6.4 Function Families & Transformations: Radical, Absolute Value, Piecewise, Logarithmic Functions and Quadratic Systems

6.4 Function Families and Transformations: Radical, Absolute Value, Piecewise, Logarithmic Functions and Quadratic Systems

Competency 004 asks you to illustrate the concept of a function with models, tables, graphs, symbols and words, and to use transformations to illustrate properties of functions. Competency 006 goes further. It expects you to use polynomial, rational, radical, absolute value, exponential, logarithmic, trigonometric and piecewise functions and relations to analyze and solve problems, to understand the effect of transformations such as $f(x \pm c)$, and to solve systems of quadratic equations and inequalities. This section brings those threads together. Trigonometric functions are developed in section 10.4.

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Relations, Functions, Domain and Range

A relation is any set of ordered pairs. A function is a relation in which each input has exactly one output.

• Mapping diagram or table: it is a function if no input value repeats with a different output.
• Graph (vertical line test): it is a function if every vertical line meets the graph at most once. The circle $x^2 + y^2 = 25$ fails, because $x = 3$ gives $y = 4$ and $y = -4$.
• Domain: the allowable inputs. Range: the resulting outputs. Context can restrict both. For example, a taxi fare function is defined only for miles $\ge 0$, and its outputs are dollar amounts.
• Function notation: $f(3)$ means the output when the input is 3. It does not mean $f$ times 3.

Inverse functions. If $f$ is one-to-one (it passes the horizontal line test), its inverse $f^{-1}$ undoes it: $f^{-1}(f(x)) = x$. The graph of $f^{-1}$ is the reflection of the graph of $f$ across $y = x$. To find it, swap $x$ and $y$ and solve for $y$. For $f(x) = 2x + 6$, swapping gives $x = 2y + 6$, so $f^{-1}(x) = \frac{x - 6}{2}$. Exponential and logarithmic functions are inverse pairs.

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One Transformation Rule for Every Family

For any parent function $f$, the function

$$g(x) = a\,f\big(b(x - h)\big) + k$$

is built from these moves:

Worked example. For $g(x) = -2\lvert x - 3 \rvert + 1$, start from the vertex of $\lvert x \rvert$ at the origin. Shift right 3 and up 1, so the new vertex is $(3, 1)$. Stretch vertically by 2 and reflect, so the V opens downward with slopes $+2$ and $-2$. The range is $y \le 1$. The $x$-intercepts solve $-2\lvert x - 3 \rvert + 1 = 0$, so $\lvert x - 3 \rvert = \frac{1}{2}$ and $x = 2.5$ or $x = 3.5$.

The same rule explains the vertex form of a quadratic from section 6.1, $a(x - h)^2 + k$, and the transformations of $\sqrt{x}$, $\frac{1}{x}$, $b^x$ and $\sin x$.

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A Gallery of Parent Functions

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Radical Functions and Extraneous Solutions

The square root function $f(x) = \sqrt{x - h} + k$ starts at $(h, k)$. Its domain is $x \ge h$ because the radicand cannot be negative in the real numbers.

Solving radical equations usually means isolating the radical and squaring. Squaring can create extraneous solutions, so always check your answers.

Solve $\sqrt{x + 7} = x - 5$:

1. Square both sides: $x + 7 = x^2 - 10x + 25$.
2. Rearrange: $x^2 - 11x + 18 = 0$, so $(x - 2)(x - 9) = 0$.
3. Check $x = 9$: $\sqrt{16} = 4$ and $9 - 5 = 4$. It works.
4. Check $x = 2$: $\sqrt{9} = 3$ but $2 - 5 = -3$. It is extraneous, because a principal square root is never negative.

The only solution is $x = 9$. Graphically, the curve $y = \sqrt{x + 7}$ and the line $y = x - 5$ meet only once.

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Piecewise and Step Functions

A piecewise function uses different rules on different parts of its domain. Many real situations are piecewise: shipping rates, parking fees and phone plans with overage charges.

Example: a parking garage. The first hour costs 4 dollars, and each additional hour (or part of an hour) costs 2 dollars, up to a daily maximum of 16 dollars. With $t$ in hours, $t > 0$:

$$C(t) = \begin{cases} 4, & 0 < t \le 1 \\ 4 + 2\lceil t - 1 \rceil, & 1 < t \le 7 \\ 16, & t > 7 \end{cases}$$

Here $\lceil t - 1 \rceil$ rounds up to the next whole hour. A 3.5-hour stay costs $4 + 2(3) = 10$ dollars. The graph is a step function, with open and closed endpoints showing which rule applies at each boundary.

