Digital SAT Quadratic Functions: Vertex, Roots and Maximum
Understand SAT quadratic functions, roots, vertex form, maxima and minima, with worked examples and graphing checks.
By Math School Editorial Team
The vertex of y = −2(x − 3)² + 18 is (3, 18).
Why quadratics matter
College Board includes quadratic equations and nonlinear functions in the digital SAT Advanced Math domain. The same quadratic can appear in different forms, each revealing a different feature. These are original practice examples, not official SAT questions.
Standard form: f(x) = ax² + bx + c, where a ≠ 0. The y-intercept is (0, c).
Factored form: f(x) = a(x − r)(x − s). The roots are r and s.
Vertex form: f(x) = a(x − h)² + k. The vertex is (h, k).
Read the maximum
For f(x) = −2(x − 3)² + 18, the vertex is (3, 18). Because −2 is negative, the parabola opens downwards. Its maximum value is 18, reached when x = 3.
“Where does the maximum occur?” asks for the x-value 3. “What is the maximum value?” asks for 18. The squared term is nonnegative, so subtracting twice that square cannot produce a value above 18.
Find the roots
Set f(x) = 0:
−2(x − 3)² + 18 = 0.
Then (x − 3)² = 9, so x − 3 = ±3. Thus the roots are x = 0 and x = 6.
Change form
Expand: −2(x − 3)² + 18 = −2x² + 12x.
Factor: −2x² + 12x = −2x(x − 6).
The factored form reveals roots 0 and 6. The vertex form reveals the maximum. The standard form reveals the y-intercept is zero. Choose the form that answers the question most directly.
Find a vertex from standard form
Suppose g(x) = 2x² − 12x + 13. The vertex has x-coordinate −b/(2a).
Here a = 2 and b = −12, so x = 12/4 = 3.
Substitute: g(3) = 2(9) − 12(3) + 13 = −5.
The vertex is (3, −5). Because a > 0, −5 is the minimum value.
Completing the square gives g(x) = 2(x − 3)² − 5, confirming the answer.
Count real roots using the discriminant
For ax² + bx + c = 0, the discriminant is D = b² − 4ac.
When D > 0, there are two distinct real roots; when D = 0, one distinct repeated real root; when D < 0, no real roots.
Example: For what value of c does x² − 10x + c = 0 have exactly one distinct real root?
Set D = 0: 100 − 4c = 0, giving c = 25. The equation becomes (x − 5)² = 0.
Graphing check
The digital SAT Bluebook app includes an embedded Desmos calculator. Graph y = −2(x − 3)² + 18 to confirm the vertex (3, 18) and roots 0 and 6. Graphs can suggest approximate values; use algebra to confirm exact answers.
Three original SAT-style questions
Q1: h(x) = −(x − 5)² + 16. Find its maximum, the x-value where it occurs, and both roots.
Answer: Maximum 16 at x = 5. Solving (x − 5)² = 16 gives roots 1 and 9.
Q2: p(x) = (x − 2)(x − 8). Find the vertex's x-coordinate and minimum value.
Answer: Roots 2 and 8 have midpoint x = 5. Then p(5) = 3(−3) = −9, the minimum.
Q3: q(x) = x² + 2x + 5. How many real zeros?
Answer: D = 2² − 4(1)(5) = −16, so no real zeros.
Fifteen-minute revision routine
Spend five minutes reading vertices from vertex form, five finding roots from factored or vertex form, and five checking answers with graphs or completing the square. Before calculating, identify whether the question asks for an x-coordinate, a function value, or the number of solutions.
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