Graphing Functions: Chapter Review

Vocabulary and notation

Graph of a function

The set of all points $(x, f(x))$: each point's first coordinate is an input, its second that input's output.

Zeros (roots) of a function

The inputs whose output is $0$, that is, the first coordinates of the $x$-intercepts.

Turning point

Where the graph stops falling and starts rising, or the reverse: a minimum at the bottom of a valley, a maximum at the top of a hill.

Shift (translation)

A move of every point the same distance in the same direction. The shape is untouched.

Stretch and compression

A rescaling of every height, or of every horizontal position, by one common factor: a stretch pulls the graph away from an axis, a compression presses it in.

Reflection

A flip across a line. Points on that line are the hinge and do not move.

Fixed point

A point a transformation leaves where it was, such as a point of the line $y = x$ under reflection across it.

Formulas and theorems

• Reading a graph in both directions

  $$\begin{gathered} (a, b) \text{ is on the graph of } f \\ \iff f(a) = b \end{gathered}$$

  Use when Needs $a$ in the domain. Read backwards, the solutions of $f(x) = b$ are the inputs directly at the crossings of the line $y = b$: several, one, or none.

• Intercepts and zeros

  The $y$-intercept is $(0, f(0))$; an $x$-intercept is a point $(a, 0)$ on the graph, so the $x$-intercepts come from solving $f(x) = 0$.

  Use when At most ONE $y$-intercept, none when $0$ is outside the domain; $x$-intercepts may be several, one, or none. Zero the coordinate of the axis crossed, the opposite letter.

  e.g. $f(x) = x^2 - x - 6$: $y$-intercept $(0, -6)$, zeros $x = 3$ and $x = -2$.

• Domain and range from a graph

  The domain is the graph's horizontal spread along the $x$-axis; the range is its vertical spread along the $y$-axis.

  Use when Arrowheads mean that spread runs on forever; a graph drawn between two endpoints stops at them.

• Increasing and decreasing

  $$\begin{gathered} \text{increasing:}\ a < b \implies f(a) < f(b) \\ \text{decreasing:}\ a < b \implies f(a) > f(b) \end{gathered}$$

  Use when Stated on an interval and always read left to right, in the direction of increasing $x$: the graph rises where $f$ increases and falls where it decreases.

• Shifts

  $$\begin{gathered} y = f(x) + k \ \text{ moves up by } k \\ y = f(x - h) \ \text{ moves right by } h \end{gathered}$$

  Text description

  A curve and its translate, with a horizontal leg h and a vertical leg k carrying the point (a, b) to the point (a + h, b + k).

  Use when INSIDE ($x - h$) is HORIZONTAL and runs OPPOSITE to the sign shown: $(a, b) \to (a + h, b)$, moving the domain and the $x$-intercepts. OUTSIDE ($+ k$) is VERTICAL and runs WITH the sign: $(a, b) \to (a, b + k)$, moving the range. A vertical shift can change HOW MANY $x$-intercepts there are.

• Stretches and compressions

  $$\begin{gathered} y = a\,f(x) \ \text{ sends } \ (x_0, y_0) \to (x_0,\ a y_0) \\ y = f(bx) \ \text{ sends } \ (x_0, y_0) \to \left(\tfrac{x_0}{b},\ y_0\right) \end{gathered}$$

  Use when $a \neq 0$ and $b \neq 0$, and a negative factor reflects as well as scales. OUTSIDE, $a$ scales heights DIRECTLY: taller when $\lvert a \rvert > 1$, shorter when $0 < \lvert a \rvert < 1$. INSIDE, $b$ scales widths by the RECIPROCAL $\tfrac{1}{b}$, so $\lvert b \rvert > 1$ COMPRESSES the graph toward the $y$-axis while $0 < \lvert b \rvert < 1$ STRETCHES it away. Vertical scaling reshapes the range and pins every $x$-intercept; horizontal scaling reshapes the domain and pins the $y$-intercept.

  e.g. $(4, 6)$ becomes $(4, -3)$ on $y = -\tfrac{1}{2}f(x)$.

• The two reflections on their own

  $$\begin{gathered} y = -f(x) \ \text{ flips across the } x\text{-axis} \\ y = f(-x) \ \text{ flips across the } y\text{-axis} \end{gathered}$$

  Text description

  One curve with two highlighted mirror images of it: the flip across the x-axis, which is the graph of minus f of x, and the flip across the y-axis, which is the graph of f of minus x.

