Thirteen questions on the nonlinear half of the Math section — the domain that, with Algebra, makes up 70% of your Math score.
Three forms of a quadratic, three different questions. This is the organising idea of the whole domain, and four of the questions here turn on it:
That last one is the question most students get wrong, and the reason is instructive: the zeros are right there in the brackets and it is very tempting to give one of them. The zeros are where the curve crosses the axis. The vertex is where it turns.
The discriminant. A question that asks for the value of a coefficient making a quadratic have exactly one real solution is asking for . Solving the quadratic instead works and takes four times as long.
Exponents. Multiplying adds the powers, dividing subtracts them, a power of a power multiplies them, and a fractional power is a root. The harder item here combines a fractional power with an outer exponent, which is the shape the real test uses when it wants this skill rather than recall.
Growth factor, not growth rate. The trap that catches more students than any other in this domain. A 12% annual increase is a multiplication by 1.12, not by 0.12; a 5% decrease is 0.95. And when the period is not one year — a colony that doubles every six hours — the factor is the doubling (2) and the exponent counts the periods (), not the hours. The explanation for that one gives you the check worth using on every exponential question: substitute one period's worth of time and see whether the expression returns what it should.
Equivalent expressions. Factoring, difference of squares, cancelling a rational expression, subtracting a bracketed polynomial with the sign error the test is waiting for. These have no numbers in them and nothing to graph — algebra is the only route, which is worth knowing when you are deciding whether to reach for the built-in calculator.
The twelve flashcards carry the facts: the three forms, the discriminant conditions, the exponent laws including negative and fractional powers, and the growth-factor conversion written as a rule you can apply on sight.
The function f is defined by f(x) = 2(x − 5)(x + 1). At what value of x does f reach its minimum? — Factored form hands you the zeros directly: f(x) = 0 at x = 5 and at x = −1. A parabola is symmetric about its vertex, so the vertex sits exactly halfway between its two zeros, at x = 2, and because the leading coefficient is positive that vertex is a minimum. The zeros themselves are where the curve crosses the axis, not where it turns.
A quadratic function is given by $f(x) = (x-3)^2 - 4$. What is the minimum value of the function?
$-4$
In vertex form $a(x-h)^2 + k$, the vertex is $(h, k)$. Here, $k = -4$, which is the minimum value since $a > 0$. Option 0 is the x-coordinate of the vertex. Option 2 is the value of the function at $x=0$. Option 3 is a distractor.