Final answer: $x = -5$ or $x = 3$ .

Step 3: Q3 - Write the exponential growth model

An investment that grows by a fixed percentage each year multiplies by a constant factor every year. A $6\%$ annual increase means the value is multiplied by $1.06$ each year (the original $100\%$ plus $6\%$ more). Starting from $\$ 500$:

V(t) = 500 \times (1.06)^t

Final answer: $V(t) = 500(1.06)^t$ ; the growth factor is $1.06$ , representing a $6\%$ annual increase.

Step 4: Q4 - Read the vertex from $f(x) = (x + 1)^2 - 9$

Vertex form is $a(x - h)^2 + k$ , where $(h, k)$ is the vertex. The key is the sign convention: the form uses $(x - h)$ , so if we see $(x + 1)$ we rewrite it as $(x - (-1))$ to identify $h = -1$ . The constant outside the square is $k$ directly.

Here $h = -1$ and $k = -9$ , so:

Final answer: Vertex $(-1, -9)$ .

Step 5: Q5 - Count intersections of $y = x^2 - 4$ and $y = 5$

To find intersections, set the two expressions for $y$ equal and solve for $x$ . This gives a single equation in $x$ :

x^2 - 4 = 5 \implies x^2 = 9 \implies x = 3 \text{ or } x = -3

Each solution gives one intersection point. Substituting back, both points have $y = 5$ .

Final answer: Two intersection points, $(3, 5)$ and $(-3, 5)$ .

Sources & how we know this

Math Specifications — College Board (2024)
What Are Content Domains? — College Board (2024)