How to solve SAT nonlinear functions questions

Nonlinear functions are functions whose graphs are not straight lines. On the Digital SAT that mostly means quadratics and exponentials, with some polynomial, radical, and rational functions mixed in. Questions ask you to read features like a vertex or a zero from an equation, interpret the numbers in a model, and work with function notation and transformations. Advanced Math is about 35% of the Math section, or 13 to 15 of the 44 questions, and has only three skills, so nonlinear functions usually account for several questions on each test.

What the questions look like

Quadratic features. You get a quadratic in one of three forms and must find its vertex, zeros, $y$-intercept, or maximum or minimum value. A common twist asks which form of a function shows a given feature as a constant or coefficient.

Exponential models. A model like $V(t) = 28{,}000(0.85)^t$ or $P(t) = 500 \cdot 3^{t/4}$ describes growth or decay. You interpret the starting value, the growth factor, or the percent change per period, or you choose the right model from a description.

Function behavior and notation. You evaluate a function from a graph or table, find the zeros of a polynomial from its factors, or describe how $g(x) = f(x - 3) + 2$ relates to $f$.

Step-by-step method

1. Identify the family. An $x^2$ term with no higher power means quadratic. A variable in the exponent means exponential.
2. Match the quadratic form to the feature.
   • Vertex form $f(x) = a(x - h)^2 + k$ shows the vertex $(h, k)$.
   • Factored form $f(x) = a(x - p)(x - q)$ shows the zeros $p$ and $q$. The vertex is halfway between them, at $x = \frac{p + q}{2}$.
   • Standard form $f(x) = ax^2 + bx + c$ shows the $y$-intercept $c$. The vertex is at $x = -\frac{b}{2a}$.
   • If $a > 0$, the parabola opens up and the vertex is a minimum. If $a < 0$, it is a maximum.
3. Decode the exponential. In $f(t) = a \cdot b^t$, $a$ is the value at $t = 0$ and $b$ is the growth factor per unit of $t$. If $b > 1$, the quantity grows by $(b - 1) \times 100\%$ per unit. If $0 < b < 1$, it decays by $(1 - b) \times 100\%$ per unit. In $a \cdot b^{t/k}$, the factor $b$ applies once every $k$ units.
4. Apply transformations. $f(x - h)$ shifts the graph right $h$ units, $f(x) + k$ shifts it up $k$ units, and $-f(x)$ reflects it across the $x$-axis.
5. Check by plugging in a value you know, such as $t = 0$ or a zero.

Worked example

Example 1. The function $f$ is defined by $f(x) = 2(x - 3)(x + 5)$. What is the minimum value of $f$?

The function is in factored form, so its zeros are $x = 3$ and $x = -5$. The vertex lies halfway between them:

$$x = \frac{3 + (-5)}{2} = -1$$

Since $a = 2 > 0$, the parabola opens up and the vertex gives the minimum. Evaluate at $x = -1$:

$$f(-1) = 2(-1 - 3)(-1 + 5) = 2(-4)(4) = -32$$

The minimum value is $-32$. Note that the question asks for the minimum value, which is the output. The input $-1$ will likely appear as a wrong choice.

Example 2. A lab starts with 500 bacteria, and the population triples every 4 hours. Write a model for the population $P$ after $t$ hours, find the population after 12 hours, and find the approximate percent increase per hour.

The population starts at 500 and multiplies by 3 once every 4 hours:

$$P(t) = 500 \cdot 3^{t/4}$$

After 12 hours, the exponent is $\frac{12}{4} = 3$, so $P(12) = 500 \cdot 27 = 13{,}500$.

To get the hourly factor, rewrite $3^{t/4}$ as $\left(3^{1/4}\right)^t$. Since $3^{1/4} \approx 1.316$, the population grows by about 31.6% per hour. It does not grow by $\frac{200\%}{4} = 50\%$ per hour, because growth compounds.

Common traps

• Sign errors in vertex form. In $f(x) = (x + 4)^2 - 7$, the vertex is $(-4, -7)$, not $(4, -7)$. Rewrite $x + 4$ as $x - (-4)$ if it helps.
• Misreading the decay factor. In $V(t) = 28{,}000(0.85)^t$, the value drops by 15% each year. The 85% is the share that remains.
• Ignoring the exponent's time unit. If the exponent is $\frac{t}{4}$ or $12t$, the growth factor does not apply once per unit of $t$. Work out how many times it applies in the time asked about.
• Confusing zeros with the $y$-intercept. Zeros are the inputs where $f(x) = 0$. The $y$-intercept is the output $f(0)$.
• Calling every percent model exponential, or none. "Increases by 5% each year" is exponential. "Increases by 5 units each year" is linear. Look for a percent or a multiplying factor.

Desmos tip

Graph the function and click on it. Desmos marks the vertex, the zeros, and the $y$-intercept as gray points, and clicking a point shows its coordinates. For transformation questions, define $f(x)$ on one line and type $f(x - 3) + 2$ on the next to see the shift. For exponential models, type $P(t)$ with $x$ in place of $t$ and add a table to read values at specific times.