Graphs of Functions: Practice

A product is zero exactly when one of its factors is zero, so set each factor to $0$.

$x - 3 = 0 \Rightarrow x = 3, \qquad x + 1 = 0 \Rightarrow x = -1$

The zeros are $x = 3$ and $x = -1$.

A point sits on the horizontal axis exactly when its height (its output) is $0$.

$y = 0 \quad \Longleftrightarrow \quad f(x) = 0$

So the x-intercepts are found by solving $f(x) = 0$, not by setting $x = 0$ (that gives the y-intercept).

For any constant $c$, the function $g(x) = f(x) + c(x - 1)(x - 2)$ agrees with $f$ at both sampled inputs, because a factor vanishes at each.

$g(1) = f(1) + c \cdot 0 = 2, \qquad g(2) = f(2) + c \cdot 0 = 5$

Different values of $c$ give different functions, so infinitely many fit the table.

A single factor $(x - r)$ is negative exactly when $x < r$.

$x - 4 < 0 \quad \Longleftrightarrow \quad x < 4$

By contrast $(x + 4)$ is negative for $x < -4$, $(4 - x)$ is negative for $x > 4$, and $(x^2 + 4)$ is always positive.

Find the zeros from the factors.

$x(x - 2)(x + 2) = 0 \Rightarrow x = -2,\ 0,\ 2$

Each factor appears to the first power, an odd power, so the sign flips at each zero and the graph crosses. That is $3$ crossings.

Factor as a difference of squares, then set each factor to $0$.

$x^2 - 16 = (x - 4)(x + 4) = 0 \Rightarrow x = 4 \text{ or } x = -4$

The zeros are $x = \pm 4$.

For $x > 5$, both factors are positive, so multiply the signs.

$(+)(+) = +$

The function is positive on the whole interval $x > 5$.

An x-intercept is a point $(a, 0)$ with $f(a) = 0$, so use the zeros $x = 2$ and $x = -5$.

$f(2) = (0)(7) = 0 \Rightarrow (2, 0)$

The point $(5, 0)$ is not an intercept because $f(5) = (3)(10) = 30 \ne 0$.

A fraction is undefined where its denominator is zero.

$x - 4 = 0 \Rightarrow x = 4$

So $x = 4$ is excluded from the domain.

For $x > 1$, the square is larger than the input itself.

$x > 1 \implies x^2 > x$

So $x^2$ passes any bound you name as $x$ grows, and the graph climbs to large positive values on the right.

Y-axis symmetry is exactly evenness, so test which rule satisfies $f(-x) = f(x)$.

$f(-x) = (-x)^2 + 1 = x^2 + 1 = f(x)$

The others ($x^3$, $2x$, $x^3 - x$) are all odd, so only $x^2 + 1$ has y-axis symmetry.

For $x \ne 2$ the fraction cancels to the line.

$\frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2} = x + 2, \qquad x \ne 2$

At $x = 2$ the fraction is undefined while $x + 2 = 4$, so they differ only at $x = 2$, where the graph has a hole.