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Nonlinear functions 3: function notation and transformations

SAT Math Β· Advanced Math Β· Week 7 of the 12-week plan
11

Nonlinear functions 3: function notation and transformations

College Board skill: Nonlinear functions
GoalEvaluate and solve with f(x) notation, read values from graphs and tables, compose functions, describe shifts and reflections, and use zeros, factors and denominators of polynomial and rational functions.
On the test

Advanced Math is about 35% of SAT Math (13–15 of 44 questions). Function notation is the language of many questions: evaluating f(a), solving f(x) = b, reading a graph or table, compositions, transformations, and the zeros of polynomials.

Key words
function Β· a rule that gives exactly one output for each inputf(a) Β· the output when the input is a; the point (a, f(a)) is on the graphcomposition Β· f(g(x)): put the output of g into f; g works firstzero Β· an input a with f(a) = 0; an x-intercept of the graphasymptote Β· a line the graph gets closer and closer to but never touches
Explanation

What f(x) means

The letter f is the name of the rule and x is the input. To find f(3), replace every x by 3 and calculate. If f(x) = 2xΒ² βˆ’ 5x + 1, then f(3) = 2(3)Β² βˆ’ 5(3) + 1 = 18 βˆ’ 15 + 1 = 4. The statement f(3) = 4 tells you two things: the output for input 3 is 4, and the point (3, 4) is on the graph. The input is always the horizontal coordinate. Put brackets around a negative input, because (βˆ’3)Β² = 9 but βˆ’3Β² = βˆ’9.

Ζ’Formula
f(3) = 4
means that the point (3, 4) is on the graph: the input is the x-coordinate and the output is the y-coordinate

Two different questions

β€œFind f(a)” gives you the input and asks for the output: substitute. β€œFor what x is f(x) = b?” gives you the output and asks for the input: set the rule equal to b and solve. Take f(x) = 3x βˆ’ 2. Substituting 4 gives f(4) = 10. Solving 3x βˆ’ 2 = 10 gives x = 4. The same point (4, 10) answers both, but the work is different.

QuestionWhat to doWith f(x) = 3x βˆ’ 2
Find f(4)substitute x = 43(4) βˆ’ 2 = 10
For what x is f(x) = 10?solve 3x βˆ’ 2 = 10x = 4

Reading graphs and tables

On a graph, f(a) is the height of the curve above (or below) x = a. To solve f(x) = b, draw a horizontal line at height b and read the x-values where it meets the curve; there can be several. The zeros of f are where the curve meets the x-axis. In the graph below, f(0) = 3, f(1) = 4, and f(x) = 0 at x = βˆ’1 and x = 3. In a table, find the input in the top row and read the output under it.

xyβˆ’3βˆ’2βˆ’112345βˆ’4βˆ’3βˆ’2βˆ’11234560y = f(x)(βˆ’1, 0)(3, 0)(1, 4)(0, 3)

Composition: inside first

The notation f(g(x)) means: put x into g first, then put that output into f. With f(x) = x + 3 and g(x) = 2x, we get g(2) = 4 and then f(4) = 7, so f(g(2)) = 7. The other order gives a different result: f(2) = 5 and g(5) = 10, so g(f(2)) = 10. Always start with the innermost brackets.

Shifts and reflections

Changing the function changes the graph in a predictable way. A change outside the brackets acts on the heights and does what you expect: f(x) + 2 moves the graph up 2. A change inside the brackets acts on x and does the opposite of what you expect: f(x βˆ’ 2) moves the graph right 2, and f(x + 2) moves it left 2. A minus sign outside, βˆ’f(x), flips the graph over the x-axis; a minus sign inside, f(βˆ’x), flips it over the y-axis. For a parabola, move the vertex the same way.

New function (h, k > 0)What happens to the graph of f
f(x) + kup k units
f(x) βˆ’ kdown k units
f(x βˆ’ h)right h units
f(x + h)left h units
βˆ’f(x)flipped over the x-axis
f(βˆ’x)flipped over the y-axis

Zeros, factors and fractions

If f(a) = 0, then a is a zero of f and (x βˆ’ a) is a factor of f(x); this works backward too. For example xΒ² βˆ’ x βˆ’ 6 = (x βˆ’ 3)(x + 2) has zeros 3 and βˆ’2. A fraction is undefined when its denominator is 0. The function f(x) = 1/(x βˆ’ 2) is undefined at x = 2, and its graph has a vertical asymptote there. A rational function equals 0 only where its numerator is 0 and its denominator is not.

