Nonlinear Functions (quadratic, exponential, polynomial)

Nonlinear functions are functions whose graphs are not straight lines. On the SAT that mostly means quadratic functions (parabolas), exponential functions (repeated growth or decay by the same percent), and occasionally higher-degree polynomials. You will evaluate them, find vertices and zeros, interpret the numbers in a model, and choose the function that matches a description.

• A quadratic can be written three ways, and each form shows something different (see the table below). Pick the form that answers the question.
• The vertex of a parabola lies exactly halfway between its two zeros.
• An exponential function multiplies by the same factor over equal intervals. Growth of $r$ percent per period means a factor of $1 + r$; decay of $r$ percent means $1 - r$ (with $r$ as a decimal).
• In $f(t) = a \cdot b^t$, $a$ is the starting value $f(0)$ and $b$ is the factor per unit of $t$.
• For a polynomial $p$, if $p(c) = 0$, then $c$ is an $x$-intercept and $(x - c)$ is a factor.

• If $a > 0$ the parabola opens up (minimum at the vertex); if $a < 0$ it opens down (maximum).
• Exponential models: $A(1 + r)^t$ for percent growth, $A(1 - r)^t$ for percent decay, and $A \cdot 2^{t/d}$ for doubling every $d$ units (use $\left(\frac{1}{2}\right)^{t/d}$ for a half-life).

For $f(x) = -x^2 + 6x - 5$, find $f(-1)$ and the vertex of the graph.

1. $f(-1) = -(-1)^2 + 6(-1) - 5 = -1 - 6 - 5 = -12$.
2. Vertex $x$-coordinate: $-\frac{b}{2a} = -\frac{6}{2(-1)} = 3$.
3. $f(3) = -9 + 18 - 5 = 4$, so the vertex is $(3, 4)$, a maximum since $a < 0$.

Check: $f(x) = -(x - 1)(x - 5)$ has zeros $1$ and $5$, and halfway between them is $x = 3$. Correct.

A machine is bought for $45,000 and loses 20% of its value each year. Write a function for its value $V$ after $t$ years, and find the first whole number of years after which the value is below $20,000.

1. Losing 20% leaves 80%, so the factor is $0.8$: $V(t) = 45{,}000(0.8)^t$.
2. Compute year by year: $V(1) = 36{,}000$, $V(2) = 28{,}800$, $V(3) = 23{,}040$, $V(4) = 18{,}432$.
3. The value first drops below $20,000 after $4$ years.

Check: $V(3) = 23{,}040$ is still above $20,000, so $3$ years is not enough.

• Using the percent as the factor. A 20% decrease is $(0.8)^t$, not $(0.2)^t$; a 3% increase is $(1.03)^t$, not $(0.03)^t$ or $(1.3)^t$.
• Sign errors in vertex form. $a(x + 3)^2 - 7$ has its vertex at $(-3, -7)$.
• Squaring a negative input without parentheses. For $f(-2)$ with $f(x) = 2x^2$, compute $2(-2)^2 = 8$, not $-8$.
• Confusing the vertex with a zero. The vertex is the turning point; zeros are where the graph crosses the $x$-axis.
• Misreading the time unit. "Doubles every $5$ hours" means the exponent is $\frac{t}{5}$, not $5t$.