Greatest integer function. $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$. So $\lfloor 3.7 \rfloor = 3$ and $\lfloor -1.2 \rfloor = -2$. Note that it is not $-1$, because $-2 \le -1.2 < -1$. This definition appears on the official Definitions and Formulas page.

Absolute value as piecewise: $\lvert x \rvert = x$ when $x \ge 0$ and $-x$ when $x < 0$. Writing $\lvert 2x - 6 \rvert$ piecewise (splitting at $x = 3$) is a strong way to show why its graph is a V.

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Logarithmic Functions

The logarithm is the exponent:

$$\log_b x = y \iff b^y = x \qquad (b > 0,\ b \ne 1,\ x > 0)$$

• $\log_2 32 = 5$ because $2^5 = 32$, and $\log_{10} 0.001 = -3$ because $10^{-3} = 0.001$.
• Properties: $\log_b(MN) = \log_b M + \log_b N$, $\log_b\frac{M}{N} = \log_b M - \log_b N$, and $\log_b(M^p) = p\log_b M$.
• Change of base: $\log_b x = \dfrac{\log x}{\log b} = \dfrac{\ln x}{\ln b}$. Scientific calculators like the TI-30XS include LOG and LN keys, so you can evaluate a logarithm in any base this way.

Solving an exponential equation. How long does it take 500 dollars to double at 6% interest compounded annually?

$$500(1.06)^t = 1000 \implies (1.06)^t = 2 \implies t = \frac{\log 2}{\log 1.06} \approx \frac{0.3010}{0.0253} \approx 11.9 \text{ years}$$

Common misconception: $\log(a + b) \ne \log a + \log b$. The product rule turns a product into a sum. It does not split a sum.

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Systems of Quadratic Equations and Inequalities

Linear–quadratic systems

Solve $y = x^2 - 4x + 3$ and $y = x - 1$. Substitute: $x^2 - 4x + 3 = x - 1$, so $x^2 - 5x + 4 = 0$ and $(x - 1)(x - 4) = 0$. The solutions are $(1, 0)$ and $(4, 3)$. A line can meet a parabola in 0, 1 or 2 points. The discriminant of the combined quadratic tells you which. When it is zero, the line is tangent to the parabola.

Quadratic–quadratic systems

Solve $y = x^2$ and $y = -x^2 + 8$. Setting them equal gives $2x^2 = 8$, so $x = \pm 2$. The solutions are $(-2, 4)$ and $(2, 4)$. Two parabolas can meet in 0, 1, 2, 3 or 4 points (3 or 4 only when the axes are not parallel), but on this exam most such systems reduce to a single quadratic.

Quadratic inequalities

Solve $x^2 - x - 6 < 0$. Factor to get $(x - 3)(x + 2) < 0$. The critical values are $-2$ and $3$. A sign chart, or picturing the upward parabola, shows the product is negative between the roots: $-2 < x < 3$. For $x^2 - x - 6 \ge 0$, the solution is outside the roots: $x \le -2$ or $x \ge 3$.

Systems of inequalities with a parabola

The solution region of $y \ge x^2 - 4$ and $y < x + 2$ lies on or above the solid parabola and below the dashed line. The boundary curves meet where $x^2 - 4 = x + 2$, that is $x^2 - x - 6 = 0$, so at $x = -2$ and $x = 3$. The region is the lens-shaped area between those intersection points. Test a point such as $(0, 0)$: $0 \ge -4$ is true and $0 < 2$ is true, so it lies in the region.

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Teaching Notes: Where Students Go Wrong

1. Shift direction. Students move $f(x - 3)$ left because they see "−3". Have them build a table: the output that $f$ produced at $x = 0$ is now produced at $x = 3$.
2. Extraneous roots. Require a substitution check whenever students square both sides.
3. Open and closed endpoints on piecewise graphs. Ask which rule owns each boundary value.
4. Logs as division. Students sometimes read $\log_2 8$ as $8 \div 2$. Anchor the meaning with the question "2 to what power is 8?"
5. Assuming every relation is a function. Present circles, sideways parabolas and mapping diagrams where one input has two outputs.