Quadratics and Nonlinear Functions

Pentru elevii români: Formula discriminantului este aceeași ca la școală ($\Delta = b^2 - 4ac$), dar pe SAT se numește discriminant și de obicei apare scrisă direct ca $b^2 - 4ac$, fără simbolul $\Delta$, iar relațiile lui Viète apar sub forma „sum and product of solutions”. Funcțiile exponențiale apar mai ales ca modele de creștere sau scădere procentuală, $a(1 + r)^t$, nu ca exerciții abstracte.

What the SAT tests here

The Advanced Math domain covers three skill areas: equivalent expressions, nonlinear equations in one variable (and systems in two variables), and nonlinear functions. In practice, that means:

• solving and analyzing quadratic equations;
• reading features of a parabola (vertex, intercepts, axis of symmetry) from its equation;
• exponential growth and decay models;
• radical and rational equations, polynomial operations, and function notation.

Key rules and formulas

Three forms of a quadratic:

If $a > 0$, the parabola opens up (minimum at the vertex); if $a < 0$, it opens down (maximum).

Vertex: $x = -\frac{b}{2a}$; substitute to find the $y$-coordinate.

Quadratic formula and discriminant:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

If $b^2 - 4ac > 0$: two real solutions; $= 0$: exactly one; $< 0$: no real solutions.

Sum and product of solutions of $ax^2 + bx + c = 0$: sum $= -\frac{b}{a}$, product $= \frac{c}{a}$.

Exponential models: $f(x) = a \cdot b^x$, where $a$ is the initial value. Growth if $b > 1$, decay if $0 < b < 1$. A change of $r$ percent per period gives $a(1 + r)^t$ or $a(1 - r)^t$.

Transformations: $f(x) + k$ shifts up $k$; $f(x - h)$ shifts right $h$; $-f(x)$ reflects across the $x$-axis.

Worked examples

Example 1 (factoring). Solve $2x^2 - 5x - 3 = 0$.

Factor: $(2x + 1)(x - 3) = 0$, so $x = -\frac{1}{2}$ or $x = 3$. Check with sum of solutions: $-\frac{1}{2} + 3 = \frac{5}{2}$, and $-\frac{b}{a} = \frac{5}{2}$. ✓

Example 2 (maximum height, US units). A ball's height, in feet, $t$ seconds after it is thrown is $h(t) = -16t^2 + 64t + 5$. What is the maximum height of the ball?

Since $a = -16 < 0$, the vertex is a maximum. $t = -\frac{64}{2(-16)} = 2$. Then $h(2) = -16(4) + 64(2) + 5 = -64 + 128 + 5 = 69$. The maximum height is 69 feet. Note: 5 is the starting height (at $t = 0$), not the maximum.

Example 3 (discriminant). For what value of $c$ does $x^2 - 6x + c = 0$ have exactly one real solution?

Set the discriminant to zero: $(-6)^2 - 4(1)(c) = 0$, so $36 = 4c$ and $c = 9$. Indeed, $x^2 - 6x + 9 = (x - 3)^2$.

Bonus (exponential). A culture starts with 2,400 bacteria and triples every 5 hours: $P(t) = 2{,}400 \cdot 3^{t/5}$. After 10 hours, $P(10) = 2{,}400 \cdot 3^2 = 21{,}600$.

Test-taking strategy

• Graph it in Desmos. Typing y=3x^2-12x+7 shows the parabola; click it to reveal the vertex and the intercepts as labeled gray points. This is often the fastest route on vertex and solution questions.
• Check equivalent expressions graphically. If two expressions are equivalent, their graphs coincide exactly. Graph the original and each answer choice; only the matching one overlaps.
• Sliders for constants. For "for what value of $c$ does the equation have exactly one solution," graph y=x^2-6x+c, add the slider, and look for the value where the parabola just touches the $x$-axis. Confirm algebraically.
• Choose the form that answers the question. Asked for a maximum or minimum? Use vertex form. Asked for zeros? Use factored form. Asked for the $y$-intercept? Use standard form.
• Translate exponential bases into percents. $1.06$ means 6% growth; $0.92$ means 8% decay.

Common traps

• Extraneous solutions. Squaring both sides can create false solutions. In $\sqrt{x + 7} = x - 5$, squaring gives $x^2 - 11x + 18 = 0$, so $x = 9$ or $x = 2$. But $x = 2$ gives $\sqrt{9} = 3$ on the left and $-3$ on the right, so only $x = 9$ works. Always check in the original equation.
• Vertex sign errors. In $y = (x + 4)^2 - 1$, the vertex is $(-4, -1)$, not $(4, -1)$.
• Linear versus exponential thinking. A 15% yearly decrease is not the same dollar amount each year.
• Dividing by a variable. From $x^2 = 5x$, dividing by $x$ loses the solution $x = 0$. Factor instead: $x(x - 5) = 0$.
• Confusing the initial value with the maximum in projectile problems.

Summary

Know the three forms of a quadratic and what each reveals, the vertex formula $x = -\frac{b}{2a}$, the discriminant test, and the sum and product of solutions. Read exponential models as initial value times a growth factor. Check every radical or rational solution in the original equation, and lean on Desmos to see vertices, intercepts, and equivalent expressions instantly.