  Use when The cases $a = -1$ and $b = -1$: the outside minus negates the output, the inside minus the input. Neither changes the shape.

  e.g. $(5, -2)$ becomes $(5, 2)$ on $y = -f(x)$ and $(-5, -2)$ on $y = f(-x)$.

• Scaling and shifting together, and vertex form

  $$\begin{gathered} y = a\,f(x - h) + k \\ \text{sends} \quad (p, q) \to (p + h,\ aq + k) \end{gathered}$$

  Use when $a \neq 0$; taking $a = 1$ leaves the pure combined shift, right $h$ and up $k$. On the output the ORDER is scale first, then add: $aq + k$, never $a(q + k)$. For a parabola this is vertex form $y = a(x - h)^2 + k$, vertex $(h, k)$, $a$ setting width and opening.

  e.g. $(4, 5)$ becomes $(5, 13)$ on $y = 3f(x - 1) - 2$.

• Graph of an inverse

  $$\begin{gathered} (a, b) \text{ on the graph of } f \\ \iff (b, a) \text{ on the graph of } f^{-1} \end{gathered}$$

  Text description

  The graph of f and the graph of its inverse as mirror images in the dashed line y equals x, with the point (a, b) on one carried to the point (b, a) on the other.

  Use when Swapping a point's coordinates reflects it across $y = x$, so the graph of $f^{-1}$ is the graph of $f$ mirrored in that diagonal, no formula needed. That mirror image is the graph of a FUNCTION exactly when $f$ is one-to-one, which is what $f^{-1}$ needs in order to exist at all, and $f(c) = c$ forces $f^{-1}(c) = c$.

  e.g. $(9, -2)$ on $f$ gives $(-2, 9)$ on $f^{-1}$.

• Domain and range swap under inversion

  The domain of $f^{-1}$ is the range of $f$, and the range of $f^{-1}$ is the domain of $f$.

  Use when Whenever $f^{-1}$ exists: reflecting across $y = x$ trades the two axes, so a horizontal spread becomes a vertical one. Never carry an interval across unchanged.

  e.g. Domain $3 \le x \le 7$, range $0 \le y \le 4$ inverts to domain $0 \le x \le 4$, range $3 \le y \le 7$.

Problem types, step by step

Exam traps

• Trap Sending $y = f(x - 4)$ LEFT because of the minus sign, so the vertex of $y = (x - 4)^2$ gets plotted at $(-4, 0)$.

  Fix An inside change runs OPPOSITE to its sign: this moves RIGHT $4$, putting that vertex at $(4, 0)$. Read $h$ as the value making the inside zero, so $y = (x + 3)^2 - 5$ has vertex $(-3, -5)$.

• Trap Reading $y = f(3x)$ as a stretch to three times the width.

  Fix An inside factor acts by its RECIPROCAL. Inputs are divided by $3$, compressing the graph toward the $y$-axis, so a zero at $x = 12$ lands at $x = 4$. Only $y = f\!\left(\tfrac{x}{3}\right)$ widens it.

• Trap Letting a vertical stretch drag the $x$-intercepts inward with it, because the stretched graph does look narrower.

  Fix Only heights change, and $a \times 0 = 0$, so every zero is pinned. Stretching $f(x) = x^2 - 4$ into $y = 3x^2 - 12$ keeps both crossings at $x = -2$ and $x = 2$; what moved is the $y$-intercept, from $(0, -4)$ to $(0, -12)$. A horizontal scaling does the reverse, pinning the $y$-intercept and sliding the zeros.

• Trap Reflecting across an axis, or negating just one coordinate, to build an inverse.

  Fix The mirror for an inverse is the diagonal $y = x$, and BOTH coordinates trade: $(a, b) \to (b, a)$. Reflecting across the $x$-axis produces $y = -f(x)$, a different graph entirely.

• Trap Reflecting the whole graph of a function that fails the horizontal line test and calling the result $f^{-1}$.

  Fix That reflection fails the vertical line test, so it is not a function. Restrict to a one-to-one branch first: all of $y = x^2$ reflects to a sideways curve the line $x = 4$ meets twice, at $(4, 2)$ and $(4, -2)$.

• Trap Assuming the graphs of $f$ and $f^{-1}$ can meet only on the line $y = x$.

  Fix Points of $y = x$ are the GUARANTEED shared ones, not the only possible ones. A decreasing function can meet its inverse off the diagonal: $f(x) = -x^3$ has the visibly different mirror curve $f^{-1}(x) = -\sqrt[3]{x}$, yet $f(1) = -1$ and $f(-1) = 1$, so $(1, -1)$ sits on BOTH graphs.