SAT Nonlinear Functions: The Complete Guide

For $q(x)=ax^2+bx+c$, the vertex input is $h=-\frac{b}{2a}$. Evaluate $q(h)$ for its output. Completing the square gives $q(x)=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}$. This is vertex form, with $k=c-\frac{b^2}{4a}$. Factor out $a$ from the quadratic and linear terms before completing the square.

For example, $2x^2-12x+23=2(x^2-6x)+23=2[(x-3)^2-9]+23=2(x-3)^2+5$. The minimum is five at $x=3$. The subtraction of nine inside the bracket becomes a subtraction of eighteen after multiplication by two; missing that scale changes the function.

A polynomial has zero $r$ exactly when it has factor $x-r$. Thus $p(x)=(x+2)(x-4)^2$ has zeros negative two and four. A repeated factor can make the graph touch the axis instead of cross it: even multiplicity gives touching behavior, while odd multiplicity gives crossing behavior for a real polynomial.

For $g(x)=A f(x-h)+k$, the input shift moves the graph right by $h$, the output shift moves it up by $k$, and multiplication by $A$ scales outputs. A negative $A$ reflects across the horizontal axis. For $g(x)=f(Bx)$ with $B>0$, horizontal distances scale by $1/B$, so an inside factor greater than one compresses the graph.

Domain restrictions travel with transformations. The square-root function $f(x)=\sqrt{x}$ needs $x\ge0$. The shifted function $g(x)=\sqrt{x-5}$ needs $x\ge5$. A rational function such as $1/(x+3)$ excludes negative three. A graph or formula must be interpreted only where the function is defined.

A quadratic's extreme value over a restricted interval may occur at an endpoint if the vertex lies outside that interval. For an upward-opening quadratic on a closed interval, compare endpoint outputs with the vertex output when the vertex is included. An unrestricted minimum and a contextual minimum need not be the same.

To make that domain check concrete, take $q(x)=(x-6)^2+2$ on $0\le x\le4$. The unrestricted vertex is $(6,2)$, but input six is not allowed. Over the stated interval the function decreases toward that vertex, so the least allowed output is $q(4)=6$. The other endpoint gives $q(0)=38$. Reporting two would describe an output the restricted model never reaches.

A quadratic can have a different maximum over a restricted interval even when it opens upward. In the same example, the largest allowed output is 38 at the left endpoint. Distinguish all-real behavior from behavior on a specified closed interval before deciding that a maximum does not exist.

If $P(t)=a b^t$ with $a>0$ and $b>0$, then $P(v)/P(u)=b^{v-u}$. A two-unit growth factor of 1.44 corresponds to one-unit factor $\sqrt{1.44}=1.2$, or 20% growth per unit. It does not mean 44% growth each unit. The factor belongs to the interval represented by the exponent difference.

Desmos is useful for checking a quadratic vertex or comparing two nonlinear models. Enter the function and inspect the vertex or an intersection. To solve $f(x)=T$ visually, graph $y=f(x)$ and $y=T$. A horizontal target line can meet a parabola twice, once, or not at all, depending on its height.

Adjust the window to the domain and relevant scale. A graph centered near the origin may hide an exponential output in the thousands. Use exact algebra for parameter conditions and exact values; use the graph to locate features and verify that your interpretation matches the shape.

Do not confuse a zero with the vertical intercept. Zeros solve $f(x)=0$ and are horizontal-axis locations. The vertical intercept is $f(0)$. In factored form, evaluate at zero if the question asks for an initial amount; do not simply select one of the factor constants.

A negative leading coefficient reverses the extreme type. The function $-2(x-4)^2+11$ has a maximum of eleven, while $2(x-4)^2+11$ has a minimum of eleven. The vertex coordinates alone do not tell you whether it is a high point or low point.

Do not assume outputs double when inputs double. For $f(x)=x^2$, $f(2x)=4x^2$, not $2x^2$. For an exponential, adding to the input multiplies the output by a fixed factor; multiplying the input changes the exponent differently.

A graph's apparent crossing may be outside the contextual interval. If a height model has a negative-time root and a positive-time root, only the allowed time may describe landing. State the domain before deciding which graphical features answer the situation.

A vertex and one extra point determine a quadratic scale. If the vertex is $(4,-6)$, write $f(x)=a(x-4)^2-6$. If $f(6)=10$, then $10=4a-6$, so $a=4$. This uses the given structure directly. Beginning with three unknown standard-form coefficients would create unnecessary algebra.

Two exponential values determine a positive factor through their ratio. If $N(2)=75$ and $N(5)=600$ for $N(t)=a b^t$, then $b^3=600/75=8$, so $b=2$. Working backward gives $a=75/4$. The initial amount need not be an integer even when both supplied measurements are integers.

A change of time unit changes the displayed base. The same model can be written $a(1.04)^t=a[(1.04)^5]^{t/5}$. The second base is the five-unit multiplier, while the first is the one-unit multiplier. An equivalent form can make the requested interval's change visible.

Polynomial zeros can determine only part of a function. Knowing zeros three and five does not prove the function is $(x-3)(x-5)$ unless the leading scale and degree are also specified. A quadratic has form $a(x-3)(x-5)$; a higher-degree polynomial can contain additional factors.

For an average rate over $[u,v]$, compute $[f(v)-f(u)]/(v-u)$. A nonlinear function can have that average rate without maintaining it inside the interval. Use the two endpoint outputs rather than treating a coefficient as a constant slope.

Use a 25-minute session. Spend five minutes looking at formulas and naming the property each form reveals without calculating. Include standard, factored, and vertex quadratics, plus exponentials with both $t$ and $t/k$ in the exponent.

Spend nine minutes in Solar on nonlinear-function questions. Before each calculation, write a short target label: output, zero, vertex input, extreme output, initial amount, or interval factor. Use a graph for one problem where it saves work and algebra for one where a useful form already reveals the answer.

Spend five minutes on transformations and domain restrictions, then six minutes correcting errors. For a missed percent interpretation, calculate one interval numerically. For a missed quadratic extreme, rewrite or graph the function and identify both vertex coordinates.

Return the next day to one quadratic model and one exponential model. Explain why the rate is constant in one multiplicative or additive sense and not the other. That explanation helps prevent fitting a line to a situation that describes repeated percentage change.