Honors Algebra II Unit 2 Review: Functions and Their Graphs

Composition means plugging one function into another. In $(f \circ g)(x) = f(g(x))$, the inner function $g$ runs first, and its output becomes the input for $f$. Order matters, so $f(g(x))$ and $g(f(x))$ are usually different.

The domain of $f(g(x))$ includes only the x-values that work in $g$ and also give an output that $f$ accepts. For example, with $f(x) = \sqrt{x}$ and $g(x) = x - 4$, you get $f(g(x)) = \sqrt{x - 4}$, so the domain is $x \ge 4$.

An inverse $f^{-1}$ undoes $f$. If $f(a) = b$, then $f^{-1}(b) = a$. A function needs to be one-to-one to have an inverse function, which means it passes the horizontal line test. If it fails, you can sometimes restrict the domain. For example, $y = x^2$ on $x \ge 0$ has the inverse $y = \sqrt{x}$. The graphs of $f$ and $f^{-1}$ are mirror images across $y = x$. The domain of $f^{-1}$ is the range of $f$, and the range of $f^{-1}$ is the domain of $f$.

Worked example. Let $f(x) = 3x - 5$. Write $y = 3x - 5$, then swap the variables to get $x = 3y - 5$. Solving gives $y = \frac{x + 5}{3}$, so $f^{-1}(x) = \frac{x + 5}{3}$. To check, compute $f(f^{-1}(x)) = 3 \cdot \frac{x + 5}{3} - 5 = x$.

A function assigns each input exactly one output. On a graph, you check this with the vertical line test. If any vertical line hits the graph twice, one input has two outputs, so the graph isn't a function. A circle fails for this reason.

The notation $f(x)$ is read "f of x." It names the output, and it does not mean $f$ times $x$. To evaluate, replace every $x$ with the input. If $f(x) = 2x + 1$, then $f(3) = 7$. Inputs can also be expressions, so $f(a + 1) = 2(a + 1) + 1 = 2a + 3$.

The domain is the set of allowed inputs, and the range is the set of outputs the function actually produces. From an equation, look for two trouble spots. A denominator can't equal zero, and the expression under an even root can't be negative.

Worked example. Find the domain of $h(x) = \frac{\sqrt{x + 1}}{x - 3}$. The square root needs $x + 1 \ge 0$, so $x \ge -1$. The denominator needs $x - 3 \neq 0$, so $x \neq 3$. Both conditions must hold, so the domain is $[-1, 3) \cup (3, \infty)$.

A function is one-to-one if different inputs always give different outputs. For example, $x^3$ is one-to-one. The function $x^2$ is not, because $f(-2)$ and $f(2)$ are both 4.

A table of values can help you plot points, but check the function's behavior before connecting them. For familiar functions, transformations can be faster. You start from a parent graph you already know, like $y = x^2$, $y = |x|$, or $y = \sqrt{x}$, and adjust it.

Changes outside the function act on y-values. $f(x) + k$ shifts the graph up $k$ units when $k > 0$ and down when $k < 0$. For $a \neq 0$, $a f(x)$ scales y-values by $|a|$ and also reflects the graph across the x-axis if $a < 0$. Changes inside the function act on x-values in the opposite way you might expect. $f(x - h)$ shifts the graph right $h$ units when $h > 0$ and left when $h < 0$. For $b \neq 0$, $f(bx)$ scales x-values by $\frac{1}{|b|}$ and also reflects the graph across the y-axis if $b < 0$.

Inside changes feel backward because you are solving for the new x. For $f(x - 3)$ to give the same output that $f$ gives at 0, you need $x = 3$. That is why the graph moves right.

Worked example. Graph $g(x) = -2(x + 1)^2 + 3$ from $y = x^2$. First shift left 1, then stretch vertically by 2, then reflect across the x-axis, then shift up 3. The vertex moves from $(0, 0)$ to $(-1, 3)$, and the parabola opens downward. To check another point, $g(0) = -2(1)^2 + 3 = 1$, so $(0, 1)$ is on the graph. That point comes from $(1, 1)$ on the parent graph.

When you combine transformations, apply stretches and reflections before vertical shifts.