xyβˆ’2βˆ’1123456βˆ’5βˆ’4βˆ’3βˆ’2βˆ’1123450y = 1/(x βˆ’ 2)
Worked examples
Example 1.
If f(x) = xΒ² βˆ’ 4x + 1, what is f(βˆ’3)?
  1. Replace every x with (βˆ’3) and keep the brackets.
  2. f(βˆ’3) = (βˆ’3)Β² βˆ’ 4(βˆ’3) + 1
  3. = 9 + 12 + 1 = 22
Trap: (βˆ’3)Β² is 9, not βˆ’9, and βˆ’4 times βˆ’3 is +12.
Example 2.
Let f(x) = 3x βˆ’ 1 and g(x) = xΒ² + 2. What is the value of g(f(2))?
  1. Work inside first: f(2) = 3(2) βˆ’ 1 = 5.
  2. Now g(5) = 5Β² + 2 = 27.
Trap: Doing g first gives f(g(2)) = f(6) = 17, which is a different quantity.
Example 3.
The graph of y = f(x) is a parabola with vertex (2, βˆ’3). What is the vertex of the graph of y = f(x + 4) βˆ’ 1?
xyβˆ’8βˆ’6βˆ’4βˆ’2246βˆ’4βˆ’224680y = f(x)f(x + 4) βˆ’ 1(2, βˆ’3)(βˆ’2, βˆ’4)
  1. (βˆ’2, βˆ’4)
  2. (6, βˆ’4)
  3. (βˆ’2, βˆ’2)
  4. (6, βˆ’2)
  1. f(x + 4) is inside the brackets: it moves the graph left 4, so the x-coordinate becomes 2 βˆ’ 4 = βˆ’2.
  2. The βˆ’ 1 is outside: it moves the graph down 1, so the y-coordinate becomes βˆ’3 βˆ’ 1 = βˆ’4.
  3. The new vertex is (βˆ’2, βˆ’4).
Trap: The choice (6, βˆ’4) moves the graph to the right. Plus inside the brackets means left.
Example 4.
The polynomial p(x) = xΒ³ βˆ’ 7x + 6 satisfies p(2) = 0. Which of the following must be a factor of p(x)?
  1. x + 2
  2. x βˆ’ 2
  3. x βˆ’ 6
  4. x + 6
  1. Check: p(2) = 8 βˆ’ 14 + 6 = 0 βœ“
  2. If p(a) = 0 then (x βˆ’ a) is a factor. Here a = 2.
  3. So x βˆ’ 2 is a factor.
Trap: The factor is x βˆ’ 2, not x + 2: for x + 2 to be a factor we would need p(βˆ’2) = 0, but p(βˆ’2) = 12.
Common traps
  • Dropping brackets around a negative inputWrite f(βˆ’3) = (βˆ’3)Β² βˆ’ 4(βˆ’3) + 1. Without brackets, βˆ’3Β² becomes βˆ’9.
  • Reading the axes backwardf(2) = 5 is the point (2, 5): the input 2 is horizontal and the output 5 is vertical.
  • Composing in the wrong orderIn f(g(x)) the function g works first. Work from the innermost brackets outward.
  • Shifting the wrong wayInside the brackets the effect is opposite: f(x + 3) moves left, f(x βˆ’ 3) moves right. Outside the brackets it is as expected.
  • Thinking f(a + b) = f(a) + f(b)This is false for most functions. If f(x) = xΒ², then f(1 + 2) = 9, not 1 + 4 = 5.
  • Forgetting that a denominator cannot be 0Set the denominator equal to 0 to find the excluded inputs. A zero of the numerator is a zero of the function, not an excluded value.
The Desmos way

Define a function by typing f(x) = 2xΒ² βˆ’ 5x + 1. Then type f(3) on a new line: Desmos shows the value. You can define g(x) too and type f(g(2)) for a composition. To see a transformation, type f(x βˆ’ 2) + 1 as a second graph and compare it with f(x). A slider for k in f(x) + k shows the shift. Click the x-axis crossings to read zeros.

  1. Type f(x) = x^2 βˆ’ 4x + 1
  2. Type f(βˆ’3) to see 22
  3. Type y = f(x + 2) βˆ’ 1 to see the shifted graph
  4. Type y = 10 and click the crossings to solve f(x) = 10

Use Desmos to check a value or to solve f(x) = b with ugly numbers. For direct substitution with small numbers, doing it by hand is faster.

Open Desmos β†—
Quick check
1
If f(x) = 2x + 7, what is f(βˆ’4)?
2
If f(x) = 4x βˆ’ 5, for what value of x is f(x) = 15?
3
How does the graph of y = f(x βˆ’ 3) compare with the graph of y = f(x)?
4
For what value of x is h(x) = 6/(x + 9) undefined?
Practice set: 10 SAT-style questionsEasy β†’ hard, with typed answers like the real test. Your score is saved in your cabinet.
Start practice β†’

More official practice

Bluebook practice tests β†—Official full-length adaptive tests from the College Board, free SAT question bank β†—Official questions you can filter by skill and difficulty Khan Academy β†—Free official SAT lessons and practice Desmos calculator β†—The same graphing calculator that is built into the test