SAT · Advanced Math · TOPIC LESSON
Connect shape, formula, table and context for quadratic, exponential and other nonlinear rules.
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BEFORE YOU START
Original educational lesson. Work through the examples before revealing the practice answers. This lesson supports SAT Math and Reading and Writing skill study; it does not reproduce official questions or guarantee a score.
LEARN THROUGH ORIGINAL PRACTICE
The teaching passages, worked examples and exercises in this lesson are original. They are not official College Board questions or past-exam reproductions. Blueprint and observed-practice analytics describe separate factual metadata; those source questions are not displayed here.
01 / UNDERSTAND & APPLY
A linear function has constant additive change per equal input step. A quadratic has constant second differences for equally spaced inputs; an exponential has a constant multiplicative ratio when outputs are nonzero. Polynomial, rational and radical functions have other characteristic structures and domains.
A finite table alone does not prove that one family is the true model. Use the model type stated in the problem or justified by its mechanism; then verify consistency with the given points.
02 / UNDERSTAND & APPLY
Standard form ax² + bx + c displays the y-intercept c. Factored form a(x − r)(x − s) displays zeros r and s. Vertex form a(x − h)² + k displays vertex (h,k), axis of symmetry x = h and an extremum k: a minimum when a > 0 and maximum when a < 0.
The leading coefficient controls the opening direction and vertical scale. It does not by itself give either intercept. A vertex output is a height; its input is the location of that height.
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ORIGINAL WORKED EXAMPLE
Rewrite f(x) = 2x² − 12x + 23 to find its minimum.
Answer Minimum 5 at x = 3
03 / UNDERSTAND & APPLY
Products of changing dimensions often produce quadratics. If a rectangle has perimeter 36 meters and width w, its length is 18 − w, so area is A(w) = w(18 − w). The model domain requires 0 < w < 18. Completing the square gives A(w) = 81 − (w − 9)², showing its maximum.
A curve drawn over all real inputs is a mathematical extension. A contextual graph may include only nonnegative time, positive lengths or the interval before an object reaches the ground.
ORIGINAL WORKED EXAMPLE
For a rectangle with perimeter 36 m, what width gives maximum area, and what is that area?
Answer Width 9 m; maximum area 81 m²
04 / UNDERSTAND & APPLY
An exponential model y = a·b^t has initial value a at t = 0 and multiplies by b for each one-unit increase in t. For a positive starting quantity, b > 1 gives growth and 0 < b < 1 gives decay. A percent increase r uses b = 1 + r; a decrease r uses b = 1 − r, with r written as a decimal.
In a·b^(t/k), b is the factor over k time units. The one-unit factor is b^(1/k). Do not confuse the factor 1.12 with a 112% increase: it represents a 12% increase.
ORIGINAL WORKED EXAMPLE
A modeled colony has 240 cells at t = 0 and 540 cells at t = 2 days. Assume N(t) = a·b^t with b > 0. Find the daily factor.
Answer Daily factor 1.5; N(t) = 240(1.5)^t
05 / UNDERSTAND & APPLY
A quantity that halves every three hours can be modeled as Q(t) = 96(1/2)^(t/3). This form displays the three-hour factor. The equivalent form Q(t) = 96[(1/2)^(1/3)]^t displays the hourly factor. The quantity remains positive for all real t in the mathematical model and approaches zero without reaching it for finite positive t.
An additive shift changes the horizontal limiting level: 10 + 96(1/2)^(t/3) tends toward 10, and its initial value is 106 rather than 96.
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ORIGINAL WORKED EXAMPLE
For Q(t) = 96(1/2)^(t/3), when is Q(t) = 12?
Answer t = 9 hours
06 / UNDERSTAND & APPLY
For g(x) = a f(x − h) + k, replacing x by x − h moves the graph right h, multiplying outside by a scales outputs and reflects across the x-axis when a < 0, and adding k shifts outputs upward k. Track domain restrictions through input changes.
For example, g(x) = −(x + 2)² + 5 has vertex (−2,5), opens downward and has maximum 5. Its y-intercept is 1, obtained by evaluating at x = 0. A shift of a graph does not mean every coefficient simply increases by that shift.
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07 / UNDERSTAND & APPLY
For y = f(x), the solutions to f(x) = 0 are x-intercepts, and f(0) is the y-intercept when zero is in the domain. In a factored polynomial, each factor reveals a root. A root with odd multiplicity crosses the axis; a root with even multiplicity touches without changing sign locally.
Evaluate a polynomial by substituting the whole input. To recover inputs from an output, set the rule equal to that output and solve, including every real solution that lies in the domain.
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ORIGINAL WORKED EXAMPLE
Let p(x) = (x − 1)²(x + 2). Find p(0) and all zeros.
Answer p(0) = 2; zeros 1 and −2
08 / UNDERSTAND & APPLY
A rational function excludes denominator zeros. For R(x) = 12/(x − 2), x = 2 is excluded; the graph has a vertical asymptote there and approaches y = 0 for large |x|. It never has output zero. In contrast, a canceled factor can create a hole rather than an asymptote.
A real square-root function requires a nonnegative radicand and returns a nonnegative principal root before any outside transformation. For S(x) = √(x − 4) + 1, the domain is x ≥ 4 and the range is y ≥ 1.
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ORIGINAL WORKED EXAMPLE
If R(x) = 12/(x − 2), find x when R(x) = 3.
Answer x = 6
ORIGINAL WORKED EXAMPLE
For S(x) = √(x − 4) + 1, find x when S(x) = 6.
Answer x = 29
09 / UNDERSTAND & APPLY
For equally spaced x-values 0,1,2,3, outputs 2,5,10,17 have differences 3,5,7 and second differences 2,2. They fit f(x) = x² + 2x + 2. Outputs 3,6,12,24 instead have ratios 2 and fit 3·2^x. Read graph points as input/output statements with units, not just plotted dots.
A rational context might be time T(v) = 180/v hours for a 180-km journey at speed v > 0 km/h. The table pairs (30,6), (60,3), (90,2) show decreasing time with increasing speed, but the decrease is not a fixed amount per equal speed change. A calculator table can test a model, while exact algebra explains its parameters.
10 / UNDERSTAND & APPLY
Keep inputs, outputs and feature locations distinct.
ORIGINAL PRACTICE · 1
(−2,−7)
Vertex form is a(x − h)² + k, so h = −2 and k = −7.
ORIGINAL PRACTICE · 2
5 and −3
Each zero makes one factor zero.
ORIGINAL PRACTICE · 3
Q(t) = 700(0.92)^t
An 8% decrease retains 92% each year, hence factor 0.92.
ORIGINAL PRACTICE · 4
Tripling every 4 time units
Increasing t by 4 raises the exponent by 1 and multiplies output by 3.
ORIGINAL PRACTICE · 5
x = −3 or 3
x² + 1 = 10 gives x² = 9 and both signs.
ORIGINAL PRACTICE · 6
x ≥ 3
The radicand must satisfy 2x − 6 ≥ 0.
ORIGINAL PRACTICE · 7
x = −4
It makes the denominator zero.
ORIGINAL PRACTICE · 8
Shift right 3 and up 2; vertex (3,2)
The input shift occurs inside the function and the output shift outside.
ORIGINAL PRACTICE · 9
3
Each output divided by the preceding output equals 3.
ORIGINAL PRACTICE · 10
35 height units at t = 4
The negative squared term is at most zero, reaching zero at t = 4.
KEEP THE BIG IDEAS
READ WITH CONTEXT
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